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

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Question

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

Answer

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 , , and then substituting in the integrand. Lastly the Theorem states you must multiply your result by (similar to the directions in using the chain rule).

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

Compare your answer with the correct one above

Question

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

What is the best interpretation of the function?

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

Which plot shows the derivative of the function ?

Answer

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

The function represents the area under the curve from to some value of .

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 with a dummy index when we write down the integral. It does not change the fundamental behavior of the function or .

The graph of the derivative of is the same as the graph for . This follows directly from the Second Fundamental Theorem of Calculus.

If the function is continuous on an interval containing , then the function defined by:

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

has for its' derivative .

Compare your answer with the correct one above

Question

Evaluate

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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 , , and then substituting in the integrand. Lastly the Theorem states you must multiply your result by (similar to the directions in using the chain rule).

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

Compare your answer with the correct one above

Question

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

What is the best interpretation of the function?

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

Which plot shows the derivative of the function ?

Answer

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

The function represents the area under the curve from to some value of .

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 with a dummy index when we write down the integral. It does not change the fundamental behavior of the function or .

The graph of the derivative of is the same as the graph for . This follows directly from the Second Fundamental Theorem of Calculus.

If the function is continuous on an interval containing , then the function defined by:

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

has for its' derivative .

Compare your answer with the correct one above

Question

Evaluate

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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 , , and then substituting in the integrand. Lastly the Theorem states you must multiply your result by (similar to the directions in using the chain rule).

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

Compare your answer with the correct one above

Question

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

What is the best interpretation of the function?

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

Which plot shows the derivative of the function ?

Answer

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

The function represents the area under the curve from to some value of .

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 with a dummy index when we write down the integral. It does not change the fundamental behavior of the function or .

The graph of the derivative of is the same as the graph for . This follows directly from the Second Fundamental Theorem of Calculus.

If the function is continuous on an interval containing , then the function defined by:

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

has for its' derivative .

Compare your answer with the correct one above

Question

Evaluate

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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 , , and then substituting in the integrand. Lastly the Theorem states you must multiply your result by (similar to the directions in using the chain rule).

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

Compare your answer with the correct one above

Question

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

What is the best interpretation of the function?

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

Which plot shows the derivative of the function ?

Answer

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

The function represents the area under the curve from to some value of .

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 with a dummy index when we write down the integral. It does not change the fundamental behavior of the function or .

The graph of the derivative of is the same as the graph for . This follows directly from the Second Fundamental Theorem of Calculus.

If the function is continuous on an interval containing , then the function defined by:

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

has for its' derivative .

Compare your answer with the correct one above

Question

Evaluate

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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 , , and then substituting in the integrand. Lastly the Theorem states you must multiply your result by (similar to the directions in using the chain rule).

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

Compare your answer with the correct one above

Question

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

What is the best interpretation of the function?

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

Which plot shows the derivative of the function ?

Answer

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

The function represents the area under the curve from to some value of .

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 with a dummy index when we write down the integral. It does not change the fundamental behavior of the function or .

The graph of the derivative of is the same as the graph for . This follows directly from the Second Fundamental Theorem of Calculus.

If the function is continuous on an interval containing , then the function defined by:

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

has for its' derivative .

Compare your answer with the correct one above

Question

Evaluate

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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 , , and then substituting in the integrand. Lastly the Theorem states you must multiply your result by (similar to the directions in using the chain rule).

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

Compare your answer with the correct one above

Question

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

What is the best interpretation of the function?

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

Which plot shows the derivative of the function ?

Answer

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

The function represents the area under the curve from to some value of .

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 with a dummy index when we write down the integral. It does not change the fundamental behavior of the function or .

The graph of the derivative of is the same as the graph for . This follows directly from the Second Fundamental Theorem of Calculus.

If the function is continuous on an interval containing , then the function defined by:

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

has for its' derivative .

Compare your answer with the correct one above

Question

Evaluate

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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 , , and then substituting in the integrand. Lastly the Theorem states you must multiply your result by (similar to the directions in using the chain rule).

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

Compare your answer with the correct one above

Question

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

What is the best interpretation of the function?

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

Which plot shows the derivative of the function ?

Answer

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

The function represents the area under the curve from to some value of .

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 with a dummy index when we write down the integral. It does not change the fundamental behavior of the function or .

The graph of the derivative of is the same as the graph for . This follows directly from the Second Fundamental Theorem of Calculus.

If the function is continuous on an interval containing , then the function defined by:

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

has for its' derivative .

Compare your answer with the correct one above

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