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AP Calculus AB : Fundamental Theorem of Calculus

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Example Questions

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Example Question #1 : Integrals

Evaluate $\frac{d}{dx}\int_4^{x^2} e^{t^2}dt$ .

Possible Answers:

$2xe^{x^4}$

$x^2e^{x^2}$

$e^{x^2}$

Does not exist

$\frac{e^{t^2}}{t^2}$

Correct answer:

$2xe^{x^4}$

Explanation:

Even though an antideritvative of $e^{t^2}$ does not exist, we can still use the Fundamental Theorem of Calculus to "cancel out" the integral sign in this expression.

$\frac{d}{dx}\int_4^{x^2} e^{t^2}dt$ . Start

$= e^{(x^2)^2} \times (x^2)'$ . You can "cancel out" the integral sign with the derivative by making sure the lower bound of the integral is a constant, the upper bound is a differentiable function of , f(x) , and then substituting t = f(x) in the integrand. Lastly the Theorem states you must multiply your result by f'(x) (similar to the directions in using the chain rule).

$=2xe^{x^4}$ .

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Example Question #2 : Integrals

The graph of a function f(x) is drawn below. Select the best answers to the following:

Pbstm

What is the best interpretation of the function?

$F(x)=\int_a^xf(t)dt$

Which plot shows the derivative of the function F(x) ?

Possible Answers:

Wrong3q10

Question 10 correct answer

Wrngan2

Wrn4

Correct answer:

Question 10 correct answer

Explanation:

$F(x)=\int_a^xf(t)dt$

The function F(x) represents the area under the curve f(x) from x=a to some value of x>a .

Do not be confused by the use of in the integrand. The reason we use is because are writing the area as a function of , which requires that we treat the upper limit of integration as a variable . So we replace the independent variable of f(x) with a dummy index when we write down the integral. It does not change the fundamental behavior of the function F(x) or f(t) .

The graph of the derivative of F(x) is the same as the graph for f(x) . This follows directly from the Second Fundamental Theorem of Calculus.

If the function f(x) is continuous on an interval containing , then the function defined by:

$F(x)=\int_a^xf(t)dt$

has for its' derivative F'(x)=f(x) .

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Example Question #1 : Integrals

Evaluate

$\int_{-(4\pi+e^4)}^{4\pi+e^4}3\sin(2x)dx$

Possible Answers:

$\frac{-3\cos(8\pi+2e^4)}{2}+\frac{3\cos(-8\pi-2e^4)}{2}$

$\frac{-3\cos(8\pi+2e^4)}{2}$

$\frac{-3\cos(\pi+e^4)}{2}$

Correct answer:

Explanation:

Here we could use the Fundamental Theorem of Calculus to evaluate the definite integral; however, that might be difficult and messy.

Instead, we make a clever observation of the graph of

$3\sin(2x)$

Namely, that

$3\sin(2x)=-3\sin(-2x)$

This means that the values of the graph when comparing x and -x are equal but opposite. Then we can conclude that

$\int_{-(4\pi+e^4)}^{4\pi+e^4}3\sin(2x)dx=\int_{-(4\pi+e^4)}^{0}3\sin(2x)dx+\int_{0}^{4\pi+e^4}3\sin(2x)dx$

=-(SOMETHING)+(SOMETHING)=0

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Example Question #111 : Integrals

Wolframalpha--plot_piecewisexxlt03-x0ltxlt213-xxgt2--2013-09-10_1008

Find $\lim _{x\rightarrow 0}f(x)$

Possible Answers:

$+\infty$

Does not exist

Correct answer:

Does not exist

Explanation:

The one side limits are not equal: left is 0 and right is 3

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Example Question #2 : Fundamental Theorem Of Calculus

Wolframalpha--plot_piecewisexxlt03-x0ltxlt213-xxgt2--2013-09-10_1008

Which of the following is a vertical asymptote?

Possible Answers:

x=0

x=3

y=0

y=3

Correct answer:

x=3

Explanation:

When approaches 3, approaches $\pm \infty$ .

Vertical asymptotes occur at values. The horizontal asymptote occurs at

y=0 .

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Example Question #112 : Integrals

$f(x) = \frac{x+1}{\sqrt{x^2 -4}}$

What are the horizontal asymptotes of f(x) ?

Possible Answers:

x=1

x=0

x=-1

y=1

y=0

Correct answer:

y=1

Explanation:

Compute the limits of f(x) as approaches infinity.

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Example Question #4 : Fundamental Theorem Of Calculus

Write the domain of the function.

$f(x)=\frac{(x^{4}-81)^{1/2}}{x-4}$

Possible Answers:

$x\neq 4\ and\ x\neq 3$

$x\neq 4\ and\ \left | x \right |\geq 3$

$x\neq -9\ and\ x\neq 9$

$x\geq 4\ and\ x\neq 3$

${x\neq 4}$

Correct answer:

$x\neq 4\ and\ \left | x \right |\geq 3$

Explanation:

The answer is ${x\mid x\neq 4\ and\ \left | x \right |\geq 3}$

The denominator must not equal zero and anything under a radical must be a nonnegative number.

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Example Question #113 : Integrals

What is the value of the derivative of $f(x)=x^3+6x^2+\frac{3}{4}x$ at x=1?

Possible Answers:

$\frac{15}{4}$

$\frac{63}{4}$

$\frac{61}{4}$

Correct answer:

$\frac{63}{4}$

Explanation:

First, find the derivative of the function, which is:

$f{}'(x)=3x^2+12x+\frac{3}{4}$

Then, plug in 1 for x:

$3(1^2)+12(1)+\frac{3}{4}$

The result is $\frac{63}{4}$ .

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Example Question #1 : Calculus 3

Evaluate the following limit:

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

Possible Answers:

$\frac{1}{2}$

Does not exist.

$\infty$

$\frac{1}{4}$

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 #1 : 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:

$\frac{1}{3}$

$-\frac{1}{2}$

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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AP Calculus AB : Fundamental Theorem of Calculus

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1 : Integrals

Example Question #2 : Integrals

Example Question #1 : Integrals

Example Question #111 : Integrals

Example Question #2 : Fundamental Theorem Of Calculus

Example Question #112 : Integrals

Example Question #4 : Fundamental Theorem Of Calculus

Example Question #113 : Integrals

Example Question #1 : Calculus 3

Example Question #1 : Fundamental Theorem Of Calculus

All AP Calculus AB Resources

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