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AP Calculus AB : Use of the Fundamental Theorem to represent a particular antiderivative, and the analytical and graphical analysis of functions so defined

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Example Questions

Example Question #1 : Integrals

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

Possible Answers:

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

$2xe^{x^4}$ $\displaystyle 2xe^{x^4}$

Does not exist

$x^2e^{x^2}$ $\displaystyle x^2e^{x^2}$

$e^{x^2}$ $\displaystyle e^{x^2}$

Correct answer:

$2xe^{x^4}$ $\displaystyle 2xe^{x^4}$

Explanation:

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

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

$= e^{(x^2)^2} \times (x^2)'$ $\displaystyle = e^{(x^2)^2} \times (x^2)'$. 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 $\displaystyle x$, f(x) $\displaystyle f(x)$, and then substituting $\displaystyle t = f(x)$ in the integrand. Lastly the Theorem states you must multiply your result by f'(x) $\displaystyle f'(x)$ (similar to the directions in using the chain rule).

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

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Example Question #2 : Integrals

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

Pbstm

$\displaystyle 1)$ What is the best interpretation of the function?

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

$\displaystyle 2)$ Which plot shows the derivative of the function F(x) $\displaystyle F(x)$?

Possible Answers:

Wrong3q10

Wrn4

Wrngan2

Question 10 correct answer

Correct answer:

Question 10 correct answer

Explanation:

$\displaystyle 1)$ $F(x)=\int_a^xf(t)dt$ $\displaystyle F(x)=\int_a^xf(t)dt$

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

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

$\displaystyle 2)$ The graph of the derivative of F(x) $\displaystyle F(x)$ is the same as the graph for f(x) $\displaystyle f(x)$. This follows directly from the Second Fundamental Theorem of Calculus.

If the function f(x) $\displaystyle f(x)$ is continuous on an interval $\displaystyle I$ containing $\displaystyle a$, then the function defined by:

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

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

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Example Question #3 : Integrals

Evaluate

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

Possible Answers:

$\displaystyle 1$

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

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

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

$\displaystyle 0$

Correct answer:

$\displaystyle 0$

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

$3\sin(2x)$ $\displaystyle 3\sin(2x)$

Namely, that

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

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

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

$\displaystyle =-(SOMETHING)+(SOMETHING)=0$

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AP Calculus AB : Use of the Fundamental Theorem to represent a particular antiderivative, and the analytical and graphical analysis of functions so defined

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 #3 : Integrals

All AP Calculus AB Resources

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