How to find area of a region - AP Calculus AB

Card 0 of 1904

Question

Find the area between and the -axis.

Answer

First we find that the equation crosses the y-axis at y=-1, 1.

Partitioning the y-axis, we get the area is equal to

$\int_{-1}^{1}(1-y^2) dy=\frac{4}{3}$ .

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Question

What is the area of the region created by the function and the -axis?

Answer

First, graph the two functions in order to identify the boundaries of the region. You will find that they are .

Therefore, when you set up your integral, it will be from to .

Then solve the integral by using the power rule

$\int x^ndx=\frac{x^{n+1}}{n+1}$ :

$\ \int_{-2}^2-x^2+4dx=-\int_{-2}^2x^2dx+\int_{-2}^24dx\ \=-\frac{x^3}{3}|_{-2}^2+4x|_{-2}^2\ \=-\left[\frac{(2)^3}{3}-\frac{(-2)^3}{3}\right]+\left[4(2)-4(-2)\right]\ \=-\left[\frac{8}{3}-\frac{-8}{3}\right]+[8+8]\ \=-\left[\frac{16}{3}\right]+16\ \=\frac{32}{3}$

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Question

What is the area of the region created by and the bounds and ?

Answer

Set up your integral using the given bounds, then solve by using the power rule

$\int x^ndx=\frac{x^{n+1}}{n+1}$ :

$\ \int_0^3x+2dx=\int_0^3xdx+\int_0^32dx\ \=\frac{x^2}{2}|_0^3+2x|_0^3\ \=\left(\frac{(3)^2}{2}-0\right)+(2(3)-0)\ \=\frac{9}{2}+6\ \=\frac{21}{2}$

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Question

Find the area created by with the boudaries and .

Answer

Set up your integral using the given bounds, then solve by using the power rule

$\int x^ndx=\frac{x^{n+1}}{n+1}$ :

$\ \int_0^6 2x+3 dx=\int_0^6 2x dx+\int_0^6 3 dx\ \=\frac{2x^2}{2}|_0^6+3x|_0^6\ \=(6)^2-(0)^2+[3(6)^2-3 (0)2]\ \=36+18\ \=54$ .

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Question

Find the area under the curve in the region bounded by the -axis, the lower bound and the upper bound

Answer

To find the area under the curve

Integrate it from the specified bounds:

$A=\int_{1}^{2}(20-3x^2)dx=20x-x^3|_1^2$

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Question

Find the area of the region bounded by the functions $\small \small f(x)=6cos(\frac{\pi}{6}x)$ , $\small \small g(x)=\frac{3}{2}x$ , and on the left by the -axis.

Answer

To find the area of the region between the functions $\small \small f(x)=6cos(\frac{\pi}{6}x)$ , $\small \small g(x)=\frac{3}{2}x$ , the first step will be to determine the lower and upper x bounds. We're told the region is bounded by the y-axis on the left, so $\small x_l=0$ . The upper x-bound will be where the two functions intersect:

$\small 6cos(\frac{\pi}{6}x_u)=\frac{3}{2}x_u$

For the value of $\small x_u=2$ , each of the functions equals $\small 3$ .

Over the interval of $\small x=[0,2]$ , $\small f(x)>g(x)$ , so the area of the region can be determined by the integral:

Use the following rules to integrate the function:

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

Applying the above rules we get,

$\small A=\int_0^2 (6cos(\frac{\pi}{6}x)-\frac{3}{2}x)dx$

$\small A=\frac{36}{\pi}sin(\frac{\pi}{6}x)-\frac{3}{4}x^2|_0^2$

$\small A=(\frac{36}{\pi}(\frac{\sqrt3}{2})-3)-(0-0)$

$\small A=\frac{18\sqrt3-3\pi}{\pi}$ .

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Question

Find the area under the curve from to $2\pi$ .

Answer

Finding the area under the curve can be understood as taking the integral of the equation and it can be rewritten into the following:

$\int_0^{2\pi}cos(x)+2xdx$

To solve:

1. Find the indefinite integral of the function

2. Plug in the upper and lower limit values and take the difference of the 2 values.

1. Using the power rule which states,

$\int x^n=\frac{x^{n+1}}{n+1}$ to the term and recalling the integral of is we find,

$\int cos(x)+2xdx = x^{2}+sin(x)+constant$ .

2. $(2\pi)^{2}+sin(2\pi) - (0^2 + sin(0)) = (2\pi)^2 = 4\pi^2$

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Question

Find the area of the region encompassed by the curves and $g(x)=\frac{1}{2}x^2$ , and the y-axis.

Answer

For this problem, we must first find the upper bound of the region, the x-value where the two curves intersect:

$10-x_u^3=\frac{1}{2}x_u^2$

This equation holds for the value of

Therefore the lower and upper bounds are . The lower bound is known since we're told the region is bounded by the y-axis.

The area of the region is thus:

$A=\int_{0}^{2}10-x^3-\frac{1}{2}x^2$

$A=10x-\frac{x^4}{4}-\frac{x^3}{6}|_0^2$

$A=(20-4-\frac{4}{3})-(0-0-0)$

$A=\frac{44}{3}$

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Question

Find the area enclosed by the lines $f(x)=\frac{1}{3}x$ , , and the x-axis.

Answer

The first step is determine the lower and upper x-values that define the area. There is a lower bound of zero that marks the transition for f(x) to move into negative y-values; however, g(x) is well into the negative at this point, so it'll be necessary to find a lower bound where it first begins to become negative. This will occur for a value of five:

This allows the creation of an initial integral:

$A_1=\int_{0}^{5}\frac{1}{3}xdx$

$A_1=\frac{x^2}{6}|_0^5$

$A_1=\frac{25}{6}-0=\frac{25}{6}$

Another upper bound can be found by determining the point where the two functions intersect:

$\frac{1}{3}x=2x-10$

$\frac{5}{3}x=10$

Now, integrate the difference of these functions over these final bounds:

$A_2=\int_{5}^{6}\frac{1}{3}x-(2x-10)dx$

$A_2=\frac{x^2}{6}-x^2+10x|_5^6$

$A_2=(\frac{36}{6}-36+60)-(\frac{25}{6}-25+50)$

$A_2=\frac{11}{6}-11+10=\frac{5}{6}$

The full area is now the sum of these two:

$A=\frac{25}{6}+\frac{5}{6}=\frac{30}{6}$

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Question

$f(x)=\frac{\pi}{3}sin(x)$

Find the area under the curve drawn by the function on the interval of $x=\frac{\pi}{4}$ to $x=\frac{3\pi}{4}$ .

Answer

In order to find the area under on the interval of $x=\frac{\pi}{4}$ to $x=\frac{3\pi}{4}$ , you must evaluate the definite integral

$\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} f(x)\ dx$

$\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi}{3}sin(x)\ dx$

First, integrate the function.

$=-\frac{\pi}{3}cos(x)\mid_{\frac{\pi}{4}}^{\frac{3\pi}{4}}$

Then, substitute values for .

$=-\frac{\pi}{3}cos\left({\frac{3\pi}{4}}\right) -\frac{-\pi}{3}cos\left({\frac{\pi}{4}}\right)$

Finally, evaluate while cancelling negative signs.

$=-\frac{\pi}{3}\frac{-\sqrt{2}}{2}} +\frac{\pi}{3}\frac{\sqrt{2}}{2}}$

$=-\frac{\pi}{3}\frac{-\sqrt{2}}{2}} +\frac{\pi}{3}\frac{\sqrt{2}}{2}}$

$=\frac{\pi}{3}\frac{\sqrt{2}}{2}} +\frac{\pi}{3}\frac{\sqrt{2}}{2}}$

$=2\frac{\pi}{3}*\frac{\sqrt{2}}{2}}$

$=\frac{\pi{\sqrt{2}}}{3}$

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Question

Find $\int \frac{3x^2-2x+1}{\sqrt{x}}dx$

Answer

$\int \frac{3x^2-2x+1}{\sqrt{x}}dx=\int (3x^{3/2}-2x^{1/2}+x^{-1/2})dx \Rightarrow$

$3(\frac{2}{5})x^{5/2}-2(\frac{2}{3})x^{3/2}+2x^{1/2}+C = \frac{6}{5}x^{5/2}-\frac{4}{3}x^{3/2}+2x^{1/2}+C$

$=\frac{2}{5}\sqrt{x}(9x^2-10x+15)+C$

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Question

Find $\int \frac{1}{sec(x)}dx.$

Answer

$\int \frac{1}{sec(c)}dx=\int cos(x)dx=sin(x)+C$

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Question

Which of the following is the correct way to evaluate the following integral?

$\int\frac{4}{x}dx$

Answer

To integrate this function, we need to know a few things.

Any constant term can be pulled out of the integral and multiplied through at the end. So our original expression becomes this:

$\int\frac{4}{x}dx=4\int\frac{1}{x}dx$

Integrating one over x doesn't follow the classic integrating rules. In this case, the integral of 1 over x is equal to the ln of x. So we do the following

$4\int\frac{1}{x}dx=4ln\left | x\right |+c$

We need to remember our plus c, otherwise we will not quite be correct.

We could then evaluate the function on any interval to find the area underneath the curve.

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Question

What is the average value of the function f(x) = 12x3 + 15x + 5 on the interval \[3, 6\]?

Answer

To find the average value, we must take the integral of f(x) between 3 and 6 and then multiply it by 1/(6 – 3) = 1/3.

The indefinite form of the integral is: 3x4 + 7.5x2 + 5x

The integral from 3 to 6 is therefore: (3(6)4 + 7.5(6)2 + 5(6)) - (3(3)4 + 7.5(3)2 + 5(3)) = (3888 + 270 + 30) – (243 + 22.5 + 15) = 3907.5

The average value is 3907.5/3 = 1302.5

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Question

Find the dot product of a = <2,2,-1> and b = <5,-3,2>.

Answer

To find the dot product, we multiply the individual corresponding components and add.

Here, the dot product is found by:

2 * 5 + 2 * (-3) + (-1) * 2 = 2.

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Question

We have the function $f(x)=\sqrt{x}$ and it is used to form a three-dimensional figure by rotating it about the line $y=4$ . Find the volume of that figure from $x=0$ to $x=5$ .

Answer

Imagine a point somewhere on the function $f(x)=\sqrt{x}$ and it rotates about $y=4$ to form a circle with a circumference of $2\pi (4-\sqrt{x})$ where $(4-\sqrt{x})$ is the radius. You may have to draw a picture/graph to make sure this is clear.

Next, pretend that this circle is a circular strip with thickness $\Delta x$ . To find the area of that circular strip, we times that thickness by the circumference so that we have

$2\pi (4-\sqrt{x})\Delta x$

Now, imagine that the three dimensional figure is made up of many of these circular strips. To find the total volume, we need to sum up the areas of all of these strips. We do this by turning $2\pi (4-\sqrt{x})\Delta x$ into an integral from 0 to 5.

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

Perform your integration

$Volume=2\pi \left [ 4(5)-\frac{2}{3}(5)^{\frac{3}{2}} - 4(0)+\frac{2}{3}(0)^{\frac{3}{2}} \right ]=2\pi\left [ 20-\frac{2}{3}\sqrt{5^3} \right ]$

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Question

Find the area of the region enclosed by the parabola and the line .

Answer

The limits of the integration are found by solving and for :

The region runs from to . The limits of the integration are , .
The area between the curves is:

$A = \int_{a}^{b} [f(x)-g(x)]dx = \int_{-1}^{2}[(2-x^2)-(-x))]dx$
$= \int_{1}^{2}(2+x-x^2)dx = \left [ 2x+x^2/2-x^3/3\right ]_{-1}^{2}$
$= (4+\frac{4}{2}-\frac{8}{3}) - (-2 +\frac{1}{2}+\frac{1}{3}) = \frac{9}{2}$

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Question

What is the area of the space below and above

Answer

is only above over the interval . Areas are given by the definite integral of each function $a_1 = \int_{0}^{1}xdx$ and $a_2 = \int_{0}^{1}x^2dx$

The area between the curves is found by subtracting the area between each curve and the -axis from each other. For this area is $\frac{1}{2}$ and for the area is $\frac{1}{3}$ giving an area between curves of $\frac{1}{6}$

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Question

What is the area below and above the -axis?

Answer

To find the area below a curve, you must find the definite integral of the function. In this case the limits of integration are where the original function intercepts the -axis at and . So you must find $\int_{-2}^{2}(-x^2+4)dx$ which is $-\frac{1}{3}x^3+4 x$ evaluated from to . This gives an answer of $\frac{32}{3}$

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Question

Find the area between the curves and .

Answer

To solve this problem, we first need to find the point where the two equations are equal. Doing this we find that

$2x=x^2\Rightarrow x^2-2x=0\Rightarrow x(x-2)=0$ .

From this, we see that the two graphs are equal at and . We also know that for , is greater than .

So to find the area between these curves we need to evaluate the integral $\int_{0}^{2} (2x-x^2)dx$ .

The solution to the integral is

$\int(2x-x^2)dx=x^2-\frac{1}{3}x^3$ .

Evaluating this at and we get

$(2*2-\frac{1}{3}2^3)-(0-0)=\frac{12}{3}-\frac{8}{3}=\frac{4}{3}.$

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