How to graph functions of area - AP Calculus AB

Card 0 of 98

Question

Find the area bounded by the curve in the first quadrant.

Answer

The curve is in quadrant one over the interval , which are the bounds of integration. To see this, note that the x-intercepts are 0 and 2 and the parabola opens downward.

The definite integral below is solved by taking the antiderivative of each term of the given polynomial function, evaluating this antiderivative at the bounds of integration, and subtracting the values.

For this particular integral use the rule, $\int x^n=\frac{x^{n+1}}{n+1}$ to solve.

$\int_{0}^{2}(2x-x^2)dx=(x^2-\frac{1}{3}x^3)|^2_0=(2^2-\frac{1}{3}\cdot2^3)-(0^2-\frac{1}{3}\cdot0^3)=\frac{4}{3}$

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Question

Find the area bounded by the curve $y=\sin x \cos x$ and the -axis over the interval $\left[0,\frac{\pi}{2}\right]$ .

Answer

The curve is in quadrant one over the given interval, which gives us the bounds of integration. Evaluating this definite integral yields the area we are after.

In order to performe the antiderivative, let $u=\sin x$ . It follows that $du = \cos x dx$ . Therefore

$\int (\sin x \cos x)dx= \int u du=\frac{1}{2}u^2 + C = \frac{1}{2} \sin^2 x + C$

so

$\int_{0}^{\frac{\pi}{2}}(\sin x \cos x)dx= \left[ \frac {1}{2}\sin^2 x \right] ^\frac{\pi}{2}_0= \left( \frac{1}{2}\sin^2\frac{\pi}{2} \right) - \left( \frac{1}{2}\sin^2 0 \right)=\frac{1}{2}$

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Question

Find the area bounded by the curve and the -axis over the interval .

Answer

The curve is positive over the given interval, so the endpoints of the interval will mark the bounds of integration. This function is very easy to integrate because the derivative of is itself!

$\int_{0}^{1} e^xdx=\left[ e^x\right]^1_0=e^1-e^0=e-1$

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Question

Find the area bounded by the curve $y=e^{-x}$ and the -axis over the interval .

Answer

This function is positve over the given interval, so the endpoints of the interval mark the bounds of integration. It is a straightforward integration that is solvable with u-substitution. Let so . This means

$\int e^{-x}dx = - \int e^udu = -e^u + C = -e^{-x} + C$

so

$\int_{0}^{1} e^{-x}dx=\left[ -e^{-x}\right]^1_0=(-e^{-1})-(-e^0)=-\frac{1}{e}+1=\frac{e-1}{e}$

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Question

Graph of a piecewise-linear function , for , is shown above.

Find

$\int_{-2}^{4}f(x)dx$ .

Answer

Find the area under the graph from . To do this break the graph up into triangles, squares, and rectangles to calculate the individual areas over smaller intervals and then add them all together.

The areas are added to be:

$A=\left[\frac{1}{2}(1)(2)+\frac{1}{2}(1)(-1)+\frac{1}{2}(1)(-1)+(1)(1)+(2)(3)\right]=7$

Therefore,

$\int_{-2}^{4}f(x)dx=7$ .

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Question

Find the area under the curve between the following bounderies of the following function.

in between the boundaries of and

Answer

We can find the area under the curve by taking the anti-derivative of the function and using the two boundaries as x values. The anti-derivative of is . If we use the two boundaries, we end up with our answer, .

Compare your answer with the correct one above

Question

Evalute the following Definite Integral:
$\int_{0}^{\pi}(Sin(x)Cos(x))dx$

Answer

For this integral one may be tempted to directly integrate; however, this is no rule of integrals which allows us to do so.

We must apply a complex method of integration, here u-subtitution works best.

We notice that Cos(x) is the derivative of Sin(x) so it may be best to let .

Now that we have found a proper u and du, we may directly substiute into our original integral

$\int_{0}^{\pi}(u)du=\frac{u^2}{2}_{x=0}^{x=\pi}$

However, our limits are in terms of x so we must substitute our u=Sin(x) back in before evaluating

$\frac{(Sin(x))^2}{2}_{x=0}^{x=\pi}=\frac{Sin(\pi)^2}{2}-\frac{Sin(0)^2}{2}=0-0=0$

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Question

Find the area bounded by the curve in the first quadrant.

Answer

The curve is in quadrant one over the interval , which are the bounds of integration. To see this, note that the x-intercepts are 0 and 2 and the parabola opens downward.

The definite integral below is solved by taking the antiderivative of each term of the given polynomial function, evaluating this antiderivative at the bounds of integration, and subtracting the values.

For this particular integral use the rule, $\int x^n=\frac{x^{n+1}}{n+1}$ to solve.

$\int_{0}^{2}(2x-x^2)dx=(x^2-\frac{1}{3}x^3)|^2_0=(2^2-\frac{1}{3}\cdot2^3)-(0^2-\frac{1}{3}\cdot0^3)=\frac{4}{3}$

Compare your answer with the correct one above

Question

Find the area bounded by the curve $y=\sin x \cos x$ and the -axis over the interval $\left[0,\frac{\pi}{2}\right]$ .

Answer

The curve is in quadrant one over the given interval, which gives us the bounds of integration. Evaluating this definite integral yields the area we are after.

In order to performe the antiderivative, let $u=\sin x$ . It follows that $du = \cos x dx$ . Therefore

$\int (\sin x \cos x)dx= \int u du=\frac{1}{2}u^2 + C = \frac{1}{2} \sin^2 x + C$

so

$\int_{0}^{\frac{\pi}{2}}(\sin x \cos x)dx= \left[ \frac {1}{2}\sin^2 x \right] ^\frac{\pi}{2}_0= \left( \frac{1}{2}\sin^2\frac{\pi}{2} \right) - \left( \frac{1}{2}\sin^2 0 \right)=\frac{1}{2}$

Compare your answer with the correct one above

Question

Find the area bounded by the curve and the -axis over the interval .

Answer

The curve is positive over the given interval, so the endpoints of the interval will mark the bounds of integration. This function is very easy to integrate because the derivative of is itself!

$\int_{0}^{1} e^xdx=\left[ e^x\right]^1_0=e^1-e^0=e-1$

Compare your answer with the correct one above

Question

Find the area bounded by the curve $y=e^{-x}$ and the -axis over the interval .

Answer

This function is positve over the given interval, so the endpoints of the interval mark the bounds of integration. It is a straightforward integration that is solvable with u-substitution. Let so . This means

$\int e^{-x}dx = - \int e^udu = -e^u + C = -e^{-x} + C$

so

$\int_{0}^{1} e^{-x}dx=\left[ -e^{-x}\right]^1_0=(-e^{-1})-(-e^0)=-\frac{1}{e}+1=\frac{e-1}{e}$

Compare your answer with the correct one above

Question

Graph of a piecewise-linear function , for , is shown above.

Find

$\int_{-2}^{4}f(x)dx$ .

Answer

Find the area under the graph from . To do this break the graph up into triangles, squares, and rectangles to calculate the individual areas over smaller intervals and then add them all together.

The areas are added to be:

$A=\left[\frac{1}{2}(1)(2)+\frac{1}{2}(1)(-1)+\frac{1}{2}(1)(-1)+(1)(1)+(2)(3)\right]=7$

Therefore,

$\int_{-2}^{4}f(x)dx=7$ .

Compare your answer with the correct one above

Question

Find the area under the curve between the following bounderies of the following function.

in between the boundaries of and

Answer

We can find the area under the curve by taking the anti-derivative of the function and using the two boundaries as x values. The anti-derivative of is . If we use the two boundaries, we end up with our answer, .

Compare your answer with the correct one above

Question

Evalute the following Definite Integral:
$\int_{0}^{\pi}(Sin(x)Cos(x))dx$

Answer

For this integral one may be tempted to directly integrate; however, this is no rule of integrals which allows us to do so.

We must apply a complex method of integration, here u-subtitution works best.

We notice that Cos(x) is the derivative of Sin(x) so it may be best to let .

Now that we have found a proper u and du, we may directly substiute into our original integral

$\int_{0}^{\pi}(u)du=\frac{u^2}{2}_{x=0}^{x=\pi}$

However, our limits are in terms of x so we must substitute our u=Sin(x) back in before evaluating

$\frac{(Sin(x))^2}{2}_{x=0}^{x=\pi}=\frac{Sin(\pi)^2}{2}-\frac{Sin(0)^2}{2}=0-0=0$

Compare your answer with the correct one above

Question

Find the area bounded by the curve in the first quadrant.

Answer

The curve is in quadrant one over the interval , which are the bounds of integration. To see this, note that the x-intercepts are 0 and 2 and the parabola opens downward.

The definite integral below is solved by taking the antiderivative of each term of the given polynomial function, evaluating this antiderivative at the bounds of integration, and subtracting the values.

For this particular integral use the rule, $\int x^n=\frac{x^{n+1}}{n+1}$ to solve.

$\int_{0}^{2}(2x-x^2)dx=(x^2-\frac{1}{3}x^3)|^2_0=(2^2-\frac{1}{3}\cdot2^3)-(0^2-\frac{1}{3}\cdot0^3)=\frac{4}{3}$

Compare your answer with the correct one above

Question

Find the area bounded by the curve $y=\sin x \cos x$ and the -axis over the interval $\left[0,\frac{\pi}{2}\right]$ .

Answer

The curve is in quadrant one over the given interval, which gives us the bounds of integration. Evaluating this definite integral yields the area we are after.

In order to performe the antiderivative, let $u=\sin x$ . It follows that $du = \cos x dx$ . Therefore

$\int (\sin x \cos x)dx= \int u du=\frac{1}{2}u^2 + C = \frac{1}{2} \sin^2 x + C$

so

$\int_{0}^{\frac{\pi}{2}}(\sin x \cos x)dx= \left[ \frac {1}{2}\sin^2 x \right] ^\frac{\pi}{2}_0= \left( \frac{1}{2}\sin^2\frac{\pi}{2} \right) - \left( \frac{1}{2}\sin^2 0 \right)=\frac{1}{2}$

Compare your answer with the correct one above

Question

Find the area bounded by the curve and the -axis over the interval .

Answer

The curve is positive over the given interval, so the endpoints of the interval will mark the bounds of integration. This function is very easy to integrate because the derivative of is itself!

$\int_{0}^{1} e^xdx=\left[ e^x\right]^1_0=e^1-e^0=e-1$

Compare your answer with the correct one above

Question

Find the area bounded by the curve $y=e^{-x}$ and the -axis over the interval .

Answer

This function is positve over the given interval, so the endpoints of the interval mark the bounds of integration. It is a straightforward integration that is solvable with u-substitution. Let so . This means

$\int e^{-x}dx = - \int e^udu = -e^u + C = -e^{-x} + C$

so

$\int_{0}^{1} e^{-x}dx=\left[ -e^{-x}\right]^1_0=(-e^{-1})-(-e^0)=-\frac{1}{e}+1=\frac{e-1}{e}$

Compare your answer with the correct one above

Question

Graph of a piecewise-linear function , for , is shown above.

Find

$\int_{-2}^{4}f(x)dx$ .

Answer

Find the area under the graph from . To do this break the graph up into triangles, squares, and rectangles to calculate the individual areas over smaller intervals and then add them all together.

The areas are added to be:

$A=\left[\frac{1}{2}(1)(2)+\frac{1}{2}(1)(-1)+\frac{1}{2}(1)(-1)+(1)(1)+(2)(3)\right]=7$

Therefore,

$\int_{-2}^{4}f(x)dx=7$ .

Compare your answer with the correct one above

Question

Find the area under the curve between the following bounderies of the following function.

in between the boundaries of and

Answer

We can find the area under the curve by taking the anti-derivative of the function and using the two boundaries as x values. The anti-derivative of is . If we use the two boundaries, we end up with our answer, .

Compare your answer with the correct one above

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