# 5.1: The Indefinite Integral

Derivatives appear in many physical phenomena, such as the motion of objects. Recall, for example, that given the position function (s(t)) of an object moving along a straight line at time (t), you could find the velocity (v(t)=s'(t)) and the acceleration (a(t)=v'(t)) of the object at time (t) by taking derivatives. Suppose the situation were reversed: given the velocity function how would you find the position function, or given the acceleration function how would you find the velocity function?

In this case calculating a derivative would not help, since the reverse process is needed: instead of differentiation you need a way of performing antidifferentiation, i.e. you would calculate an antiderivative.

Differentiation is relatively straightforward. You have learned the derivatives of many classes of functions (e.g. polynomials, trigonometric functions, exponential and logarithmic functions), and with the various rules for differentiation you can calculate derivatives of complicated expressions involving those functions (e.g. sums, powers, products, quotients). Antidifferentiation, however, is a different story.

To see some of the issues involved, consider a simple function like (f(x)=2x). Of course you know that (ddx(x^2) = 2x), so it seems that (F(x)=x^2) is the antiderivative of (f(x)=2x). But is it the only antiderivative of (f(x))? No. For example, if (F(x)=x^2+1) then (F'(x)=2x=f(x)), and so (F(x)=x^2+1) is another antiderivative of (f(x)=2x). Likewise, so is (F(x)=x^2+2). In fact, any function of the form (F(x)=x^2 + C), where (C) is some constant, is an antiderivative of (f(x)=2x).

Another potential issue is that functions of the form (F(x)=x^2 + C) are just the most obvious antiderivatives of (f(x)=2x). Could there be some other completely different function—one that cannot be simplified into the form (x^2 + C)—whose derivative also turns out to be (f(x) =2x)? The answer, luckily, is no:

To prove this, consider the function (H(x) = F(x) - G(x)), defined for all (x) in the common domain (I) of (F) and (G). Since (F'(x) = G'(x) = f(x)), then

[H'(x) ~=~ F'(x) ~-~ G'(x) ~=~ f(x) ~-~ f(x) ~=~ 0] for all (x) in (I), so (H(x)) is a constant function on (I), as was shown in Section 4.4 on the Mean Value Theorem. Thus, there is a constant (C) such that

The practical consequence of the above result can be stated as follows:

So for the function (f(x) = 2x), since (F(x) = x^2) is one antiderivative then all antiderivatives of (f(x)) are of the form (F(x) = x^2 + C), where (C) is a generic constant. Thus, functions do not have just one antiderivative but a whole family of antiderivatives, all differing only by a constant. The following notation makes all this easier to express:

The large S-shaped symbol before (f(x)) is called an integral sign. Though the indefinite integral (int f(x)~dx) represents all antiderivatives of (f(x)), the integral can be thought of as a single object or function in its own right, whose derivative is (f'(x)):

You might be wondering what the integral sign in the indefinite integral represents, and why an infinitesimal (dx) is included. It has to do with what an infinitesimal represents: an infinitesimal “piece” of a quantity. For an antiderivative (F(x)) of a function (f(x)), the infinitesimal (or differential) (d!F) is given by (d!F = F'(x),dx = f(x),dx), and so

[F(x) ~=~ int,f(x)~dx ~=~ int,d!F ~.] The integral sign thus acts as a summation symbol: it sums up the infinitesimal “pieces” (d!F) of the function (F(x)) at each (x) so that they add up to the entire function (F(x)). Think of it as similar to the usual summation symbol (Sigma) used for discrete sums; the integral sign (int) takes the sum of a continuum of infinitesimal quantities instead.

Finding (or evaluating) the indefinite integral of a function is called integrating the function, and integration is antidifferentiation.

Example (PageIndex{1}): antideriv1

Solution

Evaluate (displaystyleint,0~dx).

Solution: Since the derivative of any constant function is 0, then (int,0~dx = C), where (C) is a generic constant.

Note: From now on (C) will simply be assumed to represent a generic constant, without having to explicitly say so every time.

Example (PageIndex{1}): antideriv2

Solution

Evaluate (displaystyleint,1~dx).

Solution: Since the derivative of (F(x) = x) is (F'(x) = 1), then (int,1~dx = x + C).

Example (PageIndex{1}): antideriv3

Solution

Evaluate (displaystyleint,x~dx).

Solution: Since the derivative of (F(x) = frac{x^2}{2}) is (F'(x) = x), then (int,x~dx = frac{x^2}{2} + C).

Since (ddx,left(frac{x^{n+1}}{n+1} ight) = x^n) for any number (n e -1), and (ddx,(ln,abs{x}) = frac{1}{x} = x^{-1}), then any power of (x) can be integrated:

The following rules for indefinite integrals are immediate consequences of the rules for derivatives:

The above rules are easily proved. For example, the first rule is a simple consequence of the Constant Multiple Rule for derivatives: if (F(x) = int,f(x)~dx), then

[ddx(k,F(x)) ~=~ k,ddx(F(x)) ~=~ k,f(x) quadRightarrowquad int,k;f(x)~dx ~=~ k,F(x) ~=~ k,int,f(x)~dx ~.quadcheckmark] The other rules are proved similarly and are left as exercises. Repeated use of the above rules along with the Power Formula shows that any polynomial can be integrated term by term—in fact any finite sum of functions can be integrated in that manner:

[antideriv4] Evaluate (displaystyleint,(x^7 - 3x^4)~dx).

Solution: Integrate term by term, pulling constant multiple outside the integral:

[int,(x^7 - 3x^4)~dx ~=~ int,x^7~dx ~-~ 3int,x^4~dx ~=~ frac{x^8}{8} ~-~ frac{3x^5}{5} ~+~ C]

[antideriv5] Evaluate (displaystyleint,sqrt{x}~dx).

Solution: Use the Power Formula:

[int,sqrt{x}~dx ~=~ int,x^{1/2}~dx ~=~ frac{x^{3/2}}{3/2} ~+~ C ~=~ frac{2x^{3/2}}{3} ~+~ C]

[antideriv6] Evaluate (displaystyleint,left(dfrac{1}{x^2} + dfrac{1}{x} ight)~dx).

Solution: Use the Power Formula and integrate term by term:

[int,left(frac{1}{x^2} + frac{1}{x} ight)~dx ~=~ int,left(x^{-2} + frac{1}{x} ight)~dx ~=~ frac{x^{-1}}{-1} ~+~ ln,abs{x} ~+~ C ~=~ -frac{1}{x} ~+~ ln,abs{x} ~+~ C]

The following indefinite integrals are just re-statements of the corresponding derivative formulas for the six basic trigonometric functions:

Since (ddx(e^x) = e^x), then:

[antideriv7] Evaluate (displaystyleint,(3sin,x ~+~ 4cos,x ~-~ 5e^x)~dx).

Solution: Integrate term by term:

[egin{aligned} int,(3sin,x ~+~ 4cos,x ~-~ 5e^x)~dx ~&=~ 3int,sin,x~dx ~+~ 4int,cos,x~dx ~-~ 5int,e^x~dx

[10pt] &=~ -3cos,x ~+~ 4sin,x ~-~ 5e^x ~+~ Cend{aligned}]

Example (PageIndex{1}): gravity

Solution

Recall from Section 1.1 the example of an object dropped from a height of 100 ft. Show that the height (s(t)) of the object (t) seconds after being dropped is (s(t) = -16t^2 + 100), measured in feet.

Solution: When the object is dropped at time (t=0) the only force acting on it is gravity, causing the object to accelerate downward at the known constant rate of 32 ft/s2. The object’s acceleration (a(t)) at time (t) is thus (a(t) = -32). If (v(t)) is the object’s velocity at time (t), then (v'(t) = a(t)), which means that

[v(t) ~=~ int a(t)~dt ~=~ int -32~dt ~=~ -32t ~+~ C] for some constant (C). The constant (C) here is not generic—it has a specific
value determined by the initial condition on the velocity: the object was at rest at time (t=0). That is, (v(0) = 0), which means

[0 ~=~ v(0) ~=~ -32(0) ~+~ C ~=~ C quadRightarrowquad v(t) ~=~ -32t] for all (t ge 0). Likewise, since (s'(t) = v(t)) then

[s(t) ~=~ int v(t)~dt ~=~ int -32t~dt ~=~ -16t^2 ~+~ C] for some constant (C), determined by the initial condition that the object was 100 ft above the ground at time (t=0). That is, (s(0) = 100), which means

[100 ~=~ s(0) ~=~ -16(0)^2 ~+~ C ~=~ C quadRightarrowquad s(t) ~=~ -16t^2 ~+~ 100] for all (t ge 0).

The formula for (s(t)) in Example

Example (PageIndex{1}): gravity

Solution

can be generalized as follows: denote the object’s initial position at time (t=0) by (s_0), let (v_0) be the object’s initial velocity (positive if thrown upward, negative if thrown downward), and let (g) represent the (positive) constant acceleration due to gravity. By Newton’s First Law of motion the only acceleration imparted to the object after throwing it is due to gravity:

[a(t) ~=~ -g quadRightarrowquad v(t) ~=~ int a(t)~dt ~=~ int -g~dt ~=~ -gt ~+~ C] for some constant (C): (v_0 = v(0) = -g(0) + C = C). Thus, (v(t) = -gt + v_0) for all (t ge 0), and so

[s(t) ~=~ int v(t)~dt ~=~ int left(-gt ~+~ v_0 ight)~dt ~=~ - frac{1}{2}gt^2 ~+~ v_0t ~+~ C] for some constant (C): (s_0 = s(0) = - frac{1}{2}g(0)^2 + v_0(0) + C = C). To summarize:

Note that the units are not specified—they just need to be consistent. In metric units, (g = 9.8) m/s2, while (g = 32) ft/s2 in English units.

Thinking of an indefinite integral as the sum of all the infinitesimal “pieces” of a function—for the purpose of retrieving that function—provides a handy way of integrating a differential equation to obtain the solution. The key idea is to transform the differential equation into an equation of differentials, which has the effect of treating functions as variables. Some examples will illustrate the technique.

Example (PageIndex{1}): intdecay

Solution

For any constant (k), show that every solution of the differential equation (dydt = ky) is of the form (y = Ae^{kt}) for some constant (A). You can assume that (y(t) > 0) for all (t).

Solution: Put the (y) terms on the left and the (t) terms on the right, i.e. separate the variables:

[frac{dy}{y} ~=~ k,dt] Now integrate both sides (notice how the function (y) is treated as a variable):

[egin{aligned} int,frac{dy}{y} ~&=~ int k,dt

[6pt] ln,y + C_1 ~&=~ kt + C_2 quad ext{($C_1$ and $C_2$ are constants)} ln,y ~&=~ kt + C quad ext{(combine $C_1$ and $C_2$ into the constant $C$)} y ~&=~ e^{kt+C} ~=~ e^{kt} cdot e^C ~=~ A e^{kt}end{aligned}] where (A = e^C) is a constant. Note that this is the formula for radioactive decay from Section 2.3.

Example (PageIndex{1}): intidealgas

Solution

Recall from Section 3.6 the equation of differentials

[dfrac{dP}{P} ~+~ dfrac{dV}{V} ~=~ dfrac{dT}{T}] relating the pressure (P), volume (V) and temperature (T) of an ideal gas. Integrate that equation to obtain the original ideal gas law (PV = RT), where (R) is a constant. .

Solution: Integrating both sides of the equation yields

[egin{aligned} int,dfrac{dP}{P} ~+~ int,dfrac{dV}{V} ~&=~ int,dfrac{dT}{T}

[6pt] ln,P ~+~ ln,V ~&=~ ln,T ~+~ C quad ext{($C$ is a constant)} ln,(PV) ~&=~ ln,T ~+~ C PV ~&=~ e^{ln,T + C} ~=~ e^{ln,T} cdot e^{C} ~=~ T,e^C ~=~ RTend{aligned}] where (R = e^C) is a constant.

The integration formulas in this section depended on already knowing the derivatives of certain functions and then “working backward” from their derivatives to obtain the original functions. Without that prior knowledge you would be reduced to guessing, or perhaps recognizing a pattern from some derivative you have encountered. A number of integration techniques will be presented shortly, but there are many indefinite integrals for which no simple closed form exists (e.g. (int e^{x^2},dx) and (int sin(x^2),dx)).

[sec5dot1]

For Exercises 1-15, evaluate the given indefinite integral.

3

(displaystyleint,left(x^2 ~+~ 5x ~-~ 3 ight)~dx)

(displaystyleint,3 cos,x~dx)

(displaystyleint,4 e^x~dx)

3

(displaystyleint,left(x^5 ~-~ 8x^4 ~-~ 3x^3 ~+~ 1 ight)~dxvphantom{dfrac{3e^x}{5}})

(displaystyleint,5 sin,x~dxvphantom{dfrac{3e^x}{5}})

(displaystyleint,dfrac{3e^x}{5}~dx)

3

(displaystyleint,dfrac{6}{x}~dx)

(displaystyleint,dfrac{4}{3x}~dx)

(displaystyleint,left(-2 sqrt{x}, ight)~dxvphantom{dfrac{6}{x}})

3

(displaystyleint,dfrac{1}{3 sqrt{x}}~dx)

(displaystyleint,left(x ~+~ x^{4/3} ight)~dx)

(displaystyleint,dfrac{1}{3 sqrt[3]{x}}~dx)

3

(displaystyleint,3sec,x; an,x~dx)

(displaystyleint,5 sec^2x~dx)

(displaystyleint,7csc^2x~dx)

Prove the sum and difference rules for indefinite integrals: (int (f(x) pm g(x)),dx ;=; int f(x),dx ;pm; int g(x),dx)

Integrate both sides of the equation

[frac{dP}{P} ~+~ frac{d!M}{M} ~=~ frac{dT}{2T}] to obtain the ideal gas continuity relation: (dfrac{PM}{sqrt{T}} =) constant.

[exer:projmax0] Use the free fall motion equation for position to show that the maximum height reached by an object launched straight up from the ground with an initial velocity (v_0) is (frac{v_0^2}{2g}).

1. The function (f) is assumed to be differentiable at (x), in this case. If not then the points where (f) is not differentiable can be excluded without affecting the integral.↩
2. For a proof and fuller discussion of all this, see Ch.1-2 in Knopp, M.I., Theory of Area, Chicago: Markham Publishing Co., 1969. The book attempts to define precisely what an “area” actually means, including that of a rectangle (showing agreement with the intuitive notion of width times height).↩
3. The theorem can be proved for the weaker condition that (f) is merely continuous on (ival{a}{b}). See p.173-175 in Parzynski, W.R. and P.W. Zipse, Introduction to Mathematical Analysis, New York: McGraw-Hill, Inc., 1982.↩
4. Created by the physicist P.A.M. Dirac (1902-1984), who won a Nobel Prize in physics in 1933. The function is neither real-valued nor continuous at (x=0). The “graph” in Figure [fig:dirac] is perhaps misleading, as (infty) is not an actual point on the (y)-axis. One interpretation is that (delta) is an abstraction of an instantaneous pulse or burst of something, preceded and followed by nothing. To learn more about this fascinating and useful function, see §15 in Dirac, P.A.M., The Principles of Quantum Mechanics, 4th ed., Oxford, UK: Oxford University Press, 1958.↩
5. See pp.140-141 in Buck, R.C., Advanced Calculus, 2nd ed., New York: McGraw-Hill Book Co., 1965.↩

## Lesson Explainer: Indefinite Integrals: Exponential and Reciprocal Functions

Indefinite integrals of exponential and logarithmic functions have many real-world applications as the functions are used in mathematical models to describe population growth, cell growth, and radioactive decay.

These types of problems can be solved by using the following rules.

### Definition: Integrals of Exponential and Reciprocal Functions

to avoid dividing by zero in the first expression.

We can verify these directly using the first part of the fundamental theorem of calculus.

### Definition: The Fundamental Theorem of Calculus (FTC)

be a continuous real-valued function defined on

be the function defined, for all

is uniformly continuous on

By taking the derivative of the right-hand side for each of the statements above, we can show the result equals the integrand:

Therefore, upon integrating these, we have

Now, what if we want to integrate something like

(i.e., an exponential expression with an arbitrary base)? The trick is to use the fact that

and apply the laws of logarithms with

We can use the standard rule for exponentials (1) with base

to avoid dividing by zero. More generally, we have

. We do not need to memorize this result as we can derive it directly from applying the laws of logarithms for each exponential base.

Let’s look at an integral with both exponential and reciprocal functions given by

We know that the integral of the sum or difference of two functions is equal to the sum or difference of the integrals of those functions. In other words, we can separate out the two parts and take constant factors outside the integral. For each part, we also get a constant of integration, which we can combine into a single constant

Applying the standard rules for integrating exponential (1) and reciprocal (2) functions as given in the definition, we find

The constant of integration can also be determined if there is a boundary condition. Suppose

is a boundary condition. We can substitute this in order to determine the constant

## 5.1: The Indefinite Integral

The command to use is shown below.

Notice that Maple gives an exact answer, as a fraction. If you want a decimal approximation to an integral, you just put an evalf command around the int command, as shown below.

To compute an indefinite integral with Maple, you just leave out the range for the limits of integration, as shown below. Note that Maple does not include a constant of integration.

You can also use the Maple int command with functions or expressions you have defined in Maple. For example, suppose you wanted to find area under the curve of the function on the interval . Then you can define this function in Maple with the command and then use this definition as shown below.

You can also simply give the expression corresponding to a label in Maple, and then use that label in subsequent commands as shown below. However, notice the difference between the two methods. You are urged to choose one or the other, so you don't mix the syntax up. If you want to find the area bounded by the graph of two functions, you should first plot both functions on the same graph. You can then find the intersection points using either the solve or fsolve command. Once this is done, you can calculate the definite integral in Maple. An example below illustrates how this can be done in Maple by finding the area bounded by the graphs of and :

Note that the average value is just a number. For example, suppose you wanted to compute the average value of the function over the interval . The following Maple command would do the job.

## 2 - Indefinite Integrals of Elementary Functions

This chapter provides an introduction to indefinite integrals of elementary functions followed by its forms. Elementary functions include rational functions, algebraic functions, exponential function, hyperbolic function, trigonometric functions, logarithms and inverse-hyperbolic functions, and inverse trigonometric functions. When certain functions are integrated, the logarithm of the absolute value is obtained. In such formulas, the absolute-value bars in the argument of the logarithm are omitted for simplicity in writing. In certain cases, it is important to give the complete form of the primitive function. Such primitive functions, written in the form of definite integrals, are given. Closely related to these formulas are formulas in which the limits of integration and the integrand depend on the same parameter. A number of formulas lose their meaning for certain values of the constants (parameters) or for certain relationships among these constants. These values of the constants and the relationships among them are for the most part completely clear from the very structure of the right hand member of the formula. Therefore, throughout the chapter, remarks to this effect are omitted. The constant of integration in all the formulas of this chapter is also omitted. Therefore, the equality sign (=) means that the functions on the left and right of this symbol differ by a constant.

## Solution

If the derivative of x 2 e x is (2 x + x 2 ) e x , then an antiderivative of (2 x + x 2 ) e x is x 2 e x , so the general antiderivative or
indefinite integral of (2 x + x 2 ) e x is x 2 e x + C , where C is an arbitrary constant. Remark that it's not necessary to include
the constant C of the indefinite integral in the computation of the definite integral see Section 9.4 Evaluation.

## Properties of Indefinite Integrals

Following properties of indefinite integrals arise from the constant multiple and sum rules for derivatives.

Property 2. For functions $$<> ight)>>>$$ and $$<> ight)>>>$$ $$int ight)>>>pm<> ight)>>> ight)>=int<> ight)>>>pmint<> ight)>>>$$ . Integral of sum (difference) is sum (difference) of integrals. This formula can be easily extended to any number of functions.

Since differentiation and integration are inverse processes then we have additional two properties.

Property 3. $$int<> ight)>>>=<> ight)>>>$$ . Derivative of integral of a function is function itself.

Property 4. In notation of differentials we have that $$='=<>>>$$ , so $$int ight)>= ight)>+$$ . Integral of derivative of function is function itself plus some arbitrary constant.

Using the formula of Binom of Newton we can rewrite integrand as follows:

Using formula of Binom of Newton we have that

Note that both  and  are constants.

Example 9 . A ball is thrown upward with a speed of $$<64> frac<<>>>><>$$ from the edge of a cliff 464 ft above the ground. Find its height above the ground  seconds later. When does it reach its maximum height? When does it hit the ground?

An object near the surface of the earth is subject to a gravitational force that produces a downward acceleration denoted by $$>$$ .

For motion close to the earth we may assume that $$>$$ is constant, its value being about $$<9.8>frac<><<>^<<2>>>$$ (or $$<32>frac<<>>>><<>^<<2>>>$$ ).

The motion is vertical and we choose the positive direction to be upward. At time  the distance above the ground is $$ight)>$$ and the velocity $$ight)>$$ is decreasing.
Therefore, the acceleration $$ight)>$$ must be negative and we have $$ight)>=frac<<>><<>>=- <32>$$ .

The maximum height will be reached when $$ight)>= <0>$$ or $$=frac<<64>><<32>>=<2>$$ .

The expression for $$ight)>$$ is valid until the ball hits the ground. This happens when $$ight)>= <0>$$ , that is, when $$-<16><>^<<2>>+<64>+<464>= <0>$$ or $$<>^<<2>>-<4>-<29>= <0>$$ .

Using the quadratic formula to solve this equation, we get $$=<2>pmsqrt<<<33>>>$$ .

We reject the solution with the minus sign since it gives a negative value for time  .
Therefore, the ball hits the ground after $$<2>+sqrt<<<33>>>approx<7.7> .$$

## 2. [0/5 Points] DETAILS PREVIOUS ANSWERS LARCALCET7 5.1.018. Find the indefinite integral and check your result.

8. [0/1 Points] DETAILS PREVIOUS ANSWERS LARCALC11 6.4.021. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Find the particular solution of the first-order linear differential equation for x > 0 that satisfies the initial condition. Differential Equation Initial Condition y' + y = 0 y(2) = 2 419 y = .8 Need Help? Read It Talk to a Tutor Submit Answer

### 1. [-/1 Points] DETAILS LARCALCET7 3.5.006. Find dy/dx by implicit differentiation. 4x² – 7y²=9 dy/dx =.

1. [-/1 Points] DETAILS LARCALCET7 3.5.006. Find dy/dx by implicit differentiation. 4x² – 7y²=9 dy/dx = Need Help? Read It Talk to a Tutor 2. (-/1 Points] DETAILS LARCALCET7 3.5.010.MI. Find dy/dx by implicit differentiation. 9x²y + 4y2x = -2 dy/dx = Need Help? Read it Master it Talk to a Tutor 3. [-/1 Points] DETAILS LARCALCET7 3.5.020. Find dy/dx by implicit differentiation. cot(y) = 5x - 5y dy/dx - Need Help? Read it Watch it Talk to a Tutor.

### 5. [-/2 points) DETAILS LARCALC9 4.1.022. Find the indefinite integral and check your result by differentiation.

5. [-/2 points) DETAILS LARCALC9 4.1.022. Find the indefinite integral and check your result by differentiation. (Use C for the constant of integration.) (x+ dx Need Help? Read it | Talk to a Tutor Show My Work (Required) What steps or reasoning did you use? Your work counts towards your score. You can submit show my work an unlimited number of times.

### Help with both questions. 16. [-/1 Points] DETAILS LARCALC11 8.1.043. Find the indefinite integral. (Use C.

help with both questions. 16. [-/1 Points] DETAILS LARCALC11 8.1.043. Find the indefinite integral. (Use C for the constant of integration.) JzV292-64 dz z49z2 - 64 Need Help? Read It Watch It Talk to a Tutor 17. (-/1 Points] DETAILS LARCALC11 8.1.053.MI. Find the general solution of the differential equation. (Use C for the constant of integration.) dr 7e dt √1-e2t re Need Help? Read it Watch it Master it Talk to a Tutor

### | -/1 POINTS Find the indefinite integral and check your result by differentiation. (Use C for.

| -/1 POINTS Find the indefinite integral and check your result by differentiation. (Use C for the constant of integration.) (7-2) dr Need Help? Read It Talk to a Tutor 4. -11 POINTS Find the indefinite integral and check the result by differentiation. (Use C for the constant of integration.) 1 / (5x - sx?) dx Need Help? Read It Watch It Talk to a Tutor

### 6. [0/2 Points] DETAILS PREVIOUS ANSWERS SCALCCC4 7.3.016. MY NOTES ASK YOUR TEACHER PRAC Find the.

6. [0/2 Points] DETAILS PREVIOUS ANSWERS SCALCCC4 7.3.016. MY NOTES ASK YOUR TEACHER PRAC Find the solution of the differential equation that satisfies the given initial condition. dP = 27 Pt, P(1) = 2 dt P= Need Help? Read It Talk to a Tutor Submit Answer Viewing Saved Work Revert to Last Response

### 16. [-12 Points] DETAILS LARCALCET7 3.5.091.MI. Find all points on the circle x2 + y2 =.

16. [-12 Points] DETAILS LARCALCET7 3.5.091.MI. Find all points on the circle x2 + y2 = 676 where the slope is 5 12 (x, y) = (smaller y-value) (x, y) = ) (larger y-value) Need Help? Read It Master it Talk to a Tutor DETAILS 13. [-/1 Points] LARSONET5 3.5.049. Find an equation of the tangent line to the graph at the given point. x?y2 - 9x2 - 4y2 = 0, (-4, -2/3) y = Cruciform 2 6 4 0.

### 15. [-75 Points] DETAILS LARCALCET7 5.8.061.MI. Find the particular solution of the differential equation that satisfies.

15. [-75 Points] DETAILS LARCALCET7 5.8.061.MI. Find the particular solution of the differential equation that satisfies the initial condition. 1 dy dx = y(0) = 21 y =

### 1. [-75 Points] DETAILS LARCALCET7 5.1.044. MY NOTES ASK Find the particular solution of the differential.

1. [-75 Points] DETAILS LARCALCET7 5.1.044. MY NOTES ASK Find the particular solution of the differential equation that satisfies the initial condition(s). (Remember to use absolute values where appropriate.) 5 f"(x) x2 f'(1) = 6, f(1) = 5 = f(x) = Submit Answer

### Use a table of integrals to find the indefinite integral for botb MY 11. [-17.14 Points].

use a table of integrals to find the indefinite integral for botb MY 11. [-17.14 Points] DETAILS LARCALCET7 8.7.040. Use a table of integrals to find the indefinite integral. (Use for the constant of integration.) | Content Need Help? Read it Watch Talk to Tutor 12. [-17.14 Points] DETAILS LARCALCET7 8.7.503.XP. MY NOTES ASK YOUR TI Use a table of integrals with forms involving Valv? to find the integral. (Use C for the constant of integration.) Sla I roman Need.

## Solutions for Chapter 5.1: Antiderivatives and Indefinite Integration

Solutions for Chapter 5.1: Antiderivatives and Indefinite Integration

• 5.1.1: In Exercises 14, verify the statement by showing that the derivativ.
• 5.1.2: In Exercises 14, verify the statement by showing that the derivativ.
• 5.1.3: In Exercises 14, verify the statement by showing that the derivativ.
• 5.1.4: In Exercises 14, verify the statement by showing that the derivativ.
• 5.1.5: In Exercises 58, find the general solution of the differential equa.
• 5.1.6: In Exercises 58, find the general solution of the differential equa.
• 5.1.7: In Exercises 58, find the general solution of the differential equa.
• 5.1.8: In Exercises 58, find the general solution of the differential equa.
• 5.1.9: In Exercises 914, complete the table using Example 3 and the exampl.
• 5.1.10: In Exercises 914, complete the table using Example 3 and the exampl.
• 5.1.11: In Exercises 914, complete the table using Example 3 and the exampl.
• 5.1.12: In Exercises 914, complete the table using Example 3 and the exampl.
• 5.1.13: In Exercises 914, complete the table using Example 3 and the exampl.
• 5.1.14: In Exercises 914, complete the table using Example 3 and the exampl.
• 5.1.15: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.16: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.17: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.18: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.19: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.20: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.21: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.22: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.23: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.24: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.25: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.26: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.27: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.28: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.29: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.30: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.31: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.32: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.33: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.34: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.35: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.36: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.37: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.38: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.39: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.40: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.41: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.42: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.43: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.44: In Exercises 1544, find the indefinite integral and check the resul.
• 5.1.45: In Exercises 4548, sketch the graphs of the function for and on the.
• 5.1.46: In Exercises 4548, sketch the graphs of the function for and on the.
• 5.1.47: In Exercises 4548, sketch the graphs of the function for and on the.
• 5.1.48: In Exercises 4548, sketch the graphs of the function for and on the.
• 5.1.49: In Exercises 4952, the graph of the derivative of a function is giv.
• 5.1.50: In Exercises 4952, the graph of the derivative of a function is giv.
• 5.1.51: In Exercises 4952, the graph of the derivative of a function is giv.
• 5.1.52: In Exercises 4952, the graph of the derivative of a function is giv.
• 5.1.53: In Exercises 5356, find the equation for y, given the derivative an.
• 5.1.54: In Exercises 5356, find the equation for y, given the derivative an.
• 5.1.55: In Exercises 5356, find the equation for y, given the derivative an.
• 5.1.56: In Exercises 5356, find the equation for y, given the derivative an.
• 5.1.57: Slope Fields In Exercises 5760, a differential equation, a point, a.
• 5.1.58: Slope Fields In Exercises 5760, a differential equation, a point, a.
• 5.1.59: Slope Fields In Exercises 5760, a differential equation, a point, a.
• 5.1.60: Slope Fields In Exercises 5760, a differential equation, a point, a.
• 5.1.61: Slope Fields In Exercises 61 and 62, (a) use a graphing utility to .
• 5.1.62: Slope Fields In Exercises 61 and 62, (a) use a graphing utility to .
• 5.1.63: In Exercises 6372, solve the differential equation.fx 4x, f0 6 ,
• 5.1.64: In Exercises 6372, solve the differential equation.
• 5.1.65: In Exercises 6372, solve the differential equation.
• 5.1.66: In Exercises 6372, solve the differential equation.
• 5.1.67: In Exercises 6372, solve the differential equation.fx 2, f2 5, f2 10f
• 5.1.68: In Exercises 6372, solve the differential equation.fx x2, f0 6, f0 3f
• 5.1.69: In Exercises 6372, solve the differential equation.fx x32, f4 2, f0 0f
• 5.1.70: In Exercises 6372, solve the differential equation.fx sin x, f0 1, .
• 5.1.71: In Exercises 6372, solve the differential equation.fx ex, f0 2, f0 5SE
• 5.1.72: In Exercises 6372, solve the differential equation.fx 2x2, f1 4, f1 3f
• 5.1.73: Tree Growth An evergreen nursery usually sells a certain shrub afte.
• 5.1.74: Population Growth The rate of growth of a population of bacteria is.
• 5.1.75: Use the graph of shown in the figure to answer the following, given.
• 5.1.76: The graphs of and each pass through the origin. Use the graph of sh.
• 5.1.77: Vertical Motion In Exercises 7780, use feet per second per second a.
• 5.1.78: Vertical Motion In Exercises 7780, use feet per second per second a.
• 5.1.79: Vertical Motion In Exercises 7780, use feet per second per second a.
• 5.1.80: Vertical Motion In Exercises 7780, use feet per second per second a.
• 5.1.81: Vertical Motion In Exercises 81 and 82, use meters per second per s.
• 5.1.82: Vertical Motion In Exercises 81 and 82, use meters per second per s.
• 5.1.83: A baseball is thrown upward from a height of 2 meters with an initi.
• 5.1.84: With what initial velocity must an object be thrown upward (from a .
• 5.1.85: Lunar Gravity On the moon, the acceleration due to gravity is meter.
• 5.1.86: Escape Velocity The minimum velocity required for an object to esca.
• 5.1.87: Rectilinear Motion In Exercises 8790, consider a particle moving al.
• 5.1.88: Rectilinear Motion In Exercises 8790, consider a particle moving al.
• 5.1.89: Rectilinear Motion In Exercises 8790, consider a particle moving al.
• 5.1.90: Rectilinear Motion In Exercises 8790, consider a particle moving al.
• 5.1.91: Acceleration The maker of an automobile advertises that it takes 13.
• 5.1.92: Deceleration A car traveling at 45 miles per hour is brought to a s.
• 5.1.93: Acceleration At the instant the traffic light turns green, a car th.
• 5.1.94: Modeling Data The table shows the velocities (in miles per hour) of.
• 5.1.95: True or False? In Exercises 95100, determine whether the statement .
• 5.1.96: True or False? In Exercises 95100, determine whether the statement .
• 5.1.97: True or False? In Exercises 95100, determine whether the statement .
• 5.1.98: True or False? In Exercises 95100, determine whether the statement .
• 5.1.99: True or False? In Exercises 95100, determine whether the statement .
• 5.1.100: True or False? In Exercises 95100, determine whether the statement .
• 5.1.101: Find a function such that the graph of has a horizontal tangent at .
• 5.1.102: The graph of is shown. Sketch the graph of given that is continuous.
• 5.1.103: If f is continuous, and find f. Is f differentiable at x 2?
• 5.1.104: Let be two functions satisfying and for all If prove that sx2 cx2 1.cx
• 5.1.105: Verification Verify the natural log rule by showing that the deriva.
• 5.1.106: Verification Verify the natural log ruleby showing that the derivat.
• 5.1.107: Suppose and are nonconstant, differentiable, real-valued functions .
##### ISBN: 9780618606245

This textbook survival guide was created for the textbook: Calculus: Early Transcendental Functions , edition: 4. Since 107 problems in chapter 5.1: Antiderivatives and Indefinite Integration have been answered, more than 74678 students have viewed full step-by-step solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions. Chapter 5.1: Antiderivatives and Indefinite Integration includes 107 full step-by-step solutions. Calculus: Early Transcendental Functions was written by and is associated to the ISBN: 9780618606245.

The rate that would give the same return if interest were computed just once a year

A set of points in a plane equally distant from a fixed point called the center

Trigonometric functions when applied to real numbers are circular functions

The distance d(P, Q) between P(x, y) and Q(x, y) d(P, Q) = 2(x 1 - x 2)2 + (y1 - y2)2

For the line segment with endpoints (x 1, y1, z 1) and (x2, y2, z2), ax 1 + x 22 ,y1 + y22 ,z 1 + z 22 b

See Absolute value of a complex number.

The eight regions of space determined by the coordinate planes.

The various possible results of an experiment.

The surface created when a line is rotated about a second line that intersects but is not perpendicular to the first line.

A line joining two points of the graph of ƒ.

An end behavior asymptote that is a slant line

An ordered pair of real numbers that satisfies all of the equations or inequalities in the system

The graph of the polar curve.

The line through P0(x 0, y0, z0) in the direction of the nonzero vector V = <a, b, c> has vector equation r = r0 + tv , where r = <x,y,z>.

## [Calculus][Power Series] Evaluate the indefinite integral as a power series. Integral of x^2 * ln(1+x) dx

I have it worked out based on the answer in the book, but I don't completely understand what the reasoning is behind changing the index to get the final answer.

x 2 ⨜ 1/1+x dx = x 2 ⨜ Σ(n=1) (-1) n (x) n

x 2 ln(1+x) = x 2 Σ(n=1) (-1) n [(x) n+1 /n+1]

x 2 ln(1+x) = Σ(n=1) (-1) n [(x) n+3 /n+1]

⨜ x 2 ln(1+x) = ⨜ Σ(n=1) (-1) n [(x) n+3 /n+1]

⨜ x 2 ln(1+x) = ⨜ Σ(n=0) (-1) n-1 [(x) n+2 /n] <---------5

⨜ x 2 ln(1+x) = Σ(n=0) (-1) n-1 [(x) n+3 /n(n+3)]

Is step 5 correct? I'm starting with an index of 1 to infinity because of the ln function and changing everything in step 5 to start at 0. This is the only way I can think of to match the answer in the back of the book.

At a glance, I think that going from:

Probably not, but for your index shifting to be correct, you need to derive an equivalent sequence definition such that [ A_1 ] of the original sequence is the same as [ B_0 ] of the new sequence. Consider an analogous process with:

Where [ A ] is indexed starting at 0, and we want to re-index to a sequence definition for [ B ] , with an index starting at [ 1] So we want [ A_0 = B_1 ] , so let's assert that:

So you can see that moving the index forward by one is the equivalent of replacing the index [ i ] in the equation with [ i - 1 ] , this makes sense when you think about it on the numberline. If you move the starting position back by 1 step, but keep all the runners in the same place, then it's as if all the runners have moved forward an extra step.

In general, re-indexing ɿorward' by a step s is equivalent to replacing the index with [ i - s ] , moving backward simply occurs when you have a negative [ s ] .

EDIT: Some typos and stuff

OP, here's your actual problem. Everywhere you have n=0 should be n=1 and vice versa. Otherwise it all looks good.

Re-indexing isn't strictly necessary to solve the problem, but might be to match the given answer, as you saw.

First: Don't pull the x 2 outside of the integral. Why could you do that?

Second: You changed ln(1+x) to ∫1/(1+x)dx. In the context of the problem, then, you would need to be taking the integral OF that integral. something you probably haven't ever seen, and a very bad sign for the problem.

Instead: Rewrite ln(1+x) using its power series representation. Multiply everything by x 2 . Then take the integral of THAT form.

I'm very confused by your comment and its upvotes. It's possible OP edited, but I don't see the * so I don't think this happened.

First: Don't pull the x 2 outside of the integral. Why could you do that?

What? Look again. This didn't happen. I guess I see why you said it, but OP didn't do anything wrong here. x 2 never got pulled out of the appropriate integral.

Second: You changed ln(1+x) to ∫1/(1+x)dx. In the context of the problem, then, you would need to be taking the integral OF that integral. something you probably haven't ever seen, and a very bad sign for the problem.

Again, what's wrong with this? They are equivalent, and I don't see what your integral of an integral or that not being very common have to do with anything.

Instead: Rewrite ln(1+x) using its power series representation. Multiply everything by x 2. Then take the integral of THAT form.

This is what OP did. And here is why I'm super confused about your comment. What are you even saying? OP found the power series representation of ln(1+x) BY rewriting is as ∫1/(1+x)dx which is exactly what I would expect any student to do.

(I should note I agree that they started with the wrong index, should be n=0.)

## 5. (-/1 Points) DETAILS LARCALC9 8.8.002. Decide whether the integral is improper. dx X4 proper improper.

5. [-/2 points) DETAILS LARCALC9 4.1.022. Find the indefinite integral and check your result by differentiation. (Use C for the constant of integration.) (x+ dx Need Help? Read it | Talk to a Tutor Show My Work (Required) What steps or reasoning did you use? Your work counts towards your score. You can submit show my work an unlimited number of times.

### Question 2 please 1 and 2, determine whether or not the integral is In exercises improper. If it is improper, explain why 12. (a) 12 x-2/5 dx 「x-2/5 dx 「x2/5 dx (b) (c) I. (a) 0 13. (a 40 1 dx 2 x.

Question 2 please 1 and 2, determine whether or not the integral is In exercises improper. If it is improper, explain why 12. (a) 12 x-2/5 dx 「x-2/5 dx 「x2/5 dx (b) (c) I. (a) 0 13. (a 40 1 dx 2 x 14. (a In exercises 3-18, determine whether the integral converges or diverges. Find the value of the integral if it converges. 15. (a (b)人1x-4/3 dr 3, (a) l.lyMdx (b) x43 dx 16. (a 4. (a) 45 dx.

### 25. [-13 Points] DETAILS SCALCETS 11.3.008. Use the Integral Test to determine whether the series is.

25. [-13 Points] DETAILS SCALCETS 11.3.008. Use the Integral Test to determine whether the series is convergent or divergent. I den n = 1 Evaluate the following integral. 5. 3 хе-х5 dx he Since the integral ---Select--- finite, the series is --Select-- V

### 4. (1.02/7.14 Points] DETAILS PREVIOUS ANSWERS SCALCETS 11.3.512.XP.MI.SA. This question has several parts that must be.

4. (1.02/7.14 Points] DETAILS PREVIOUS ANSWERS SCALCETS 11.3.512.XP.MI.SA. This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Use the Integral Test to determine whether the series is convergent or divergent. ne-108 Step 1 In order to apply the Integral Test for the function f(x) on the.

### DETAILS SCALCET8 5. 8. [-/1 Points] Evaluate the integral. So'x60Vx + 74x) dx Need Help? Read.

DETAILS SCALCET8 5. 8. [-/1 Points] Evaluate the integral. So'x60Vx + 74x) dx Need Help? Read It Watch It Show My Work (Optional) 9. [-11 Points] DETAILS SCALCET8 5.4.03 Evaluate the integral. DETAILS SC 9. (-/1 Points] Evaluate the integral. 1 (x! + 9x)dx JO Need Help? Read It И Show My Work (Optional) The elevation at x = 2 Need Help? Read It Talk to Tutor Show My Work (Optional) DETAILS 14. (-12 points) SCALCET8 5.4.061. MY NOTES ASK.

### [1.02/7.14 Points] DETAILS PREVIOUS ANSWERS SCALCETS 11.3.512.XP.MI.SA. MY NOT This question has several parts that must.

[1.02/7.14 Points] DETAILS PREVIOUS ANSWERS SCALCETS 11.3.512.XP.MI.SA. MY NOT This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Use the Integral Test to determine whether the series convergent or divergent. Step In order to apply the Integral Test for the function f(x) on the interval (a.

### 0 points) The numbers 5 = xo <*, < x2 <<Xn-1 < Xn = 8 divide.

0 points) The numbers 5 = xo <*, < x2 <<Xn-1 < Xn = 8 divide the closed interval [5, 8) inton ny intervals of equal length Ax. Show your work and justify your answers. a) if n = 12, what is the value of Ax? 5) For n = 12, write a formula for Xi, the right endpoint of the i-th interval. =) Express lim 27-11(5 + )". (E)1 as a definite integral, identifying the integrand, the lower limit.

### 7. [-17 Points) DETAILS LARCALCET7 4.2.043.EP. Emal. Campbell, Outlook MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER.

7. [-17 Points) DETAILS LARCALCET7 4.2.043.EP. Emal. Campbell, Outlook MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Consider the following function and closed interval FX)-3X, [1, 2] Is i continuous on the closed interval [1, 2]? Yes NO Iff is differentiable on the open interval (1, 2), find ). (If it is not differentiable on the open interval, enter ONE.) Find (1) and R2). (1) - R2) - Find for (a, b) - (1,21 b) - Determine whether the Mean Value.

### 1. 5 points Let G, H, J, and P denote the following propositional variables: G: "Jo knows where the gold is hidden.

1. 5 points Let G, H, J, and P denote the following propositional variables: G: "Jo knows where the gold is hidden." J : "Jo is a knight." H: "Pat knows where the gold is hidden." P: "Pat is a knight." One fine day, while we are strolling along on The Island of Knights & Knaves, we meet Jo and Pat, each of whom is an inhabitant of The Island of Knights & Knaves Jo says "We both know where.