# 4.12: Triple Integrals (Exercises 2)

## Terms and Concepts

1. The strategy for establishing bounds for triple integrals is "from ________ to ________, then from ________ to ________ and then from ________ to ________."

We integrate from surface to surface, then from curve to curve and then from point to point.

2. Give an informal interpretation of what (intintint_Q ,dV) means.

(intintint_Q ,dV) = Volume of the solid region (Q)

3. Give two uses of triple integration.

To compute total mass or average density of a solid object, given a density function or to compute the average temperature in a solid region or object.

4. If an object has a constant density (delta) and a volume (V), what is its mass?

It's mass is (delta V).

## Volume of Solid Regions

In Exercises 5-8, two surfaces (f_1(x,y)) and (f_2(x,y)) and a region (R) in the (xy)-plane are given. Set up and evaluate the triple integral that represents the volume between these surfaces over (R).

5. (f_1(x,y) = 8-x^2-y^2,,f_2(x,y) =2x+y;)
(R) is the square with corners ((-1,-1)) and ((1,1)).

6. (f_1(x,y) = x^2+y^2,,f_2(x,y) =-x^2-y^2;)
(R) is the square with corners ((0,0)) and ((2,3)).

7. (f_1(x,y) = sin x cos y,,f_2(x,y) =cos x sin y +2;)
(R) is the triangle with corners ((0,0), ,(pi , 0)) and ((pi,pi)).

8. (f_1(x,y) = 2x^2+2y^2+3,,f_2(x,y) =6-x^2-y^2;)
(R) is the circle (x^2+y^2=1).

In Exercises 9-16, a domain (D) is described by its bounding surfaces, along with a graph. Set up the triple integral that gives the volume of (D) in the indicated order of integration, and evaluate the triple integral to find this volume.

9. (D) is bounded by the coordinate planes and (z=2-frac{2}{3}x-2y).
Evaluate the triple integral with order (dz,dy,dx).

10. (D) is bounded by the planes (y=0,y=2,x=1,z=0) and (z=(2-x)/2).
Evaluate the triple integral with order (dx,dy,dz).

11. (D) is bounded by the planes (x=0,x=2,z=-y) and by (z=y^2/2).
Evaluate the triple integral with order (dy,dz,dx).

12. (D) is bounded by the planes (z=0,y=9, x=0) and by (z=sqrt{y^2-9x^2}).
Do not evaluate any triple integral. Just set this one up.

13. (D) is bounded by the planes (x=2,y=1,z=0) and (z=2x+4y-4).
Evaluate the triple integral with order (dx,dy,dz).

14. (D) is bounded by the plane (z=2y) and by (y=4-x^2).
Evaluate the triple integral with order (dz,dy,dx).

15. (D) is bounded by the coordinate planes and (y=1-x^2) and (y=1-z^2).
Do not evaluate any triple integral. Which order would be easier to evaluate: (dz,dy,dx) or (dy,dz,dx)? Explain why.

16. (D) is bounded by the coordinate planes and by (z=1-y/3) and (z=1-x).
Evaluate the triple integral with order (dx,dy,dz).

In Exercises 17-20, evaluate the triple integral.

17. (displaystyle int_{-pi/2}^{pi/2}int_{0}^{pi}int_{0}^{pi} (cos x sin y sin z ),dz,dy,dx)

18. (displaystyle int_{0}^{1}int_{0}^{x}int_{0}^{x+y} (x+y+z ),dz,dy,dx)

19. (displaystyle int_{0}^{pi}int_{0}^{1}int_{0}^{z} (sin (yz)),dx,dy,dz)

20. (displaystyle int_{pi}^{pi^2}int_{x}^{x^3}int_{-y^2}^{y^2} (cos x sin y sin z ),dz,dy,dx)

In the following exercises, evaluate the triple integrals over the rectangular solid box (B).

[iiint_B (2x + 3y^2 + 4z^3) space dV,] where (B = {(x,y,z) | 0 leq x leq 1, space 0 leq y leq 2, space 0 leq z leq 3})

[Hide Solution]

(192)

[iiint_B (xy + yz + xz) space dV,] where (B = {(x,y,z) | 1 leq x leq 2, space 0 leq y leq 2, space 1 leq z leq 3})

[iiint_B (x space cos space y + z) space dV,] where (B = {(x,y,z) | 0 leq x leq 1, space 0 leq y leq pi, space -1 leq z leq 1})

[Hide solution]

(0)

[iiint_B (z space sin space x + y^2) space dV,] where (B = {(x,y,z) | 0 leq x leq pi, space 0 leq y leq 1, space -1 leq z leq 2})

In the following exercises, change the order of integration by integrating first with respect to (z), then (x), then (y).

[int_0^1 int_1^2 int_2^3 (x^2 + ln space y + z) space dx space dy space dz]

[Hide Solution]

[int_0^1 int_1^2 int_2^3 (x^2 + ln space y + z) space dx space dy space dz = frac{35}{6} + 2 space ln 2]

[int_0^1 int_{-1}^1 int_0^3 (ze^x + 2y) space dx space dy space dz]

[int_{-1}^2 int_1^3 int_0^4 left(x^2z + frac{1}{y} ight) space dx space dy space dz]

[Hide solution]

[int_{-1}^2 int_1^3 int_0^4 left(x^2z + frac{1}{y} ight) space dx space dy space dz = 64 + 12 space ln space 3]

[int_1^2 int_{-2}^{-1} int_0^1 frac{x + y}{z} space dx space dy space dz]

Let (F), (G), and (H) be continuous functions on ([a,b]), ([c,d]), and ([e,f]), respectively, where (a, space b, space c, space d, space e), and (f) are real numbers such that (a < b, space c < d), and (e < f). Show that

[int_a^b int_c^d int_e^f F (x) space G (y) space H(z) space dz space dy space dx = left(int_a^b F(x) space dx ight) left(int_c^d G(y) space dy ight) left(int_e^f H(z) space dz ight).]

Let (F), (G), and (H) be differential functions on ([a,b]), ([c,d]), and ([e,f]), respectively, where (a, space b, space c, space d, space e), and (f) are real numbers such that (a < b, space c < d), and (e < f). Show that

[int_a^b int_c^d int_e^f F' (x) space G' (y) space H'(z) space dz space dy space dx = [F (b) - F (a)] space [G(d) - G(c)] space H(f) - H(e)].]

In the following exercises, evaluate the triple integrals over the bounded region

(E = {(x,y,z) | a leq x leq b, space h_1 (x) leq y leq h_2 (x), space e leq z leq f }.)

[iiint_E (2x + 5y + 7z) space dV, ] where (E = {(x,y,z) | 0 leq x leq 1, space 0 leq y leq -x + 1, space 1 leq z leq 2})

[Hide solution]

(frac{77}{12})

[iiint_E (y space ln space x + z) space dV,] where (E = {(x,y,z) | 1 leq x leq e, space 0 leq y ln space x, space 0 leq z leq 1})

[iiint_E (sin space x + sin space y) dV,] where (E = {(x,y,z) | 0 leq x leq frac{pi}{2}, space -cos space x leq y cos space x, space -1 leq z leq 1 })

[Hide Solution]

(2)

[iiint_E (xy + yz + xz ) dV] where (E = {(x,y,z) | 0 leq x leq 1, space -x^2 leq y leq x^2, space 0 leq z leq 1 })

In the following exercises, evaluate the triple integrals over the indicated bounded region (E).

[iiint_E (x + 2yz) space dV,] where (E = {(x,y,z) | 0 leq x leq 1, space 0 leq y leq x, space 0 leq z leq 5 - x - y })

[Hide Solution]

(frac{430}{120})

[iiint_E (x^3 + y^3 + z^3) space dV,] where (E = {(x,y,z) | 0 leq x leq 2, space 0 leq y leq 2x, space 0 leq z leq 4 - x - y })

[iiint_E y space dV,] where (E = {(x,y,z) | -1 leq x leq 1, space -sqrt{1 - x^2} leq y leq sqrt{1 - x^2}, space 0 leq z leq 1 - x^2 - y^2 })

[Hide Solution]

(0)

[iiint_E x space dV,] where (E = {(x,y,z) | -2 leq x leq 2, space -4sqrt{1 - x^2} leq y leq sqrt{4 - x^2}, space 0 leq z leq 4 - x^2 - y^2 })

In the following exercises, evaluate the triple integrals over the bounded region (E) of the form

(E = {(x,y,z) | g_1 (y) leq x leq g_2(y), space c leq y leq d, space e leq z leq f }).

[iiint_E x^2 space dV,] where (E = {(x,y,z) | 1 - y^2 leq x leq y^2 - 1, space -1 leq y leq 1, space 1 leq z leq 2 })

[Hide Solution]

(-frac{64}{105})

[iiint_E (sin space x + y) space dV,] where (E = {(x,y,z) | -y^4 leq x leq y^4, space 0 leq y leq 2, space 0 leq z leq 4})

[iiint_E (x - yz) space dV,] where (E = {(x,y,z) | -y^6 leq x leq sqrt{y}, space 0 leq y leq 1x, space -1 leq z leq 1 })

[Hide Solution]

(frac{11}{26})

[iiint_E z space dV,] where (E = {(x,y,z) | 2 - 2y leq x leq 2 + sqrt{y}, space 0 leq y leq 1x, space 2 leq z leq 3 })

In the following exercises, evaluate the triple integrals over the bounded region

(E = {(x,y,z) | g_1(y) leq x leq g_2(y), space c leq y leq d, space u_1(x,y) leq z leq u_2 (x,y) })

[iiint_E z space dV,] where (E = {(x,y,z) | -y leq x leq y, space 0 leq y leq 1, space 0 leq z leq 1 - x^4 - y^4 })

[Hide Solution]

(frac{113}{450})

[iiint_E (xz + 1) space dV,] where (E = {(x,y,z) | 0 leq x leq sqrt{y}, space 0 leq y leq 2, space 0 leq z leq 1 - x^2 - y^2 })

[iiint_E (x - z) space dV,] where (E = {(x,y,z) | - sqrt{1 - y^2} leq x leq y, space 0 leq y leq frac{1}{2}x, space 0 leq z leq 1 - x^2 - y^2 })

[Hide Solution]

(frac{1}{160}(6 sqrt{3} - 41))

[iiint_E (x + y) space dV,] where (E = {(x,y,z) | 0 leq x leq sqrt{1 - y^2}, space 0 leq y leq 1x, space 0 leq z leq 1 - x })

In the following exercises, evaluate the triple integrals over the bounded region

(E = {(x,y,z) | (x,y) in D, space u_1 (x,y) x leq z leq u_2 (x,y) }), where (D) is the projection of (E) onto the (xy)-plane

[iint_D left(int_1^2 (x + y) space dz ight) space dA,] where (D = {(x,y) | x^2 + y^2 leq 1})

[Hide Solution]

(frac{3pi}{2})

[iint_D left(int_1^3 x (z + 1)space dz ight) space dA,] where (D = {(x,y) | x^2 -y^2 geq 1, space x leq sqrt{5}})

[iint_D left(int_0^{10-x-y} (x + 2z) space dz ight) space dA,] where (D = {(x,y) | y geq 0, space x geq 0, space x + y leq 10})

[Hide Solution]

(1250)

[iint_D left(int_0^{4x^2+4y^2} y space dz ight) space dA,] where (D = {(x,y) | x^2 + y^2 leq 4, space y geq 1, space x geq 0})

The solid (E) bounded by (y^2 + z^2 = 9, space z = 0), and (x = 5) is shown in the following figure. Evaluate the integral [iiint_E z space dV] by integrating first with respect to (z), then (y), and then (x).

[Hide Solution]

[int_0^5 int_{-3}^3 int_0^{sqrt{9-y^2}} z space dz space dy space dx = 90]

The solid (E) bounded by (y = sqrt{x}, space x = 4, space y = 0), and (z = 1) is given in the following figure. Evaluate the integral [iiint_E xyz space dV] by integrating first with respect to (x), then (y), and then (z).

[T] The volume of a solid (E) is given by the integral [int_{-2}^0 int_x^0 int_0^{x^2+y^2} dz space dy space dx.] Use a computer algebra system (CAS) to graph (E) and find its volume. Round your answer to two decimal places.

[Hide Solution]

(V = 5.33)

[T] The volume of a solid (E) is given by the integral [int_{-1}^0 int_{-x^3}^0 int_0^{1+sqrt{x^2+y^2}} dz space dy space dx.] Use a CAS to graph (E) and find its volume (V). Round your answer to two decimal places.

In the following exercises, use two circular permutations of the variables (x, space y,) and (z) to write new integrals whose values equal the value of the original integral. A circular permutation of (x, space y), and (z) is the arrangement of the numbers in one of the following orders: (y, space z,) and (x) or (z, space x,) and (y).

[int_0^1 int_1^3 int_2^4 (x^2z^2 + 1) dx space dy space dz]

[Hide Solution]

[int_0^1 int_1^3 int_2^4 (y^2z^2 + 1) dz space dx space dy;] [int_0^1 int_1^3 int_2^4 (x^2z^2 + 1) dx space dy space dz]

[int_0^3 int_0^1 int_0^{-x+1} (2x + 5y + 7z) dy space dx space dz]

[int_0^1 int_{-y}^y int_0^{1-x^4-y^4} ln space x dz space dx space dy]

[int_{-1}^1 int_0^1 int_{-y^6}^{sqrt{y}} (x + yz) dx space dy space dz]

Set up the integral that gives the volume of the solid (E) bounded by (y^2 = x^2 + z^2) and (y = a^2), where (a > 0).

[Hide Solution]

[V = int_{-a}^a int_{-sqrt{a^2-z^2}}^{sqrt{a^2-z^2}} int_{sqrt{x^2+z^2}}^{a^2} dy space dx space dz]

Set up the integral that gives the volume of the solid (E) bounded by (x = y^2 + z^2) and (x = a^2), where (a > 0).

Find the average value of the function (f(x,y,z) = x + y + z) over the parallelepiped determined by (x + 0, space x = 1, space y = 0, space y = 3, space z = 0), and (z = 5).

[Hide Solution]

(frac{9}{2})

Find the average value of the function (f(x,y,z) = xyz) over the solid (E = [0,1] imes [0,1] imes [0,1]) situated in the first octant.

Find the volume of the solid (E) that lies under the plane (x + y + z = 9) and whose projection onto the (xy)-plane is bounded by (x = sin^{-1} y, space y = 0), and (x = frac{pi}{2}).

Consider the pyramid with the base in the (xy)-plane of ([-2,2] imes [-2,2]) and the vertex at the point ((0,0,8)).

a. Show that the equations of the planes of the lateral faces of the pyramid are (4y + z = 8, space 4y - z = -8, space 4x + z = 8), and (-4x + z = 8).

b. Find the volume of the pyramid.

[Hide Solution]

a. Answers may vary; b. (frac{128}{3})

Consider the pyramid with the base in the (xy)-plane of ([-3,3] imes [-3,3]) and the vertex at the point ((0,0,9)).

a. Show that the equations of the planes of the side faces of the pyramid are (3y + z = 9, space 3y + z = 9, space y = 0) and (x = 0).

b. Find the volume of the pyramid.

The solid (E) bounded by the sphere of equation (x^2 + y^2 + z^2 = r^2) with (r > 0) and located in the first octant is represented in the following figure.

a. Write the triple integral that gives the volume of (E) by integrating first with respect to (z), then with (y), and then with (x).

b. Rewrite the integral in part a. as an equivalent integral in five other orders.

[Hide Solution]

[a. space int_0^4 int_0^{sqrt{r^2-x^2}} int_0^{sqrt{r^2-x^2-y^2}} dz space dy space dx; space b. space int_0^2 int_0^{sqrt{r^2-x^2}} int_0^{sqrt{r^2-x^2-y^2}} dz space dx space dy,]

[int_0^r int_0^{sqrt{r^2-x^2}} int_0^{sqrt{r^2-x^2-y^2}} dy space dx space dz, space int_0^r int_0^{sqrt{r^2-x^2}} int_0^{sqrt{r^2-x^2-y^2}} dy space dz space dx,]

[int_0^r int_0^{sqrt{r^2-x^2}} int_0^{sqrt{r^2-x^2-y^2}} dx space dy space dz, space int_0^r int_0^{sqrt{r^2-x^2}} int_0^{sqrt{r^2-x^2-y^2}} dx space dz space dy,]

The solid (E) bounded by the sphere of equation (9x^2 + 4y^2 + z^2 = 1) and located in the first octant is represented in the following figure.

a. Write the triple integral that gives the volume of (E) by integrating first with respect to (z) then with (y) and then with (x).

b. as an equivalent integral in five other orders.

Find the volume of the prism with vertices ((0,0,0), space (2,0,0), space (2,3,0), space (0,3,0), space (0,0,1)), and ((2,0,1)).

[Hide Solution]

(3)

Find the volume of the prism with vertices ((0,0,0), space (4,0,0), space (4,6,0), space (0,6,0), space (0,0,1)), and ((4,0,1)).

The solid (E) bounded by (z = 10 - 2x - y) and situated in the first octant is given in the following figure. Find the volume of the solid.

[Hide Solution]

(frac{250}{3})

The solid (E) bounded by (z = 1 - x^2) and situated in the first octant is given in the following figure. Find the volume of the solid.

The midpoint rule for the triple integral [iiint_B f(x,y,z) dV] over the rectangular solid box (B) is a generalization of the midpoint rule for double integrals. The region (B) is divided into subboxes of equal sizes and the integral is approximated by the triple Riemann sum [sum_{i=1}^l sum_{j=1}^m sum_{k=1}^n f(ar{x_i}, ar{y_j}, ar{z_k}) Delta V,] where ((ar{x_i}, ar{y_j}, ar{z_k})) is the center of the box (B_{ijk}) and (Delta V) is the volume of each subbox. Apply the midpoint rule to approximate [iiint_B x^2 dV] over the solid (B = {(x,y,z) | 0 leq x leq 1, space 0 leq y leq 1, space 0 leq z leq 1 }) by using a partition of eight cubes of equal size. Round your answer to three decimal places.

[Hide Solution]

(frac{5}{16} approx 0.313)

[T]

a. Apply the midpoint rule to approximate [iiint_B e^{-x^2} dV] over the solid (B = {(x,y,z) | 0 leq x leq 1, space 0 leq y leq 1, space 0 leq z leq 1 }) by using a partition of eight cubes of equal size. Round your answer to three decimal places.

b. Use a CAS to improve the above integral approximation in the case of a partition of (n^3) cubes of equal size, where (n = 3,4, ..., 10).

Suppose that the temperature in degrees Celsius at a point ((x,y,z)) of a solid (E) bounded by the coordinate planes and (x + y + z = 5) is (T (x,y,z) = xz + 5z + 10). Find the average temperature over the solid.

[Hide Solution]

(frac{35}{2})

Suppose that the temperature in degrees Fahrenheit at a point ((x,y,z)) of a solid (E) bounded by the coordinate planes and (x + y + z = 5) is (T(x,y,z) = x + y + xy). Find the average temperature over the solid.

Show that the volume of a right square pyramid of height (h) and side length (a) is ( v = frac{ha^2}{3}) by using triple integrals.

Show that the volume of a regular right hexagonal prism of edge length (a) is (frac{3a^3 sqrt{3}}{2}) by using triple integrals.

Show that the volume of a regular right hexagonal pyramid of edge length (a) is (frac{a^3 sqrt{3}}{2}) by using triple integrals.

If the charge density at an arbitrary point ((x,y,z)) of a solid (E) is given by the function ( ho (x,y,z)), then the total charge inside the solid is defined as the triple integral [iiint_E ho (x,y,z) dV.] Assume that the charge density of the solid (E) enclosed by the paraboloids (x = 5 - y^2 - z^2) and (x = y^2 + z^2 - 5) is equal to the distance from an arbitrary point of (E) to the origin. Set up the integral that gives the total charge inside the solid (E).

One of the essential tools in Calculus is Integration. We use it to find anti-derivatives, the area of two-dimensional regions, volumes, central points, among many other ways. Knowing how to use integration rules is, therefore, key to being good at Calculus.
But, let’s start with the basics Integrals. What is Integrals?

There are two types of integrals: The indefinite integral and the definite integral. The indefinite integral f(x) , which is denoted by f(x) dx , is the anti-derivative of f(x) . The derivative of f(x) dx is, therefore, f(x) . As you might already be aware, the derivative of a constant is always 0 . For this reason, indefinite integrals are only defined up to some arbitrary constant. Consider sin(x)dx = -cos (x) + constant . The derivative of –cos(x) + constant is sin (x) .

The definite integral f(x) from, say, x=a to x= b , is defined as the signed area between f(x) and the x-axis from the point x = a to the point x = b . The definite integral is denoted by a f(x) d(x) .

It is important to note that both the definite and indefinite integrals are interlinked by the fundamental theorem of calculus. The theorem states that if f(x) is continuous on [a,b] , and F(x) is its continuous indefinate integral, then a f(x) dx= F(b) – F(a) . This, therefore, means that 0 sin(x) dx = <-cos(π)>– <-cos(0)>= 2 .

Note, however, that sometimes, you may be required to find an approximation to a definite integral. The most common way to do this is to have several thin rectangles under the curve from the initial point x = a to the last point x = b . Add the signed areas (areas of the rectangles) together, and there you go! You have the desired definite integral.

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## Integration by Substitution Calculator

### Difficult Problems

Solved example of integration by substitution

We can solve the integral $int xcosleft(2x^2+3 ight)dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $2x^2+3$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

Differentiate both sides of the equation $u=2x^2+3$

The derivative of a sum of two functions is the sum of the derivatives of each function

The derivative of the constant function ($3$) is equal to zero

The derivative of a function multiplied by a constant ($2$) is equal to the constant times the derivative of the function

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^$

Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

## 4.12: Triple Integrals (Exercises 2)

### Problem 1: The blood groups of 30 students of Class VIII are recorded as follows:

A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,

A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.

### Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?

Rarest blood group (least frequency blood group) :- AB.

### Problem 2: The distance (in km) of 40 engineers from their residence to their place of work were found as follows:

5 3 10 20 25 11 13 7 12 31

19 10 12 17 18 11 32 17 16 2

7 9 7 8 3 5 12 15 18 3

12 14 2 9 6 15 15 7 6 12

### Problem 3: The relative humidity (in %) of a certain city for a month of 30 days was as follows:

98.1 98.6 99.2 90.3 86.5 95.3 92.9 96.3 94.2 95.1

89.2 92.3 97.1 93.5 92.7 95.1 97.2 93.3 95.2 97.3

96.2 92.1 84.9 90.2 95.7 98.3 97.3 96.1 92.1 89

(i) Construct a grouped frequency distribution table with classes 84 – 86, 86 – 88, etc.

(ii) Which month or season do you think this data is about?

(iii) What is the range of this data?

### Problem 4: The heights of 50 students, measured to the nearest centimetres, have been found to be as follows:

161 150 154 165 168 161 154 162 150 151

162 164 171 165 158 154 156 172 160 170

153 159 161 170 162 165 166 168 165 164

154 152 153 156 158 162 160 161 173 166

161 159 162 167 168 159 158 153 154 159

(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 – 165, 165 – 170, etc.

(ii) What can you conclude about their heights from the table?

### Problem 5: A study was conducted to find out the concentration of sulphur dioxide in the air in

2020-21 parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:

0.03 0.08 0.08 0.09 0.04 0.17

0.16 0.05 0.02 0.06 0.18 0.20

0.11 0.08 0.12 0.13 0.22 0.07

0.08 0.01 0.10 0.06 0.09 0.18

0.11 0.07 0.05 0.07 0.01 0.04

(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08, and so on.

(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?

### Problem 6: Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:

0 1 2 2 1 2 3 1 3 0

1 3 1 1 2 2 0 1 2 1

3 0 0 1 1 2 3 2 2 0

### Problem 7: The value of π upto 50 decimal places is given below:

3.14159265358979323846264338327950288419716939937510

(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.

(ii) What are the most and the least frequently occurring digits?

### Problem 8: Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:

1 6 2 3 5 12 5 8 4 8

10 3 4 12 2 8 15 1 17 6

3 2 8 5 9 6 8 7 14 12

(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5 – 10.

(ii) How many children watched television for 15 or more hours a week?

### Problem 9: A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:

2.6 3.0 3.7 3.2 2.2 4.1 3.5 4.5

3.5 2.3 3.2 3.4 3.8 3.2 4.6 3.7

2.5 4.4 3.4 3.3 2.9 3.0 4.3 2.8

3.5 3.2 3.9 3.2 3.2 3.1 3.7 3.4

4.6 3.8 3.2 2.6 3.5 4.2 2.9 3.6

Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 – 2.5.

## 9.3. NUMERICAL INTEGRATION

Midpoint rule. Goal: given continuous function f(x) of one variable, compute &int f(x) dx over interval from a to b. Estimate integral by M = (b-a) * f(c), where c = (a + b)/2.

Trapezoidal rule. Goal: given continuous function f(x) of one variable, compute &int f(x) dx over interval from a to b. TrapezoidalRule.java numerically integrates a function of one variable using the trapezoidal rule. We can estimate the integral of f(x) from a to b using the formula T = (b-a)/2 (f(a) + f(b)). Breaking the interval from a to b up into N equally spaced intervals (and combining common terms) we obtain the formula:

where the interval [a, b] is broken up into N subintervals of uniform size h = (b - a) / N. Under certain technical conditions, if N is large then the formula above is a good estimate of the integral.

Simpson's rule. The trapezoidal rule is rarely used to integrate in practice. For smooth f, the midpoint rule is approximately twice as accurate as the trapezoidal rule, and the errors have different signs. By combining the two expressions, we obtain a more accurate estimate of f: S = 2/3*M + 1/3*T. This combination is known as Simpson's 1/3 rule. S = (b-a)/6 (f(a) + 4f(c) + f(b)), where c = (a + b)/2. Breaking the interval from a to b up into N equally spaced intervals (and combining common terms) we obtain the formula:

where the interval [a, b] is broken up into N subintervals of uniform size h = (b - a) / (N - 1) and the 2/3 and 4/3 coefficients alternate throughout the interior. Here are some nice animations of numerical quadrature. Under certain technical conditions, if N is large then the formula above is a good estimate of the integral. The program SimpsonsRule.java numerically integrates x^4 log (x + sqrt(x^2 + 1)) from a = 0 to b = 2.

Program organization. To write a reusable intergration routine, we would like to be able to create a class TrapezoidalRule and pass an arbitrary continuous function to be integrated. One way of accomplishing this is to declare an interface for continuous functions that has a single method eval, which takes a real argument x and returns f(x).

Then, we could implement the error function as

Adaptive quadrature. The Matlab function quad uses adapative quadrature with extrapolated Simpson's rule. [Reference Chapter 6, Cleve Moler.]

If we are careful, we can save function evaluations from one iteration to the next and get away with just two function evaluations per recursive invocation. (See exercise.)

Our method fails spectacularly when integrating 1 / (3x - 1) from x = 0 to 1. Integral has a nonsingularity and our function will go into an infinite loop. An industrial strength implementation would diagnose such a situation and report an appropriate error message.

Monte Carlo integration. Theory based on the Law of Large Numbers. Estimating area of circle by rejection method. Dartboard. Throwing darts at Irish pub after several pints. Program Dartboard.java reads in a command line integer N, throws N darts uniformly distributed in the unit box, and plots the results. The pictures below show sample results for N = 1,000, 10,000, and 100,000.

Alternate viewpoint: 2D Monte Carlo integration of f(x, y) = 1 if x 2 + y 2 &le 1. Show histogram of plot - normal with standard deviation.

To estimate the integral of f over a multi-dimensional volume V, we select N points x1, x2, . xN in the volume uniformly at random. Our estimate of the integral is the fraction of random points that fall below f multiplied by the volume = V <f>, where

The fundamental theorem of Monte Carlo integration asserts that the integral of f over V equals V <f> +- V sqrt((<f 2 > - <f> 2 ) / N). The key observation is that the error goes as 1 / sqrt(N). This means that you have to quadruple the number of simulations to double the accuracy of your approximation. However, there is no dependence on dimension! Not competitive in one or two dimensions, but its power shines in higher dimensions.

Center of mass of a torus. As an example (borrowed from p. 221 Numerical Recipes), suppose we want to estimate the weight and center of mass of the intersection of a torus and two planes, defined by points (x, y, z) satisfying

We must compute the integrals f(x, y, z) = &rho, fx(x, y, z) = x&rho, fy(x, y, z) = y&rho, and fz(x, y, z) = z&rho. The center of mass is then (fx/f, fy/f, fz/f). To sample a point (x, y, z) uniformly from the torus, we use the rejection method. In this case, the torus in enclosed in the rectangular box with 1 &le x &le 4, -3 &le y &le 4, and -1 &le z &le 1. Program Torus.java takes a command line parameter N and computes these quantities using N sample points.

Gaussian cdf. Estimate the Gaussian cdf Phi(z) by generating random numbers from a Gaussian distribution and recording the fraction of values less than z. This is an example of importance sampling. Note: can't sample uniformly between -infinity and infinity anyway.

## General Form of the Length of a Curve

If the horizontal distance is "dx" (or "a small change in x") and the vertical height of the triangle is "dy" (or "a small change in y") then the length of the curved arc "dr" is approximated as:

Now, if we move point E very close to point F, we will have a very good approximation for the length of the curve in that local region.

We need to add all those infinitesimally small lengths. We use integration, as it represents the sum of such infinitely small distances. We have for the distance between where x = a to x = b:

By performing simple surd manipulation, we can express this in a more familiar form as follows.

The arc length of the curve y = f(x) from x = a to x = b is given by:

Of course, we are assuming the function y = f(x) is continuous in the region x = a to x = b (otherwise, the formula won't work).

### Back to the Width of the Sheet example

Use the above formula to find the required width of the metal sheet in our example.

We can now use this formula to find the required width of our flat sheet of iron. Remember, we're finding the width needed for one wave, then we'll multiply by the number of waves.

For one period, the lower limit is x = 0 and the upper limit is x = 10.67. Substituting these into our formula gives us one wavelength of the curved material:

(You can see where this answer comes from in Wolfram|Alpha.)

The corrugated sheet has (106.7)/(10.67)=10 complete wavelengths across its width, so to give a corrugated width of 106.7 "cm", the original flat sheet will need to be:

### Example 2 - Parabolic dish application

The cross section of a parabolic dish is modelled by y = 0.04x 2 (in meters), from x = &minus5 m to x = 5 m. We need to construct a steel support band which is to be placed around the outer edge of the dish. How long should we make the steel band?

We need to find the curved distance AB.

Estimate: Looking at the graph (which has equal scaling along the x- and y-axes), we can see the final answer should be a little more than 10 m, somewhere between 10 m and 11 m.

Now to find the exact length:

The lower limit is x = &minus5 and the upper limit is x = 5. Substituting these into our formula gives:

[Note: I used a computer algebra system to find the above integral. Many arc length problems lead to impossible integrals. (This example does have a solution, but it is not straightforward.) Often the only way to solve arc length problems is to do them numerically, or using a computer. You can see the answer in Wolfram|Alpha.]

So the length of the steel supporting band should be 10.26 m.

This is consistent with our earlier estimate.

### Example 3 - Golden Gate Bridge cables

The central span of the Golden Gate Bridge in San Francisco, USA, is 1280 "m" long.

Central span of the Golden Gate Bridge. [Image source]

The height of the tower is 152 "m" from the roadway. What is the length of the main suspension cable between the 2 towers?

We first need to model the curve (find an equation that represents the curve accurately).

A freely hanging cable takes the form of a catenary. In the case of a suspension bridge, we need to also consider the mass of the roadway, which will give us a flattened catenary. The general form for a flattened catenary is the sum of 2 exponential functions, based on the hyperbolic function y=cosh x:

The Golden Gate Bridge cable is almost a catenary and almost a parabola, but not quite either (because of the weight of the cables, the suspender ropes and the roadway). For the sake of this discussion, we'll assume it is a flattened catenary.

For convenience, we'll place the origin at the lowest point of the cable.

The required curve (after initial analysis and some guess and check) passing through (-640, 152) , (0, 0) and (640, 152) is given by:

Here is the graph of the above equation. We can see it passes through the required points.

The derivative of our function is

Using the length of a curve formula, with start point x = -640 and end point x = 640, we have:

So the length of the central span of the main cable is 1327.0 "m". (You can see the answer in Wolfram|Alpha here.)

Of course, the cable continues on both sides of the towers. The total length of each cable is 2,332 "m".

Interesting trivia: The main cables of the Golden Gate Bridge are nearly one meter in diameter (actually, 0.91 m) and the "total length of galvanized steel wire used in both main cables is 129,000 km". [Source]

## Business Calculus with Excel

At the end of the last section we warned that the symbolic integration techniques we have developed only work for problems that exactly fit our formulas. When we tried integrating an exponential function where the exponent was a constant times t, we had to change the base to get a function with only t in the exponent. We want to develop one more technique of integration, that of change of variables or substitution, to handle integrals that are pretty close to our stated rules. This technique is often called u-substitution and is related to the chain rule for differentiation.

We start by exploring some examples where we can get the desired result by the guess and check technique.

###### Example 7.4.1 . Power of a linear by guess and check.

: We could do this problem by rewriting the integrand as an explicit seventh degree polynomial and then using the power and sum rules, but that is too much work. Instead I will notice the integrand looks almost like a power, and thus guess an answer of (frac<1> <8>(3x+5)^8+C ext<.>) I then check by differentiating. Using the chain rule

Thus our guess was off by a factor of 3 and the correct antiderivative is

We can easily use the same trick to produce a rule for powers of a linear polynomial.

###### Example 7.4.2 . Power of a generic linear by guess and check.

: As we did in example 1, we first guess the antiderivative to be (frac<1> (ax+b)^+C ext<.>) We then take the derivative of that expression and obtain (a(ax+b)^n ext<.>) This misses our integrand by a factor of (a ext<.>) We adjust by that factor and find the antiderivative is (frac<1> frac<1> (ax+b)^+C ext<.>)

We can easily use the same trick to produce a rule for functions that are the exponential of a linear function.

###### Example 7.4.3 . Antidifferentiation of an exponential function by guess and check.

: As we did in example 2, our first guess uses the basic rule without worrying about the linear term, so we guess (e^+C ext<.>) We then take the derivative of that expression and obtain (ae^ ext<.>) This misses our integrand by a factor of (a ext<.>) We adjust by that factor and find the antiderivative is (frac<1> e^+C ext<.>)

We run into a problem if we try to extend this method with quadratic terms. If we start with ((x^2+5)^3) and guess an antiderivative of (frac<1> <4>(x^2+5)^4 ext<,>) when we differentiate we get ((x^2+5)^3 2x) and are off by a factor of (8x ext<.>) However when we divide by that factor to get (frac<(x^2+5)^4><8x>) as a proposed antiderivative, and then differentiate again we get

which is not what we want. The key is to start by recalling the chain rule

We want to use the same rule with a different notation, using implicit differentiation and a new variable (u ext<:>)

By the fundamental theorem of calculus, we can convert this to an integration formula:

We will generally simplify (frac dx) to (du ext<,>) so our substitution rule is

Let us rework some earlier examples with this method and then illustrate the method with a more difficult problem.

###### Example 7.4.4 . Power of linear example redone with change of variables.

: The obvious candidate for (u) is (3x+5 ext<.>) Then (du=3dx ext<.>) Thus

This is easy to generalize for a power of a linear term.

###### Example 7.4.5 . Power of generic linear example redone with change of variables.

: The obvious candidate for (u) is (ax+b ext<.>) Then (du=a dx ext<.>) Hence

To use this method with u replacing something more complicated than a linear term, we need to have (du) available, with the possible addition of multiplying by a scalar constant.

###### Example 7.4.6 . Power of quadratic function with change of variables.

Find (int (2x^3+11)^7 x^2 dx ext<.>)

: The obvious candidate for (u) is (2x^3+11 ext<.>) Then (du=6x^2d)x. Thus

By convention, (u) is often used the new variable used with this change of variables technique, so the technique is often called u-substitution.

In the definite integral, we understand that a and b are the (x)-values of the ends of the integral. We could be more explicit and write (x=a) and (x=b ext<.>) The last step in solving a definite integral is to substitute the endpoints back into the antiderivative we have found. We can either change the variables for the endpoints as well, or we can convert the antiderivative back to the original variables before substituting. Consider the following example.

###### Example 7.4.7 . A definite integral with change of variables.

: (Convert everything to (u ext<.>)) The obvious candidate for (u) is (2x+5 ext<.>) Then (du=2dx ext<.>) For the lower endpoint, (x=1) becomes (u=7 ext<.>) For the upper endpoint (x=3) becomes (u=11 ext<.>) Substituting,

: (Keeping, but labeling the endpoints.) We have the same u and du, but do not convert the endpoints. To reduce confusion we make sure to label the variable when we are using both (x) and (u ext<.>) Thus,

It should be noted that when we change variables we may find ourselves looking at an integral from (a) to (b) where the (b lt a ext<.>) We do not change the order of the endpoints.

###### Example 7.4.8 . A second definite integral with change of variables.

: (Convert everything to (u ext<.>)) The obvious candidate for (u) is (x^2 ext<.>) Then (du=2x dx ext<.>) For the lower endpoint, (x=-2) becomes (u=4 ext<.>) For the upper endpoint (x=1) becomes (u=1 ext<.>) Substituting,

### Exercises Exercises: Integration by Change of Variables or Substitution Problems

Evaluate the following integrals. In each case identify the term that will be treated as u.

## General Form

Generally, one uses differentiation under the integral sign to evaluate integrals that can be thought of as belonging to some family of integrals parameterized by a real variable. To better understand this statement, consider the following example:

In the example, part of the integrand was replaced with a variable and the resultant function was studied using differentiation under the integral sign. This is a good illustration of the problem-solving principle: if stuck on a specific problem, try solving a more general problem.

Compute the definite integral ∫ 0 1 ( x ln ⁡ x ) 50 d x . int_<0>^ <1>(xln x)^ <50>, dx. ∫ 0 1 ​ ( x ln x ) 5 0 d x .

Another example illustrates the power of this technique in its general form, as one may use it to compute the Gaussian integral.

Hint: Consider the function f ( t ) = ∫ 0 2 π e t cos ⁡ θ cos ⁡ ( t sin ⁡ θ ) d θ f(t) = int_<0>^ <2pi>e^ cos(tsin heta) , d heta f ( t ) = ∫ 0 2 π ​ e t cos θ cos ( t sin θ ) d θ

### Objective

Endurance exercise training reduces insulin resistance, adipose tissue inflammation and non-alcoholic fatty liver disease (NAFLD), an effect often associated with modest weight loss. Recent studies have indicated that high-intensity interval training (HIIT) lowers blood glucose in individuals with type 2 diabetes independently of weight loss however, the organs affected and mechanisms mediating the glucose lowering effects are not known. Intense exercise increases phosphorylation and inhibition of acetyl-CoA carboxylase (ACC) by AMP-activated protein kinase (AMPK) in muscle, adipose tissue and liver. AMPK and ACC are key enzymes regulating fatty acid metabolism, liver fat content, adipose tissue inflammation and insulin sensitivity but the importance of this pathway in regulating insulin sensitivity with HIIT is unknown.

### Methods

In the current study, the effects of 6 weeks of HIIT were examined using obese mice with serine–alanine knock-in mutations on the AMPK phosphorylation sites of ACC1 and ACC2 (AccDKI) or wild-type (WT) controls.

### Results

HIIT lowered blood glucose and increased exercise capacity, food intake, basal activity levels, carbohydrate oxidation and liver and adipose tissue insulin sensitivity in HFD-fed WT and AccDKI mice. These changes occurred independently of weight loss or reductions in adiposity, inflammation and liver lipid content.

### Conclusions

These data indicate that HIIT lowers blood glucose levels by improving adipose and liver insulin sensitivity independently of changes in adiposity, adipose tissue inflammation, liver lipid content or AMPK phosphorylation of ACC.