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Calculus AB : Calculus AB

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Example Questions

Example Question #1 : Find Average Value

Find the average value of the function f(x)=2x^5-6x^3+7 over the interval [1,2] .

Possible Answers:

7.5

5.5

Correct answer:

5.5

Explanation:

Because the objective of this problem is to find the average value of the function, the formula f $f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx$ will be useful. Since the interval and function are provided, this problem consists of recognizing the base components and making the appropriate substitutions:

$f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx=\frac{1}{2-1}\int_{1}^{2}2x^5-6x^3+7 dx =[\frac{1}{3}x^6-\frac{3}{2}x^4+7x] |_1^2$ $=\frac{64}{3}-24+14-\frac{1}{3}+\frac{3}{2}-7=5.5$

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Example Question #4 : Find Average Value

Identify the correct integral expression for the average value of the function f(x)=1x+2x over the interval [1,4] .

Possible Answers:

$\frac{1}{3}\int_{1}^{4}\frac{1}{t}+\sqrt{2t} dx$

$\frac{1}{4}\int_{1}^{4}\frac{1}{x}+\sqrt{2x} dx$

$\int_{1}^{4}\frac{1}{x}+\sqrt{2x} dx$

$\frac{1}{3}\int_{1}^{4}\frac{1}{x}+\sqrt{2x} dx$

Correct answer:

$\frac{1}{3}\int_{1}^{4}\frac{1}{x}+\sqrt{2x} dx$

Explanation:

When finding the average value of a function, it is useful to keep the following formula in mind: $f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx$ . This equation allows the substitution of the function and interval to solve for the average value. Because the function indicated in the problem is in terms of , the definite integral expression should also be in terms of .

$f_{avg}=\frac{1}{4-1}\int_{1}^{4}\frac{1}{x}+\sqrt{2x} dx =\frac{1}{3}\int_{1}^{4}\frac{1}{x}+\sqrt{2x} dx$

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Example Question #5 : Find Average Value

Find the average value of the function f(x)=20x+x^3 over the interval [-1,1] .

Possible Answers:

Correct answer:

Explanation:

$f_{avg}=\frac{1}{1-(-1)}\int_{-1}^{1}20x+x^3 dx$ $=\frac{1}{2}[10x^2+(\frac{1}{4})(x)^4] |_-1^1=\frac{1}{2}(10+\frac{1}{4}-10-\frac{1}{4})=0$

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Example Question #2 : Find Average Value

Let f(x)=4x+11 . What value of c allows the average value of f(x) over the interval [0,c] to be ?

Possible Answers:

Correct answer:

Explanation:

$27=\frac{1}{c-0}\int_{c}^{0}4x +11 dx$

Next, the definite integral can be taken to continue solving for .

$27=\frac{1}{c}[2x^2+11x] |_0^c$

$27=(\frac{1}{c})(2c^2+11c)$

27=2c+11

c=8

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Example Question #2 : Find Average Value

Let f(x)=3x2+4x-3 . What value of allows the average value of f(x) over the interval [0,c] to be ?

Possible Answers:

Correct answer:

Explanation:

$12=\frac{1}{c-0}\int_{0}^{c}3x^2+4x-3 dx$

Next, the definite integral can be taken to continue solving for .

$12=\frac{1}{c}[x^3+2x^2-3x] |_0^c$

$12=(\frac{1}{c})(c^3+2c^2-3c)$

12=c^2+2c-3

0=c^2+2c-15

0=(c+5)(c-3)

c={-5,3}

Because the problem states that c>0 , the answer c=-5 can be eliminated. Therefore, the correct answer is c=3 .

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Example Question #1 : Find Cross Sections: Squares & Rectangles

Find the volume of a pyramid whose base is a square with sides of length and whose height is .

Possible Answers:

Correct answer:

Explanation:

First, it is important to consider the shape of this solid. This solid is a pyramid, with one square face and four triangular faces. Through a relationship of similar triangles, we are able to relate the known information (a height of and a base side length of ) to our general variables for side length (x) and height of the pyramid (y) . We can think of this plotted on the coordinate plane, with the width of the pyramid solid being in the direction.

$\frac{x}{3}=\frac{y}{10} \Rightarrow x=\frac{3y}{10}$

Since the side length can be squared to find a general formula for the area of the pyramid’s base (A=x^2) , this can be applied to the volume using cross-sections formula $(V=\int_{a}^{b}A(x) dx)$ as the next step. Note that the equation $x=\frac{3y}{10}$ solved for above is in terms of . Our new volume function, therefore, is also in terms of $y: V=\int_{a}^{b}(\frac{3y}{10})^2 \: dy$

Because the pyramid reaches a maximum height of , and we assume the pyramid’s starting height is at y=0 , the appropriate bounds for the integral expression are: $0\leq y\leq 10$ .

To wrap up this problem, combine the above information into one cohesive expression:

$V=\int_{0}^{10}(\frac{3y}{10})^2 dy =\frac{9}{100} \int_{0}^{10}y^2dy =\frac{9}{100}[\frac{1}{3}y^3] |_0^{10}=\frac{9}{100}*\frac{1}{3}*1000=30$

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Example Question #1 : Find Cross Sections: Squares & Rectangles

Find the volume of the solid whose base is bounded by y=x+4 and y=x^2-2 and whose cross-sections are rectangles of height and perpendicular to the axis.

Possible Answers:

50.5

62.5

Correct answer:

62.5

Explanation:

The cross-sections are perpendicular to the axis; therefore, the volume expression will be in terms of .

The area of a rectangle is A=l*w . By applying this formula to our general volume formula $(V=\int_{a}^{b}A(x) dx)$ , we get the following: $V=\int_{a}^{b}lw \: dx$ .

Next, an expression for must be determined. The problem specifies the length (or height) of the rectangle cross-sections is . This just leaves the value of w to be found. The width of the rectangle will vary as the region creating the base of the solid varies. The region is defined by the area enclosed between y=x+4 and y=x^2-2 , therefore, w=(x+4)-(x^2-2)=6+x-x^2 .

Since the region is bounded by y=x+4 and y=x^2-2 , the base has the following domain: $-2\leq x\leq 3$ .

Putting this all together, we find the following:

$V=\int_{-2}^{3}(3)(6+x-x^2) dx$ $= 3[6x+\frac{1}{2}x^2-\frac{1}{3}x^3] |_{-2}^3=3(18+\frac{9}{2}-9+12-2-\frac{8}{3})=62.5$

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Example Question #3 : Finding Volume Using Integration

Identify the correct expression for the volume of a solid whose base is bounded by y=x-3 , y=0 and y=5-x and whose cross-sections are squares perpendicular to the axis.

Possible Answers:

$V=\int_{3}^{5}4y^2-32y+64 \: dy$

$V=\int_{3}^{5}x^2 \: dx$

$V=\int_{3}^{5}y^2 \: dy$

$V=\int_{3}^{5}4x^2-32x+64 \: dx$

Correct answer:

$V=\int_{3}^{5}4x^2-32x+64 \: dx$

Explanation:

The cross-sections are perpendicular to the axis; therefore, the volume expression will be in terms of .

The area of a square is A=s^2 , where s is the side length of the square. By applying this formula to our general volume formula $(V=\int_{a}^{b}A(x) \: dx)$ , we get the following: $V=\int_{a}^{b}s^2 \: dx$ .

Next, an expression for s2 must be determined. Because s should be the width of the solid’s base, the expression for that length can be used to solve for s2.

s=(x-3)-(5-x)=2x-8

s^2=(2x-8)^2=4x^2-32x+64

Since the region is bounded by y=x-3 and y=5-x , the base has the following domain: 3x5 .

Putting this all together, we find the following:

$V=\int_{3}^{5}4x^2-32x+64 \: dx$

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Example Question #1 : Find Cross Sections: Squares & Rectangles

Identify the correct expression for the volume of a solid whose base is bounded by a disk of radius and whose cross-sections are squares parallel to the axis.

Possible Answers:

$4\int_{-R}^{R}y^2 \: dy$

$4\int_{-R}^{R}R^2-y^2 \: dy$

$\int_{-R}^{R}R^2-x^2 \: dx$

$\int_{-R}^{R}R^2 \: dy$

Correct answer:

$4\int_{-R}^{R}R^2-y^2 \: dy$

Explanation:

The cross-sections are parallel to the axis; in other words, they are perpendicular to the axis. This indicates that the expression should be in terms of .

Because the disk is of radius , the base is defined by the following formula: x^2+y^2=R^2 .

The area of a square is A=s^2 , with s being the side length. By applying this formula to our general volume formula $(V=\int_{a}^{b}A(y) \: dy)$ , we get the following: $V=\int_{a}^{b}s^2 \: dy$ .

The radius defines the bounds as being $-R\leq y\leq R$ . Next, can be found by understanding that differs as the width of the circle changes. The value of is the distance from one side of the circle to the other at any given point along $-R\leq y\leq R$ . The length of one side of the square, therefore, is $s=2\sqrt{R^2-y^2}$ .

Putting it all together, the following is obtained:

$V=\int_{-R}^{R}(2\sqrt{R2-y^2})^2\: dy= 4\int_{-R}^{R}R^2-y^2 dy$

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Example Question #5 : Finding Volume Using Integration

Let be the region bounded by y=e^x , x=0 , and x=4 . Find the volume of the solid whose base is region and whose cross-sections are squares perpendicular to the axis.

Possible Answers:

$\frac{e^8-1}{2}$

$\frac{e^4-1}{2}$

e^8

e^4

Correct answer:

$\frac{e^8-1}{2}$

Explanation:

The cross-sections are perpendicular to the axis; therefore, the volume expression will be in terms of .

Next, an expression for s^2 must be determined. Because s should be the width of the solid’s base, the expression for that length can be used to solve for s^2 .

s=e^x-0

$s^2=e^{2x}$

Since the region is bounded by x=0 and x=4 , the base has the following domain: $0\leq x\leq 4$ .

Putting this all together, we find the following:

$V=\int_{0}^{4}e^{2x} \: dx =[\frac{1}{2}e^{2x}] |_0^4=\frac{1}{2}(e^8-e^0)=\frac{e^8-1}{2}$

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Calculus AB : Calculus AB

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #1 : Find Average Value

Example Question #4 : Find Average Value

Example Question #5 : Find Average Value

Example Question #2 : Find Average Value

Example Question #2 : Find Average Value

Example Question #1 : Find Cross Sections: Squares & Rectangles

Example Question #1 : Find Cross Sections: Squares & Rectangles

Example Question #3 : Finding Volume Using Integration

Example Question #1 : Find Cross Sections: Squares & Rectangles

Example Question #5 : Finding Volume Using Integration

All Calculus AB Resources

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