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AP Calculus AB : AP Calculus AB

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Example Questions

Example Question #5 : Fundamental Theorem Of Calculus

Evaluate the following limit:

$\lim_{x\rightarrow \infty }\frac{1-x^4}{x^2-4x^4}$

Possible Answers:

$\frac{1}{4}$

$\infty$

$\frac{1}{2}$

Does not exist.

Correct answer:

$\frac{1}{4}$

Explanation:

First, let's multiply the numerator and denominator of the fraction in the limit by $\frac{1}{x^{4}}$ .

$\lim_{x\rightarrow \infty }\frac{1-x^4}{x^2-4x^4}$ $=\lim_{x\rightarrow \infty }\frac{(\frac{1}{x^4})(1-x^4)}{\frac{1}{x^{4}}(x^2-4x^4)}$

$=\lim_{x\rightarrow \infty }\frac{\frac{1}{x^4}-1}{\frac{1}{x^2}-4}$

As becomes increasingly large the $\frac{1}{x^{4}}$ and $\frac{1}{x^{2}} ^{}$ terms will tend to zero. This leaves us with the limit of $\frac{1}{4}$ .

$\lim_{x\rightarrow \infty }\frac{-1}{-4}=\frac{1}{4}$ .

The answer is $\frac{1}{4}$ .

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Example Question #6 : Fundamental Theorem Of Calculus

Let and be inverse functions, and let

$\\f(3)=4 \\g(3)=-5 \\g(4)=3 \\f'(3)=-2 \\f'(4)=4 \\g'(3)=-1$ .

What is the value of g'(4) ?

Possible Answers:

$-\frac{1}{2}$

$\frac{1}{3}$

Correct answer:

$-\frac{1}{2}$

Explanation:

Since and are inverse functions, g(f(x))=f(g(x))=x . We can differentiate both sides of the equation g(f(x)) with respect to to obtain the following:

$g'(f(x))\cdot f'(x)=1$

We are asked to find g'(4) , which means that we will need to find such that f(x)=4 . The given information tells us that f(3)=4 , which means that x=3 . Thus, we will substitute 3 into the equation.

$g'(f(3))\cdot f'(3)=1$

The given information tells us that f'(3)=-2 .

The equation then becomes $g'(4)\cdot (-2)=1$ .

We can now solve for g'(4) .

$g'(4)=-\frac{1}{2}$ .

The answer is $-\frac{1}{2}$ .

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Example Question #11 : Fundamental Theorem Of Calculus

Using the fundamental theorem of calculus, find the integral of the function x^3+2 x+9 from x=0 to x=1 .

Possible Answers:

$\frac{13}{4}$

$\frac{43}{4}$

$\frac{23}{4}$

$\frac{39}{4}$

$\frac{41}{4}$

Correct answer:

$\frac{41}{4}$

Explanation:

The fundamental theorem of calculus is, $\int_{a}^{b} f(x)\ dx=F(b)-F(a)$ , now lets apply this to our situation.

$\int_{0}^{1} x^3+2x+9\ dx$

We can use the inverse power rule to solve the integral, which is $\int x^n\ dx =\frac{x^{n+1}}{n+1}+C$ .

$\int_{0}^{1} x^3+2x+9\ dx=\frac{x^4}{4}+x^2+9x \Big|^{1}_{0}$

$=\left[ \frac{1^4}{4}+1^2+9(1)\right]-\left[\frac{0^4}{4}+0^2+9(0) \right ]$

$=\frac{1}{4}+1+9-0=\frac{41}{4}$

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Example Question #1 : How To Find Volume Of A Region

Consider the region bounded by the -axis, the line x=4 , and the function f(x) , where $f(x) = \sqrt{x}$ . Find the volume of the solid generated by rotating the region described about the line y=5 .

Possible Answers:

$\frac{136\pi }{3}$

$\frac{160\pi }{3}$

$\frac{160\pi ^{2}}{3}$

$\frac{136\pi ^{2}}{3}$

$8\pi$

Correct answer:

$\frac{136\pi }{3}$

Explanation:

To find the volume, one would use the washer method:

$V = \pi\int [R^{2}(x) - r^{2}(x)]dx.$

For this problem: $R(x) = 5, r(x) = 5-\sqrt{x}$ . So, the following integral would give the volume:

$V = \pi\int_{0}^{4} [5^{2} - (5-\sqrt{x})^{2} ]dx$

$V = \pi\int_{0}^{4} [25 - (25 - 10\sqrt{x}+ x)] dx$

$V = \pi\int_{0}^{4} (10\sqrt{x} - x) dx$

$V = \pi[\frac{20}{3}x^\frac{3}{2}} - \frac{x^{2}}{2}]$ , evaluated from x=0 to x=4 .

$V = \pi[\frac{20}{3}(4)\tfrac{3}{2} -\frac{4^{2}}{2} - (\frac{20}{3}(0)\tfrac{3}{2}-\frac{0^{2}}{2})]$

$V = \pi[(\frac{160}{3} - 8) - (0-0)]$

$V = \pi(\frac{160 - 24}{3}) = \frac{136\pi }{3}$

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Example Question #51 : Derivative As A Function

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

Possible Answers:

$\frac{301}{6}$

$\frac{434}{6}$

$\frac{401}{6}$

$\frac{334}{6}$

$\frac{343}{6}$

Correct answer:

$\frac{343}{6}$

Explanation:

The two curves intersect in between x=-3 and x=4 , which can be found by solving the quadratic equation $12-x^2=-x$ .

To solve for the area between curves, y=f(x) and y=g(x) , 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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Example Question #1 : Mean Value Theorem

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

Possible Answers:

$1$

$4$

$3$

$2$

$5$

Correct answer:

$2$

Explanation:

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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Example Question #2 : Mean Value Theorem

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

Possible Answers:

2.5

3.5

1.5

Correct answer:

Explanation:

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 x=3 and y=2 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 y=x is 2. Therefore, the area bounded is 6-1-2=3 .

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Example Question #1 : Mean Value Theorem

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

Possible Answers:

Correct answer:

Explanation:

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

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Example Question #4 : Mean Value Theorem

Consider the region bounded by the functions

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

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

between x = 0 and x = 2 . What is the area of this region?

Possible Answers:

$\frac{3+\pi }{4}$

$\frac{\pi }{4}$

$\frac{4}{3} +\frac{4}{\pi }$

$\frac{4}{3}$

$\frac{3\pi }{4}$

Correct answer:

$\frac{4}{3} +\frac{4}{\pi }$

Explanation:

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 x = 0 to x = 2 .

$F(2) = \frac{-8}{3} + 4 + \frac{2 }{\pi }$ , 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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Example Question #671 : Ap Calculus Ab

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

True or false: As a consequence of Rolle's Theorem, has a zero on the interval (0, 2) .

Possible Answers:

True

False

Correct answer:

False

Explanation:

By Rolle's Theorem, if is continuous on [a,b] and differentiable on (a,b) , and f(a)= f(b) , then there must be $c \in \left ( a,b \right )$ such that f'(c) = 0 . 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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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #5 : Fundamental Theorem Of Calculus

Example Question #6 : Fundamental Theorem Of Calculus

Example Question #11 : Fundamental Theorem Of Calculus

Example Question #1 : How To Find Volume Of A Region

Example Question #51 : Derivative As A Function

Example Question #1 : Mean Value Theorem

Example Question #2 : Mean Value Theorem

Example Question #1 : Mean Value Theorem

Example Question #4 : Mean Value Theorem

Example Question #671 : Ap Calculus Ab

All AP Calculus AB Resources

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