Mean Value Theorem - AP Calculus AB

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Question

Find the area of the region enclosed by the parabola $y=12-x^2$ and $y=-x$ .

Answer

The two curves intersect in between and , which can be found by solving the quadratic equation $12-x^2=-x$ .

To solve for the area between curves, and , we use the formula $A=\int^b_a(f(x)-g(x))dx$

For our problem:

$A=\int^4_{-3}(12-x^2-(-x))dx$

$A=12x-\frac{x^3}{3}+\frac{x^2}{2}$ evaluated from to which yields $\frac{343}{6}$ .

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Question

Find the area under the curve $f(x)=\frac{1}{\sqrt{x+2}}$ between $2\leq x\leq 7$ .

Answer

To find the area under the curve, we need to integrate. In this case, it is a definite integral.

$\int_{2}^{7}\frac{1}{\sqrt{x+2}}dx=2\sqrt{x+2}\Big|_2^7=2$

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Question

Find the area bounded by $y=2, y=x, y=\frac{1}{9}x^2, x=3$

Answer

The easiest way to look at this is to plot the graphs. The shaded area is the actual area that we want to compute. We can first find area bounded by and in the first quadrant and subtract the excessive areas. The area of that rectangle box is 6. The area under the curve $\frac{1}{9}x^{2}$ is $\int_{0}^{3}\frac{1}{9}x^2=1$ .

The area of the triangle above the curve is 2. Therefore, the area bounded is .

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Question

$\int _0^\pi sin(x) dx$

Answer

$\int sin(x) dx = -cos(x) + C$ $\int _0^\pi sin(x) dx = -cos(\pi) - -cos(0)= 1-(-1) = 2$

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Question

Consider the region bounded by the functions

$f(x) = -x^{2} + 2x$ and

$g(x) = -sin(\frac{\pi x}{2})$

between and . What is the area of this region?

Answer

The area of this region is given by the following integral:

$Area = \int_{0}^{2} f(x) - g(x) = \int_{0}^{2} -x^{2} + 2x - (-sin(\frac{\pi x}{2}))$ or

$Area = \int_{0}^{2} -x^{2} + 2x + sin(\frac{\pi x}{2})$

Taking the antiderivative gives

$-\frac{x^{3}}{3} + x^{2} -\frac{2}{\pi }cos(\frac{\pi x}{2})$ , evaluated from to .

$F(2) = \frac{-8}{3} + 4 + \frac{2 }{\pi }$ , and

$F(0) = -\frac{2}{\pi }$ .

Thus, the area is given by:

$F(2) - F(0) =- \frac{8}{3} + 4 + \frac{2}{\pi } - (-\frac{2}{\pi }) =\frac{4}{3} +\frac{4}{\pi }$

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Question

Let $f(x) = x^{4}+ 6x - 17$ .

True or false: As a consequence of Rolle's Theorem, has a zero on the interval .

Answer

By Rolle's Theorem, if is continuous on and differentiable on , and , then there must be $c \in \left ( a,b \right )$ such that . Nothing in the statement of this theorem addresses the location of the zeroes of the function itself. Therefore, the statement is false.

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Question

is continuous and differentiable on $\mathbb{R}$ .

The values of for five different values of are as follows:

Which of the following is a consequence of Rolle's Theorem?

Answer

By Rolle's Theorem, if is continuous on and differentiable on , and , then there must be $c \in \left ( a,b \right )$ such that .

is given to be continuous. Also, if we set , we note that . This sets up the conditions for Rolle's Theorem to apply. As a consequence, there must be $c \in (1,2)$ such that .

Incidentally, it does follow from the given information that must have a zero on the interval , but this is due to the Intermediate Value Theorem, not Rolle's Theorem.

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Question

$f(x) = x^{3} - 7x + 16$

As a consequence of the Mean Value Theorem, there must be a value $c \in (-2, 2 )$ such that:

Answer

By the Mean Value Theorem (MVT), if a function is continuous and differentiable on , then there exists at least one value $c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$ . , a polynomial, is continuous and differentiable everywhere; setting , it follows from the MVT that there is $c \in (-2, 2 )$ such that

$f'(c) = \frac{f(2)-f(-2)}{2-(-2)}$

Evaluating and :

$f(x) = x^{3} - 7x + 16$

$f(2) = 2^{3} - 7 (2) + 16 = 8 -14+16 = 10$

$f(-2) = (-2) ^{3} - 7 (-2) + 16 = -8 +14+16 = 22$

The expression for is equal to

$f'(c) = \frac{10-22}{2-(-2)} = \frac{-12}{4} = -3$ ,

the correct choice.

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Question

Find the mean value of the function over the interval $(0,\frac{\pi}{2})$ .

Answer

To find the mean value of a function over some interval , one mus use the formula: .

Plugging in

$(f(\frac{\pi}{2})-f(0))=(\frac{\pi}{2})(-Sin(c))$

Simplifying

$-1=\frac{\pi}{2}(-Sin(c))$

$\frac{2}{\pi}=Sin(c)$

One must then use the inverse Sine function to find the value c:

$c=sin^{-1}\frac{2}{\pi}$

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Question

Find the area of the region enclosed by the parabola $y=12-x^2$ and $y=-x$ .

Answer

The two curves intersect in between and , which can be found by solving the quadratic equation $12-x^2=-x$ .

To solve for the area between curves, and , we use the formula $A=\int^b_a(f(x)-g(x))dx$

For our problem:

$A=\int^4_{-3}(12-x^2-(-x))dx$

$A=12x-\frac{x^3}{3}+\frac{x^2}{2}$ evaluated from to which yields $\frac{343}{6}$ .

Compare your answer with the correct one above

Question

Find the area under the curve $f(x)=\frac{1}{\sqrt{x+2}}$ between $2\leq x\leq 7$ .

Answer

To find the area under the curve, we need to integrate. In this case, it is a definite integral.

$\int_{2}^{7}\frac{1}{\sqrt{x+2}}dx=2\sqrt{x+2}\Big|_2^7=2$

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Question

Find the area bounded by $y=2, y=x, y=\frac{1}{9}x^2, x=3$

Answer

The easiest way to look at this is to plot the graphs. The shaded area is the actual area that we want to compute. We can first find area bounded by and in the first quadrant and subtract the excessive areas. The area of that rectangle box is 6. The area under the curve $\frac{1}{9}x^{2}$ is $\int_{0}^{3}\frac{1}{9}x^2=1$ .

The area of the triangle above the curve is 2. Therefore, the area bounded is .

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Question

$\int _0^\pi sin(x) dx$

Answer

$\int sin(x) dx = -cos(x) + C$ $\int _0^\pi sin(x) dx = -cos(\pi) - -cos(0)= 1-(-1) = 2$

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Question

Consider the region bounded by the functions

$f(x) = -x^{2} + 2x$ and

$g(x) = -sin(\frac{\pi x}{2})$

between and . What is the area of this region?

Answer

The area of this region is given by the following integral:

$Area = \int_{0}^{2} f(x) - g(x) = \int_{0}^{2} -x^{2} + 2x - (-sin(\frac{\pi x}{2}))$ or

$Area = \int_{0}^{2} -x^{2} + 2x + sin(\frac{\pi x}{2})$

Taking the antiderivative gives

$-\frac{x^{3}}{3} + x^{2} -\frac{2}{\pi }cos(\frac{\pi x}{2})$ , evaluated from to .

$F(2) = \frac{-8}{3} + 4 + \frac{2 }{\pi }$ , and

$F(0) = -\frac{2}{\pi }$ .

Thus, the area is given by:

$F(2) - F(0) =- \frac{8}{3} + 4 + \frac{2}{\pi } - (-\frac{2}{\pi }) =\frac{4}{3} +\frac{4}{\pi }$

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Question

Let $f(x) = x^{4}+ 6x - 17$ .

True or false: As a consequence of Rolle's Theorem, has a zero on the interval .

Answer

By Rolle's Theorem, if is continuous on and differentiable on , and , then there must be $c \in \left ( a,b \right )$ such that . Nothing in the statement of this theorem addresses the location of the zeroes of the function itself. Therefore, the statement is false.

Compare your answer with the correct one above

Question

is continuous and differentiable on $\mathbb{R}$ .

The values of for five different values of are as follows:

Which of the following is a consequence of Rolle's Theorem?

Answer

By Rolle's Theorem, if is continuous on and differentiable on , and , then there must be $c \in \left ( a,b \right )$ such that .

is given to be continuous. Also, if we set , we note that . This sets up the conditions for Rolle's Theorem to apply. As a consequence, there must be $c \in (1,2)$ such that .

Incidentally, it does follow from the given information that must have a zero on the interval , but this is due to the Intermediate Value Theorem, not Rolle's Theorem.

Compare your answer with the correct one above

Question

$f(x) = x^{3} - 7x + 16$

As a consequence of the Mean Value Theorem, there must be a value $c \in (-2, 2 )$ such that:

Answer

By the Mean Value Theorem (MVT), if a function is continuous and differentiable on , then there exists at least one value $c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$ . , a polynomial, is continuous and differentiable everywhere; setting , it follows from the MVT that there is $c \in (-2, 2 )$ such that

$f'(c) = \frac{f(2)-f(-2)}{2-(-2)}$

Evaluating and :

$f(x) = x^{3} - 7x + 16$

$f(2) = 2^{3} - 7 (2) + 16 = 8 -14+16 = 10$

$f(-2) = (-2) ^{3} - 7 (-2) + 16 = -8 +14+16 = 22$

The expression for is equal to

$f'(c) = \frac{10-22}{2-(-2)} = \frac{-12}{4} = -3$ ,

the correct choice.

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Question

Find the mean value of the function over the interval $(0,\frac{\pi}{2})$ .

Answer

To find the mean value of a function over some interval , one mus use the formula: .

Plugging in

$(f(\frac{\pi}{2})-f(0))=(\frac{\pi}{2})(-Sin(c))$

Simplifying

$-1=\frac{\pi}{2}(-Sin(c))$

$\frac{2}{\pi}=Sin(c)$

One must then use the inverse Sine function to find the value c:

$c=sin^{-1}\frac{2}{\pi}$

Compare your answer with the correct one above

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