multiple texts

#$&*

course MTH 173

12/18 23:50i had submitted all the texts much long time ago. But they do not show on my access page so i am submitting them as a whole along with their confirmations

they include

text_16,17,18,19,20,21,22,23 and 29" "Text_16

Text Section 5.4

Problem 5.4.3

Given that the value of the integral of the square of the function f(x) from a to b is 12 and that of the square of the function g(t) is 3. The integral of square of the function g(x) with respect to x will be same as that of the g(t) function with respect to t, it is just changing the variables.

Thus the value of the integral of the function (f(x))^2 - (g(x))^2 from a to b will be same as value of the integral of f(x)^2 - value of the integral of g(x)^2 from a to b. (using property of sum and differences of the integrand. Thus the value of the integral will be 12 - 3 = 9.

Problem 5.4.6

In the given integral the function has been shifted by 5 units in the direction of 5 units by using f(x-5) and at the same time the limits used are a + 5 and b + 5. This indicates that even the limits are shifted by 5 units in the same direction as that of the function. On shifting the shape of the function does not and thus the value of the integral also does not change, thus it is same as the value of the integral of f(x) with respect to x with limits a to b. thus the value of the integral is 8.

Another way to say the same is, assume t = x - 5, thus dx = dt and a + 5 - 5 = a1 and b + 5 - 5 = b1, thus the new function is f(t), the new limits are a1 = a and b1 = b and the integral is with respect to t. Now if the variable is changed from t to x, the integral will again become integral of the function f(x) from a to b which is 8.

Problem 5.4.9

The average value of a function f(x) over the interval [a,b] will be the value of the integral over the same interval divided by the length of the interval. The function f(x) is f(x) = 2 and the value of the integral over[a,b] is 2(b-a) and the length of the interval is (b-a) and thus the average value will be

= 2(b-a)/(b-a) = 2.

Problem 5.4.12

The average value will be the integral of the function over an interval divided by the length of the same interval. The unit for the integral of the function when f(x) is integrated with respect to x will be unit of f(x) * unit of x. The unit for the duration of the interval will be the unit of x. Thus the unit of the average value will be (unit of f(x)* unit of x) / (unit of x) = unit of f(x).

Thus unit of the average value is same as the unit of f(x).

Problem 5.4.15

Given y = x^2 and y = x^3 over the interval [0,1]

For this interval x^2 function lies above x^3 and thus the area of the region between this 2 curves will be equal to the value of the integral of x^2 - x^3 from x = 0 to 1. This integral is same as integral of x^2 from 0 to 1 - integral of x^3 from 0 to 1.

The value of the integral of x^2 from x = 0 to x = 1 will be 1/3

The value of the integral of x^3 from x = 0 to x = 1 will be 1/4

Thus the value of the area of the region between the graphs x^2 and x^3 will be 1/3 - Ľ = 1/12 square units.

Problem 5.4.18

Given y = cos(t) and y = sin(t) for t belongs to [0,pi]

To calculate the area of the region between the 2 curves we break the interval into 3 parts, the 1st part from t [0,pi/4], next from t [pi/4,pi/2] and from t [pi/2,pi]

For t [0,pi/4]

cos(t) is above sin(t) and thus the value of the area of the region between the 2 curves will be equal to the difference of the integral of the function cos(t) and the integral of sin(t) from [0,pi/4].

The value of the integral of the function cos(t) from [0,pi/4] = 1/sqrt(2).

The value of the integral of the function sin(t) from [0,pi/4] = -1/sqrt(2) + 1

Thus the area of the region between t = 0 to t = pi/4 will be sqrt(2) - 1.

For t [pi/4,pi/2]

sin(t) is above cos(t) and thus the value of the area of the region between the 2 curves will be equal to the difference of the integral of the function sin(t) and the integral of cos(t) from [pi/4,pi/2].

The value of the integral of the function cos(t) from [pi/4,pi/2] = 1 - 1/sqrt(2)

The value of the integral of the function sin(t) from [pi/4,pi/2] = 1/sqrt(2)..

Thus the area of the region between t = pi/4 to t = pi/2 will be sqrt(2) - 1.

For t [pi/2,pi]

The value of the integral of the function cos(t) from [pi/2,pi] = -1

The value of the integral of the function sin(t) from [pi/2,pi] = 1.

Thus the area of the region between t = pi/2 to t = pi will be 2.

Thus the total area between the curve sin(t) and cos(t) will be 2sqrt(2) - 2 + 2 = 2sqrt(2).

Problem 5.4.21

(a) The average value is defined as the value of the integral over an interval divided by the length of the same interval which is 6/3 = 2.

(b) Since the function is even the function is symmetric about y axis and thus the value of the integral of the function f(x) for the interval 0 to 3 will be same as the value of the integral from -3 to 0 and thus the value of the integral of the function from x = -3 to x = 3 will be 12 and thus the average value will be 12/6 = 2 same as the one above. This is also because the function is even.

(c) If the function is odd the function is symmetric about origin and thus the value of the integral of the function f(x) from x = -3 to x = 0 is negative to the value of the integral of the function f(x) from 0 to 3 and thus the integral of f(x) from x = -3 to x = 3 will be 0 and thus the average value will also be 0 as the average value is the value of the integral divided by the length of the interval. Since the integral is 0 thus the average value is also 0.

Problem 5.4.24

Given that the value of the integral of 2f(x) + 3 from x = 2 to x = 5 is equal to 17. This integral is same as the value of 2 * the value of the integral of f(x) from x = 2 to x = 5 + the value of the integral of 3 from x = 2 to x = 5. The value of the integral of 3 from x = 2 to x = 5 will be 3*3 = 9. Thus 2*the value of the integral of f(x) will be equal to 8 and the value of the integral of f(x) from x = 2 to x = 5 will be 8/2 = 4.

Thus the value of the integral of the function f(x) from x = 2 to x = 5 will be 4.

Problem 5.4.27

Given that the function is odd, thus the function is symmetric about origin. Thus the value of the integral of f(x) from x = -2 to x = 3 can be broken to the integral from x = -2 to x = 2 + the integral of the function from x = 2 to x = 3. Since the function is odd the value of the integral form x = -2 to x = 2 will be 0

Thus value of the integral of the function from x = - 2 to x = 3 will be equal to the value of the integral of the function from x = 2 till x = 3 = 30.

Thus the value of the integral of the function f(x) from x = 2 to x = 3 is 30

Problem 5.4.30

Given that the average value of the function f(x) over the interval x = 2 to x= 5 is 4. The length of the interval is 3, thus the value of the integral of the function f(x) from x = 2 to x = 5 will be 12.

The value of the integral of (3f(x) + 2) from x = 2 till x = 5 will be 3*(value of the integral of the function f(x) from x = 2 to x = 5 ) + the value of the integral of the function 2 from x = 2 to x = 5.

Thus the value of the integral of (3f(x) + 2) from x = 2 to x = 5 will be 3*(12) + 6 = 42.

Problem 5.4.33

(a)

From the graph it can be seen that for x values from x = -3 to x = 1 the value of the integral is negative and that for x values from x = 1 to x = 3 the value of the integral will be positive. The value of the integral is the area under the curve which is nothing but the area of the squares. Since each square is of unit length, the area of each square will be equal to 1 unit square and thus the area under the curve will be the number of squares under the curve. The area under the curve for x values from x = -3 to x = = 1 can be approximated to 6.25 square unit and thus the value of the integral can be approximated to -6.25.

The area under curve for x values from x = 1 to x = 3 can be approximated to 1.6 square units and thus the value of the integral can be approximated to 1.6. Thus the total value of the integral of the curve from x = -3 to x = 3 will be -6.25 + 1.6 = -4.65.

(b)

(I) The average value of integral of the function from x = -3 to x = 3 will be the ratio of the integral from x = -3 to x = 3 to the total duration of the integral = -4.65/6 = 0.775.

(II) The value of the integral from x = 1 to x = 3 is 1.6 and that from x = 0 to x = 1 is - 0.6. The value of the integral from x = 0 to x = 3 will be 1.6 - 0.6 = 1. The duration of the x interval is 3 and thus the average value of the function from x = 0 to x = 3 will be 1/3 = 0.334.

Thus we can see that the average value of the function over the interval x = -3 to x = 3 is greater than the average value of the function over the interval x = 0 to x = 3.

Problem 5.4.35

Given the population growth function is P = 112*(1.011) ^ t where t is the number of years since 2010.

(a) The average population between t = 2010 to t = 2050 will be equal to the value of the integral of the function from t = 0 to t = 40 / the length of the interval that is 40.

The integral function will be = f(t) = (112/ln(1.011))*((1.011)^t).

The value of the integral from t = 0 to t = 40 will be f(40) - f(0).

Which is = (112/ln(1.011))*(1.011^40 - 1) = 5620.316462

Thus the average of the function will be = 5620.316462 / 40 = 140.5079116.

(b) Population at 2010 according to the population function will be P(0) = 112.

Population at 2050 according to the population function will be P(40) = 112*(1.011)^40 = 173.4859251

Thus average population in 2010 and the predicted population in 2050 will be 142.74296255.

(c) My answer in part (b) is larger than my answer in part (a). In terms on concavity the function is concave up graph, and thus if we use trapezoidal approximation graphs to calculate the integral we can see that the average value of each trapezoid will be on the right hand side of the average value of the function for the same interval. And since the function is increasing the function will have a lower average value as compared to that calculated from the average of the end values.

Problem 5.4.36

(a) The graph of e^(-(x^2)/2) is a curve whose maximum value is 1 and the decreases on either side of x. If we draw a vertical line y = 1 for x values 0 to 1 we see that the area that is the value of the integral is 1. For e^(-(x^2)/2) the curve starts with the value of 1 at x = 0 and then is always below y = 1. Using the theorem we can say that if a function f(x) is equal to or less than g(x) then the value of the integral of f(x) will be less than the value of the integral of g(x) over the same intervals. Here f(x) = e^(-(x^2)/2) and g(x) = 1 and thus the value of the integral of e^^(-(x^2)/2) will be less than 1.

(b) The area can be found by integrating the given exponential function

The value of this integral using a calculator is approximately 0.855624.

Problem 5.4.39

A length representing f(b) - f(a) will be equal to the length on the y axis between f(b) and f(a). Draw vertical lines from x = a and x = b, reach them to the graph and then horizontal lines reaching the y axis. The distance between this parallel horizontal lines will be the length representing f(b) - f(a)

Problem 5.4.42

F(b) - F(a) / (b-a) is the average value of the function f(x) over the interval x = a to x = b. The average value can be interpreted as the height of the rectangle with the same base as the interval and the same area as the area under the curve for the function f(x) from x = a to x = b. This length can also be found out by joining f(a) and f(b) and finding the value of x = (a + b)/2 on that line. The height corresponding to that x value will be equal to the average height and will also be the length roughly approximating F(b) - F(a) / (b-a). On the graph this value will corresponding to an x value slightly less than (a + b)/2 since it is a concave down graph.

Form Confirmation

Thank you for submitting the following information:

identifyingInfo: Submitting Assignment

yourCourse: MTH 173

Contact_FirstName: Kevin

Contact_LastName: Shah

firstPendLength: text_16

email: krs2070@email.vccs.edu

11/10 21:20

response

Text_16 Text Section 5.4 Problem 5.4.3 Given that the value of the integral of the square of the function f(x) from a to b is 12 and that of the square of the function g(t) is 3. The integral of square of the function g(x) with respect to x will be same as that of the g(t) function with respect to t, it is just changing the variables. Thus the value of the integral of the function (f(x))^2 - (g(x))^2 from a to b will be same as value of the integral of f(x)^2 - value of the integral of g(x)^2 from a to b. (using property of sum and differences of the integrand. Thus the value of the integral will be 12 - 3 = 9. Problem 5.4.6 In the given integral the function has been shifted by 5 units in the direction of 5 units by using f(x-5) and at the same time the limits used are a + 5 and b + 5. This indicates that even the limits are shifted by 5 units in the same direction as that of the function. On shifting the shape of the function does not and thus the value of the integral also does not change, thus it is same as the value of the integral of f(x) with respect to x with limits a to b. thus the value of the integral is 8. Another way to say the same is, assume t = x - 5, thus dx = dt and a + 5 - 5 = a1 and b + 5 - 5 = b1, thus the new function is f(t), the new limits are a1 = a and b1 = b and the integral is with respect to t. Now if the variable is changed from t to x, the integral will again become integral of the function f(x) from a to b which is 8. Problem 5.4.9 The average value of a function f(x) over the interval [a,b] will be the value of the integral over the same interval divided by the length of the interval. The function f(x) is f(x) = 2 and the value of the integral over[a,b] is 2(b-a) and the length of the interval is (b-a) and thus the average value will be = 2(b-a)/(b-a) = 2. Problem 5.4.12 The average value will be the integral of the function over an interval divided by the length of the same interval. The unit for the integral of the function when f(x) is integrated with respect to x will be unit of f(x) * unit of x. The unit for the duration of the interval will be the unit of x. Thus the unit of the average value will be (unit of f(x)* unit of x) / (unit of x) = unit of f(x). Thus unit of the average value is same as the unit of f(x). Problem 5.4.15 Given y = x^2 and y = x^3 over the interval [0,1] For this interval x^2 function lies above x^3 and thus the area of the region between this 2 curves will be equal to the value of the integral of x^2 - x^3 from x = 0 to 1. This integral is same as integral of x^2 from 0 to 1 - integral of x^3 from 0 to 1. The value of the integral of x^2 from x = 0 to x = 1 will be 1/3 The value of the integral of x^3 from x = 0 to x = 1 will be 1/4 Thus the value of the area of the region between the graphs x^2 and x^3 will be 1/3 - Ľ = 1/12 square units. Problem 5.4.18 Given y = cos(t) and y = sin(t) for t belongs to [0,pi] To calculate the area of the region between the 2 curves we break the interval into 3 parts, the 1st part from t [0,pi/4], next from t [pi/4,pi/2] and from t [pi/2,pi] For t [0,pi/4] cos(t) is above sin(t) and thus the value of the area of the region between the 2 curves will be equal to the difference of the integral of the function cos(t) and the integral of sin(t) from [0,pi/4]. The value of the integral of the function cos(t) from [0,pi/4] = 1/sqrt(2). The value of the integral of the function sin(t) from [0,pi/4] = -1/sqrt(2) + 1 Thus the area of the region between t = 0 to t = pi/4 will be sqrt(2) - 1. For t [pi/4,pi/2] sin(t) is above cos(t) and thus the value of the area of the region between the 2 curves will be equal to the difference of the integral of the function sin(t) and the integral of cos(t) from [pi/4,pi/2]. The value of the integral of the function cos(t) from [pi/4,pi/2] = 1 - 1/sqrt(2) The value of the integral of the function sin(t) from [pi/4,pi/2] = 1/sqrt(2).. Thus the area of the region between t = pi/4 to t = pi/2 will be sqrt(2) - 1. For t [pi/2,pi] sin(t) is above cos(t) and thus the value of the area of the region between the 2 curves will be equal to the difference of the integral of the function sin(t) and the integral of cos(t) from [pi/4,pi/2]. The value of the integral of the function cos(t) from [pi/2,pi] = -1 The value of the integral of the function sin(t) from [pi/2,pi] = 1. Thus the area of the region between t = pi/2 to t = pi will be 2. Thus the total area between the curve sin(t) and cos(t) will be 2sqrt(2) - 2 + 2 = 2sqrt(2). Problem 5.4.21 (a) The average value is defined as the value of the integral over an interval divided by the length of the same interval which is 6/3 = 2. (b) Since the function is even the function is symmetric about y axis and thus the value of the integral of the function f(x) for the interval 0 to 3 will be same as the value of the integral from -3 to 0 and thus the value of the integral of the function from x = -3 to x = 3 will be 12 and thus the average value will be 12/6 = 2 same as the one above. This is also because the function is even. (c) If the function is odd the function is symmetric about origin and thus the value of the integral of the function f(x) from x = -3 to x = 0 is negative to the value of the integral of the function f(x) from 0 to 3 and thus the integral of f(x) from x = -3 to x = 3 will be 0 and thus the average value will also be 0 as the average value is the value of the integral divided by the length of the interval. Since the integral is 0 thus the average value is also 0. Problem 5.4.24 Given that the value of the integral of 2f(x) + 3 from x = 2 to x = 5 is equal to 17. This integral is same as the value of 2 * the value of the integral of f(x) from x = 2 to x = 5 + the value of the integral of 3 from x = 2 to x = 5. The value of the integral of 3 from x = 2 to x = 5 will be 3*3 = 9. Thus 2*the value of the integral of f(x) will be equal to 8 and the value of the integral of f(x) from x = 2 to x = 5 will be 8/2 = 4. Thus the value of the integral of the function f(x) from x = 2 to x = 5 will be 4. Problem 5.4.27 Given that the function is odd, thus the function is symmetric about origin. Thus the value of the integral of f(x) from x = -2 to x = 3 can be broken to the integral from x = -2 to x = 2 + the integral of the function from x = 2 to x = 3. Since the function is odd the value of the integral form x = -2 to x = 2 will be 0 Thus value of the integral of the function from x = - 2 to x = 3 will be equal to the value of the integral of the function from x = 2 till x = 3 = 30. Thus the value of the integral of the function f(x) from x = 2 to x = 3 is 30 Problem 5.4.30 Given that the average value of the function f(x) over the interval x = 2 to x= 5 is 4. The length of the interval is 3, thus the value of the integral of the function f(x) from x = 2 to x = 5 will be 12. The value of the integral of (3f(x) + 2) from x = 2 till x = 5 will be 3*(value of the integral of the function f(x) from x = 2 to x = 5 ) + the value of the integral of the function 2 from x = 2 to x = 5. Thus the value of the integral of (3f(x) + 2) from x = 2 to x = 5 will be 3*(12) + 6 = 42. Problem 5.4.33 (a) From the graph it can be seen that for x values from x = -3 to x = 1 the value of the integral is negative and that for x values from x = 1 to x = 3 the value of the integral will be positive. The value of the integral is the area under the curve which is nothing but the area of the squares. Since each square is of unit length, the area of each square will be equal to 1 unit square and thus the area under the curve will be the number of squares under the curve. The area under the curve for x values from x = -3 to x = = 1 can be approximated to 6.25 square unit and thus the value of the integral can be approximated to -6.25. The area under curve for x values from x = 1 to x = 3 can be approximated to 1.6 square units and thus the value of the integral can be approximated to 1.6. Thus the total value of the integral of the curve from x = -3 to x = 3 will be -6.25 + 1.6 = -4.65. (b) (I) The average value of integral of the function from x = -3 to x = 3 will be the ratio of the integral from x = -3 to x = 3 to the total duration of the integral = -4.65/6 = 0.775. (II) The value of the integral from x = 1 to x = 3 is 1.6 and that from x = 0 to x = 1 is - 0.6. The value of the integral from x = 0 to x = 3 will be 1.6 - 0.6 = 1. The duration of the x interval is 3 and thus the average value of the function from x = 0 to x = 3 will be 1/3 = 0.334. Thus we can see that the average value of the function over the interval x = -3 to x = 3 is greater than the average value of the function over the interval x = 0 to x = 3. Problem 5.4.35 Given the population growth function is P = 112*(1.011) ^ t where t is the number of years since 2010. (a) The average population between t = 2010 to t = 2050 will be equal to the value of the integral of the function from t = 0 to t = 40 / the length of the interval that is 40. The integral function will be = f(t) = (112/ln(1.011))*((1.011)^t). The value of the integral from t = 0 to t = 40 will be f(40) - f(0). Which is = (112/ln(1.011))*(1.011^40 - 1) = 5620.316462 Thus the average of the function will be = 5620.316462 / 40 = 140.5079116. (b) Population at 2010 according to the population function will be P(0) = 112. Population at 2050 according to the population function will be P(40) = 112*(1.011)^40 = 173.4859251 Thus average population in 2010 and the predicted population in 2050 will be 142.74296255. (c) My answer in part (b) is larger than my answer in part (a). In terms on concavity the function is concave up graph, and thus if we use trapezoidal approximation graphs to calculate the integral we can see that the average value of each trapezoid will be on the right hand side of the average value of the function for the same interval. And since the function is increasing the function will have a lower average value as compared to that calculated from the average of the end values. Problem 5.4.36 (a) The graph of e^(-(x^2)/2) is a curve whose maximum value is 1 and the decreases on either side of x. If we draw a vertical line y = 1 for x values 0 to 1 we see that the area that is the value of the integral is 1. For e^(-(x^2)/2) the curve starts with the value of 1 at x = 0 and then is always below y = 1. Using the theorem we can say that if a function f(x) is equal to or less than g(x) then the value of the integral of f(x) will be less than the value of the integral of g(x) over the same intervals. Here f(x) = e^(-(x^2)/2) and g(x) = 1 and thus the value of the integral of e^^(-(x^2)/2) will be less than 1. (b) The area can be found by integrating the given exponential function The value of this integral using a calculator is approximately 0.855624. Problem 5.4.39 A length representing f(b) - f(a) will be equal to the length on the y axis between f(b) and f(a). Draw vertical lines from x = a and x = b, reach them to the graph and then horizontal lines reaching the y axis. The distance between this parallel horizontal lines will be the length representing f(b) - f(a) Problem 5.4.42 F(b) - F(a) / (b-a) is the average value of the function f(x) over the interval x = a to x = b. The average value can be interpreted as the height of the rectangle with the same base as the interval and the same area as the area under the curve for the function f(x) from x = a to x = b. This length can also be found out by joining f(a) and f(b) and finding the value of x = (a + b)/2 on that line. The height corresponding to that x value will be equal to the average height and will also be the length roughly approximating F(b) - F(a) / (b-a). On the graph this value will corresponding to an x value slightly less than (a + b)/2 since it is a concave down graph.

Return to the form.

Text_16

Text Section 5.4

Problem 5.4.3

Problem 5.4.6

Problem 5.4.9

= 2(b-a)/(b-a) = 2.

Problem 5.4.12

Thus unit of the average value is same as the unit of f(x).

Problem 5.4.15

Given y = x^2 and y = x^3 over the interval [0,1]

The value of the integral of x^2 from x = 0 to x = 1 will be 1/3

The value of the integral of x^3 from x = 0 to x = 1 will be 1/4

Thus the value of the area of the region between the graphs x^2 and x^3 will be 1/3 - Ľ = 1/12 square units.

Problem 5.4.18

Given y = cos(t) and y = sin(t) for t belongs to [0,pi]

To calculate the area of the region between the 2 curves we break the interval into 3 parts, the 1st part from t [0,pi/4], next from t [pi/4,pi/2] and from t [pi/2,pi]

For t [0,pi/4]

The value of the integral of the function cos(t) from [0,pi/4] = 1/sqrt(2).

The value of the integral of the function sin(t) from [0,pi/4] = -1/sqrt(2) + 1

Thus the area of the region between t = 0 to t = pi/4 will be sqrt(2) - 1.

For t [pi/4,pi/2]

The value of the integral of the function cos(t) from [pi/4,pi/2] = 1 - 1/sqrt(2)

The value of the integral of the function sin(t) from [pi/4,pi/2] = 1/sqrt(2)..

Thus the area of the region between t = pi/4 to t = pi/2 will be sqrt(2) - 1.

For t [pi/2,pi]

The value of the integral of the function cos(t) from [pi/2,pi] = -1

The value of the integral of the function sin(t) from [pi/2,pi] = 1.

Thus the area of the region between t = pi/2 to t = pi will be 2.

Thus the total area between the curve sin(t) and cos(t) will be 2sqrt(2) - 2 + 2 = 2sqrt(2).

Problem 5.4.21

(a) The average value is defined as the value of the integral over an interval divided by the length of the same interval which is 6/3 = 2.

Problem 5.4.24

Thus the value of the integral of the function f(x) from x = 2 to x = 5 will be 4.

Problem 5.4.27

Thus value of the integral of the function from x = - 2 to x = 3 will be equal to the value of the integral of the function from x = 2 till x = 3 = 30.

Thus the value of the integral of the function f(x) from x = 2 to x = 3 is 30

Problem 5.4.30

Thus the value of the integral of (3f(x) + 2) from x = 2 to x = 5 will be 3*(12) + 6 = 42.

Problem 5.4.33

(a)

(b)

(I) The average value of integral of the function from x = -3 to x = 3 will be the ratio of the integral from x = -3 to x = 3 to the total duration of the integral = -4.65/6 = 0.775.

Thus we can see that the average value of the function over the interval x = -3 to x = 3 is greater than the average value of the function over the interval x = 0 to x = 3.

Problem 5.4.35

Given the population growth function is P = 112*(1.011) ^ t where t is the number of years since 2010.

(a) The average population between t = 2010 to t = 2050 will be equal to the value of the integral of the function from t = 0 to t = 40 / the length of the interval that is 40.

The integral function will be = f(t) = (112/ln(1.011))*((1.011)^t).

The value of the integral from t = 0 to t = 40 will be f(40) - f(0).

Which is = (112/ln(1.011))*(1.011^40 - 1) = 5620.316462

Thus the average of the function will be = 5620.316462 / 40 = 140.5079116.

(b) Population at 2010 according to the population function will be P(0) = 112.

Population at 2050 according to the population function will be P(40) = 112*(1.011)^40 = 173.4859251

Thus average population in 2010 and the predicted population in 2050 will be 142.74296255.

Problem 5.4.36

(b) The area can be found by integrating the given exponential function

The value of this integral using a calculator is approximately 0.855624.

Problem 5.4.39

Problem 5.4.42

Form Confirmation

Thank you for submitting the following information:

identifyingInfo: Submitting Assignment

yourCourse: MTH 173

Contact_FirstName: Kevin

Contact_LastName: Shah

firstPendLength: text_16

email: krs2070@email.vccs.edu

11/10 21:20

response

Return to the form.

Text_18

Text Section 3.3

• Section 3.3, Problems 1-3, 6, 9, 14, 16, 19, 23, 25, 28, 30-32, 36-40, 44, 47, 51, 54, 56, 61

Text Problem 3.3.1

Given f(x) = x^2(x^3 + 5)

Using the product rule of differentiation we get

f’(x) = x^2(3x^2) + (x^3 + 5)2x = 3x^4 + 2x^4 + 10x = 5x^4 + 10x

for the second method we multiply the factors first

thus f(x) = x^5 + 5x^2

using the rule of additions we get

f’(x) = 5x^4 + 10x

We do get the same result with both the methods. We should indeed get same results as it is the same function which is being differentiated and thus the slopes indeed will remain the same and so will f’(x) function.

Text Problem 3.3.2

Given function f(x) = 2^x*3^x

Using the product rule of differentiation we get

f’(x) = 2^x(3^x ln(3)) + 3^x(2^xln(2)) = 2^x*3^x*(ln(2) + ln(3)) = 6^x(ln(6))

by using the second method by multiplying before differentiating we get

f(x) = 6^x

using the a^x rule of differentiation we get f’(x) = 6^x(ln(6))

Text Problem 3.3.3

Given function f(x) = x*e^x

Thus if g(x) = x and h(x) = e^x

Using the product rule of differentiation we get

f’(x) = g(x)*h’(x) + h(x)*g’(x)

f’(x) = e^x + (x*e^x) = e^x*(x+1)

Text Problem 3.3.6

Given function y = (t^2 + 3)e^t

Thus if f(t) = (t^2 + 3) and g(t) = e^t

Using the product rule of differentiation we get

y’ = f’(t)g(t) + g’(t)f(t)

y’ = (2t)e^t + (t^2 + 3)e^t = e^t(t^2 + 2t + 3)

Text Problem 3.3.9

Given function f(x) = x/e^x

Where g(x) = x and h(x) = e^x

Using the quotient rule of differentiation we get

f’(x) = (h(x)g’(x) - g(x)h’(x)) / h^2(x) = (e^x - xe^x)/e^2(x) = (1-x)/e^(x)

Text Problem 3.3.14

Given function g(t) = (t-4)/(t+4)

Where f(t) = (t-4) and h(t) = (t+4)

Using the quotient rule of differentiation we get

g’(t) = (h(t)f’(t) - h’(t)f(t)) / h^2(t) = ((t+4) - (t-4))/(t+4)^2 = 8/(t+4)^2.

Text Problem 3.3.16

Given function z = (t^2 + 5t + 2)/(t+3)

Using the quotient rule of differentiation and the method explained in earlier problems in this assignment we get

z’ = ((t+3)(2t + 5) - (t^2 + 5t + 2))/(t+3)^2 = (2t^2 + 11t + 15 - t^2 - 5t - 2)/(t+3)^2 = (t^2 + 6t + 13)/(t^2 + 6t + 9)

Text Problem 3.3.19

Given function w = (y^3 - 6y^2 + 7y)/y = y^2 - 6y + 7

Thus using the rules of differentiation we get

w’ = 2y - 6.

Text Problem 3.3.23

Given function h(r) = r^2 / (2r + 1)

Using the quotient rule of differentiation and the previous method of differentiation we get

h’(r) = ((2r + 1)(2r) - r^2(2))/(2r + 1)^2 = (4r^2 + 2r - 2r^2)/(4r^2 + 4r + 1) = (2r^2 + 2r)/(4r^2 + 4r + 1)

Text Problem 3.3.25

Given function w(x) = 17e^x / 2^x

Using the quotient rule of differentiation and the previous methods of differentiation we get

w’(x) = 17*(2^x*e^x - e^x*(2^x*ln(2)))/4^x = 17e^x(1 - ln(2))/2^x.

Text Problem 3.3.28

Given function f(x) = (ax + b)/(cx + k)

Using the quotient rule of differentiation and the previous method of differentiation we get

f’(x) = ((cx + k)a - (ax+b)c)/(cx+k)^2 = (acx + ak - acx - cb)/(cx + k)^2 = (ak - cb)/(cx + k)^2.

Text Problem 3.3.30

Given function f(x) = (2 - 4x - 3x^2)*(6x^e - 3pi)

Using the product rule of differentiation and the previous method of differentiation we get

f’(x) = (2 - 4x - 3x^2)(6ex^(e-1)) + (6x^e - 3pi)*(-4-6x)

f’(x) = 12ex^(e-1) - 24ex^e - 18ex^(e+1) - 24x^e + 12pi - 36x^(e+1) + 18xpi

f’(x) = - 54ex^(e+1) - 48ex^e + 12ex^(e-1) + 18xpi + 12pi

Text Problem 3.3.31

Given function h(x) = f(x).g(x)

Using the product rule of differentiation

The derivative of the function h(x) will be h’(x) = f’(x)g(x) + g’(x)f(x)

(a)

For h’(1), we find the derivative of f(x) and g(x) at x = 1

Derivative of f(x) will be the slope of the line f(x) at x = 1 which is 2

Derivative of g(x) will be the slope of the line g(x) at x = 1 which is -1.

The value of the function f(x) and g(x) at x = 1 is 2 and 3 respectively.

Thus the value of the derivative at x = 1 will be h’(1) = 2(3) - 1(2) = 6 - 2 = 4.

(b)

For h’(2), we need derivative of the function at x = 2 for f(x) and g(x). Since the function f(x) has a sharp point at x = 2, the function is not differentiable at x = 2. Since f(x) is not differentiable at x = 2, the product of f(x)g(x) will also be not differentiable at x = 2 and h’(2) is actually not defined. However this rule is not provided within this section of the text.

(c)

For h’(3), we find the derivative of f(x) and g(x) at x = 3

Derivative of f(x) will be the slope of the line f(x) at x = 3 which is -2

Derivative of g(x) will be the slope of the line g(x) at x = 3 which is -1.

The value of the function f(x) and g(x) at x = 3 is 2 and 1 respectively.

Thus the value of the derivative at x = 3 will be h’(3) = -2(1) - 1(2) = -2 - 2 = -4.

Text Problem 3.3.32

Given function k(x) = f(x)/g(x)

Using the quotient rule of differentiation we get k’(x) = (g(x)f’(x) - f(x)g’(x))/g^2(x)

(a)

To calculate k’(1) we calculate the slope of the function g(x) and f(x) at x = 1.

From the graph, we get f’(1) = slope of graph at x = 1 will be = 2

g’(1) = slope of graph at x = 1 will be = -1.

The value of the function at x = 1 will be, f(1) = 2 and g(1) = 3.

We get the value of k’(x) at x = 1 will thus be = (3*2 - 2*(-1))/9 = (9+2)/9 = 11/9.

(b)

Similar to the product function when f(x) is not differentiable at x = 2, the function f(x)/g(x) will also not be differentiable as f’(x) for x = 2 is not defined, due to the sharp corner of the f(x) function at x = 2.

(c)

To calculate k’(3) we calculate the slope of the function g(x) and f(x) at x = 3.

From the graph, we get f’(3) = slope of graph at x = 3 will be = -2

g’(3) = slope of graph at x = 3 will be = -1.

The value of the function at x = 3 will be, f(3) = 2 and g(3) = 1.

We get the value of k’(x) at x = 3 will thus be = (1*(-2) - 2*(-1))/1 = (-2 + 2)/1 = 0.

Thus k’(3) = 0.

Text Problem 3.3.36

From the given graphs and using the grid squares we approximate f’(2) = 1 and g’(2) = -2.

f(2) = 0.17 and g(2) is approximately = 1.83.

Given, to calculate h’(2). Given h(x) = f(x).g(x), thus h’(x) = f’(x)g(x) + g’(x)f(x).

Thus h’(2) = f’(2)g(2) + g’(2)f(2) = 1(1.83) - 2(0.17) = 1.83 - 0.34 = 1.49.

Text Problem 3.3.37

From the given graphs and using the grid squares we approximate f’(2) = 1 and g’(2) = -2.

f(2) = 0.17 and g(2) is approximately = 1.83.

Given, to calculate k’(2). Given k(x) = f(x)/g(x), thus k’(x) = (g(x)f’(x) - f(x)g’(x))/g^2(x).

Thus k’(2) = ( 1.83 + 0.34)/(1.83*1.83) = 0.6479739 approximately.

Text Problem 3.3.38

From the given graphs and using the grid squares we approximate f’(1) = 0.34 and g’(1) = 0.84

f(1) = -0.34 approximately and g(1) = 2.

Given to calculate l’(1). Given l(x) = g(x)/f(x), thus l’(x) = (f(x)g’(x) - g(x)f’(x))/f^2(x)

Thus l’(1) = (-0.2856 - 0.68)/0.1156 = -8.35294 approximately.

Text Problem 3.3.39

From the given graphs and using the grid squares we approximate f’(2) = 1 and g’(2) = -2.

f(2) = 0.17 and g(2) is approximately = 1.83.

Given to calculate l’(2). Given l(x) = g(x)/f(x), thus l’(x) = (f(x)g’(x) - g(x)f’(x))/f^2(x)

Thus l’(2) = (-0.34 - 1.83)/1 = -2.17

Text Problem 3.3.40

Given function f(t) = e^(-t) = 1 / e^t.

Let h(t) = 1 and g(t) = e^t, and thus h’(t) = 0 and g’(t) = e^t.

f(t) = h(t) / g(t), using the differentiation rule of quotients f’(t) will be

f’(t) = (g(t)h’(t) - h(t)g’(t))/g^2(t)

f’(t) = (e^t*0 - 1(e^t))/e^(2t) = - 1/e^t = -e^(-t)

Text Problem 3.3.44

Given function g(x) = 1 / (x^2 + 1) = (x^2 + 1)^-1.

On differentiating the given g(x) function using the power rule we get

g’(x) = (-1)*(x^2 + 1)^-2 *(2x) = (-2x)*(x^2 + 1)^-2.

On further differentiating g’(x) for the second time we get g”(x).

Using product and power rule we get

g”(x) = (-4x)*(x^2 + 1)^-3*(2x) + (x^2 + 1)^-2*(-2) = - 8x^2*(x^2 + 1) - 2(x^2 + 1)^-2.

Since g”(x) sign talks about the concavity of the graph and x^2 is positive for all values of x, g”(x) will be negative for all values of x. Since g”(x) is always negative, the function g(x) will be concave down for all values of x.

Text Problem 3.3.47

(a)

Given function y = e^x/x.

We have f(x) = e^x and f’(x) = e^x and g(x) = x and g’(x) = 1.

Differentiating the given function using quotient rule of differentiation we get y’ = (xe^x - e^x)/x^2.

Thus y’ = e^x(x - 1)/x^2.

Given function y = e^x/x^2.

We have f(x) = e^x and f’(x) = e^x and g(x) = x^2 and g’(x) = 2x.

Differentiating the given function using quotient rule of differentiation we get y’ = (x^2e^x-e^x(2x))/x^4

Thus y’ = e^x(x -2)/x^3

Given function y = e^x/x^3.

We have f(x) = e^x and f’(x) = e^x and g(x) = x^3 and g’(x) = 3x^2.

Differentiating the given function using quotient rule of differentiation we get y’ = (x^3e^x-e^x(3x^2))/x^6

And thus y’ = e^x(x - 3)/x^4

(b)

On seeing the trend for x, x^2 and x^3 we can estimate the value of y’ for x^n as

If y = e^x/x^n,

We get y’ = e^x(x - n)/x^(n+1).

To confirm this guess

We have f(x) = e^x, thus f’(x) = e^x and g(x) = x^n, thus g’(x) = n(x^(n-1)).

We get y’ = (x^n(e^x) - e^x(n(x^(n-1)))/x^2n = e^x(x - n)/x^(n+1).

Text Problem 3.3.51

Given function y(x) = (f(x)g(x))/h(x), with a restriction that h(x) is not 0 for any x considered.

To find y’(x), we first find the derivative of f(x)g(x).

The derivative of f(x)g(x) is f(x)g’(x) + g(x)f’(x).

Using the quotient rule of differentiation the value of the differentiation would be

y’(x) = (h(x)(f’(x)g(x) + g’(x)f(x)) - f(x)g(x)h’(x))/h^2(x)

thus the simplest form of differentiation in terms of f(x), g(x) and h(x) would be

y’(x) = (h(x)f’(x)g(x) + h(x)g’(x)f(x) - f(x)g(x)h’(x))/h^2(x).

Text Problem 3.3.54

Given that f(3) = 6, g(3) = 12, f’(3) = ˝ and g’(3) = 4/3.

Given function y = (f(x)g(x))’ - (g(x) - 4f’(x))

y = f(x)g’(x) + g(x)f’(x) - g(x) + 4f’(x).

to evaluate the value of the function at x = 3 we evaluate y by substituting x = 3, we get

y(3) = f(3)g’(3) + g(3)f’(3) - g(3) + 4f’(3) = 6*(4/3) + 12*(1/2) - 12 + 4(1/2) = 8 + 6 - 12 + 2 = 4.

Text Problem 3.3.56

Given that p is selling price in dollars, and q = f(p) is a quantity function with respect to price.

Given that f(140) = 15,000 and f’(140) = -100.

(a)

f(140) = 15,000 means that a quantity of 15,000 skateboard is sold for a cost of 140 dollars.

f’(140) = -100 means that the rate at which the quantity function changes at a price of 140 dollars is -100 skateboards/dollar. This is close vicinity of $140, it can be considered that a change 1 unit in the dollars would cause a change of - 100 skateboards.

Given that the total revenue earned R is given by the product of p and q, thus R = pq.

The derivative of this rate function would thus be R’ = p(q’) + q(p’).

And thus dR/dp = q(dp/dp) + p(dq/dp) = q + p(f’(p))

Thus dR/dp|p = 140 = q(140) + 140(f’(140)) = f(140) + 140(f’(140)) = 15,000 + 140(-100) = 15000 - 14000 = 1,000.

(c) The sign of dR/dp|p = 140 will be positive. Since the revenue function is R = p*q. Current selling price increased from $140 to $141, and thus the value of p increased, plus the quantity function q can be seen to have a negative rate when p = $140, thus when p increases, whereas q will decrease. For p = $140/quantity, q = 15000 quantity. Thus R = 140*15,000 = $2,100,000. The estimate change in q from p = $140 to p = $141 is - 100 quantities, which is estimated by the rate function. Thus f(141) = 15000 - 100 = 14,900. The revenue will thus be R(141) = 14900*141 = $2,100,900.

On changing p from $140 to $141, revenue R increases by an approximate value of $900.

Text Problem 3.3.61

Given function

(a) f(x) = (x -1)(x - 2)

Using the product rule of differentiation the derivative of the function will be

f’(x) = (x - 1) + (x - 2) = 2x - 3.

(b) f(x) = (x - 1)(x - 2)(x - 3)

We can assume g(x) = (x - 2)(x - 3)

f(x) = (x - 1)g(x)

Using the product rule of differentiation, the derivative of the function g(x) will be

g’(x) = (x - 2) + (x - 3) = 2x - 5.

Thus f’(x) = g(x) + (x - 1)g’(x) = (x - 2)(x - 3) + (x - 1)(2x - 5) = (x - 2)(x - 3) + (x - 1)(x - 2) + (x - 1)(x - 3).

The derivative is equivalent to function which says, the derivative of one factor*(the other 2 remaining factors). This also leads to the same conclusion of the derivative which is

f’(x) = (x - 1)(x - 2) + (x - 2)(x - 3) + (x - 3)(x - 1).

Using the same derivative rules used in the previous question we get the derivative function,

f’(x) = (x - 1)(x - 2)(x - 3) + (x - 2)(x - 3)(x - 4) + (x - 3)(x - 4)(x - 1) + (x - 4)(x - 1)(x - 2)

Form Confirmation

Thank you for submitting the following information:

identifyingInfo: Submitting Assignment

yourCourse: MTH 173

Contact_FirstName: Kevin

Contact_LastName: Shah

firstPendLength: text_18

email: krs2070@email.vccs.edu

11/13 00:10

response

Text_18 Text Section 3.3 Text Problem 3.3.1 Given f(x) = x^2(x^3 + 5) Using the product rule of differentiation we get f’(x) = x^2(3x^2) + (x^3 + 5)2x = 3x^4 + 2x^4 + 10x = 5x^4 + 10x for the second method we multiply the factors first thus f(x) = x^5 + 5x^2 using the rule of additions we get f’(x) = 5x^4 + 10x We do get the same result with both the methods. We should indeed get same results as it is the same function which is being differentiated and thus the slopes indeed will remain the same and so will f’(x) function. Text Problem 3.3.2 Given function f(x) = 2^x*3^x Using the product rule of differentiation we get f’(x) = 2^x(3^x ln(3)) + 3^x(2^xln(2)) = 2^x*3^x*(ln(2) + ln(3)) = 6^x(ln(6)) by using the second method by multiplying before differentiating we get f(x) = 6^x using the a^x rule of differentiation we get f’(x) = 6^x(ln(6)) We do get the same result with both the methods. We should indeed get same results as it is the same function which is being differentiated and thus the slopes indeed will remain the same and so will f’(x) function. Text Problem 3.3.3 Given function f(x) = x*e^x Thus if g(x) = x and h(x) = e^x Using the product rule of differentiation we get f’(x) = g(x)*h’(x) + h(x)*g’(x) f’(x) = e^x + (x*e^x) = e^x*(x+1) Text Problem 3.3.6 Given function y = (t^2 + 3)e^t Thus if f(t) = (t^2 + 3) and g(t) = e^t Using the product rule of differentiation we get y’ = f’(t)g(t) + g’(t)f(t) y’ = (2t)e^t + (t^2 + 3)e^t = e^t(t^2 + 2t + 3) Text Problem 3.3.9 Given function f(x) = x/e^x Where g(x) = x and h(x) = e^x Using the quotient rule of differentiation we get f’(x) = (h(x)g’(x) - g(x)h’(x)) / h^2(x) = (e^x - xe^x)/e^2(x) = (1-x)/e^(x) Text Problem 3.3.14 Given function g(t) = (t-4)/(t+4) Where f(t) = (t-4) and h(t) = (t+4) Using the quotient rule of differentiation we get g’(t) = (h(t)f’(t) - h’(t)f(t)) / h^2(t) = ((t+4) - (t-4))/(t+4)^2 = 8/(t+4)^2. Text Problem 3.3.16 Given function z = (t^2 + 5t + 2)/(t+3) Using the quotient rule of differentiation and the method explained in earlier problems in this assignment we get z’ = ((t+3)(2t + 5) - (t^2 + 5t + 2))/(t+3)^2 = (2t^2 + 11t + 15 - t^2 - 5t - 2)/(t+3)^2 = (t^2 + 6t + 13)/(t^2 + 6t + 9) Text Problem 3.3.19 Given function w = (y^3 - 6y^2 + 7y)/y = y^2 - 6y + 7 Thus using the rules of differentiation we get w’ = 2y - 6. Text Problem 3.3.23 Given function h(r) = r^2 / (2r + 1) Using the quotient rule of differentiation and the previous method of differentiation we get h’(r) = ((2r + 1)(2r) - r^2(2))/(2r + 1)^2 = (4r^2 + 2r - 2r^2)/(4r^2 + 4r + 1) = (2r^2 + 2r)/(4r^2 + 4r + 1) Text Problem 3.3.25 Given function w(x) = 17e^x / 2^x Using the quotient rule of differentiation and the previous methods of differentiation we get w’(x) = 17*(2^x*e^x - e^x*(2^x*ln(2)))/4^x = 17e^x(1 - ln(2))/2^x. Text Problem 3.3.28 Given function f(x) = (ax + b)/(cx + k) Using the quotient rule of differentiation and the previous method of differentiation we get f’(x) = ((cx + k)a - (ax+b)c)/(cx+k)^2 = (acx + ak - acx - cb)/(cx + k)^2 = (ak - cb)/(cx + k)^2. Text Problem 3.3.30 Given function f(x) = (2 - 4x - 3x^2)*(6x^e - 3pi) Using the product rule of differentiation and the previous method of differentiation we get f’(x) = (2 - 4x - 3x^2)(6ex^(e-1)) + (6x^e - 3pi)*(-4-6x) f’(x) = 12ex^(e-1) - 24ex^e - 18ex^(e+1) - 24x^e + 12pi - 36x^(e+1) + 18xpi f’(x) = - 54ex^(e+1) - 48ex^e + 12ex^(e-1) + 18xpi + 12pi Text Problem 3.3.31 Given function h(x) = f(x).g(x) Using the product rule of differentiation The derivative of the function h(x) will be h’(x) = f’(x)g(x) + g’(x)f(x) (a) For h’(1), we find the derivative of f(x) and g(x) at x = 1 Derivative of f(x) will be the slope of the line f(x) at x = 1 which is 2 Derivative of g(x) will be the slope of the line g(x) at x = 1 which is -1. The value of the function f(x) and g(x) at x = 1 is 2 and 3 respectively. Thus the value of the derivative