Area - 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

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

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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Question

Find the value of $\int_{0}^{\infty}xe^{-x^2}dx.$

Answer

To solve this problem, we will need to do a -substitution. Letting

$u=-x^2\Rightarrow \frac{du}{dx}=-2x\Rightarrow \frac{-du}{2}=xdx$ .

Substituting our function back into the integral, we get

$\int \frac{-1}{2}e^udu=\frac{-1}{2}e^u=\frac{-1}{2}e^{-x^2}.$

Evaluating this at $x=\infty$ and we get

$\frac{-1}{2}e^{- \infty^2}-(\frac{-1}{2}e^0)=0-(\frac{-1}{2})=\frac{1}{2}$

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Question

Find the average value of on the interval

Answer

The average is given by integration as:

$\text{avg}_{A, B}[ f ] = \frac{1}{B - A} \int_A^B f(x)~dx$

This means that:

$\text{avg}_{0, 1}[x \to x^3 - 2x^2 + x ] = \frac{1}{1 - 0} \int_0^1 [x^3 - 2x^2 + x]~dx$

$= \left[\frac{x^4}{4} - \frac{2 x^3}{3} + \frac{x^2}{2}\right]_0^1 = \frac{1}{4} - \frac{2}{3} + \frac{1}{2} = \frac{1}{12}$

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Question

Calculate the area between the parabola and the line .

Answer

To calculate the area between these two functions, we are going to need to set up a definite integral, so our first step is to see where our two boundaries intersect. We do this by setting our two functions equal to each other, and solving for the x values at which they intersect:

Here we can see that our two functions intersect at x=5 and x= -1, so these will be the limits for our definite integral. Now that we know our limits, we set up our integral of the function of the upper boundary minus the function for the lower boundary:

$\int_{-1}^{5}((4+5x-x^2)-(x-1))dx=\int_{-1}^{5}(-x^2+4x+5)dx$

$=(-\frac{1}{3}x^3+2x^2+5x)_{-1}^{5}$

$=(-\frac{1}{3}(5)^3+2(5)^2+5(5))-(-\frac{1}{3}(-1)^3+2(-1)^2+5(-1))$

$=(-\frac{125}{3}+50+25)-(\frac{1}{3}+2-5)=(\frac{100}{3})-(-\frac{8}{3})=36$

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Question

Find the area under the curve of from .

Answer

To find the area under the curve, integrate the function and evaluate at the bounds.

$\int_{-1}^{1} 1 dx= x|_{-1}^{1} = 1-(-1) = 2$

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Question

Find the area bounded by and $y=\sqrt x$ .

Answer

The top curve will be $y=\sqrt x$ , and the bottom curve will be . The area is the integral of the top minus the bottom curve.

First, determine the bounds of integration by setting both equations equal to each other, and solve for x. These values are where both graphs intersect.

$x= \sqrt x$

The bounds of integration will be from 0 to 1. Setup and evaluate the integral.

$\int_{0}^{1}\sqrt(x)-x:dx= \left[\frac{2}{3}x^\frac{3}{2} -\frac{1}{2}x^2\right]_{0}^{1} = \frac{2}{3}-\frac{1}{2}=\frac{1}{6}$

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Question

What is the area under the curve bounded by from ?

Answer

The area under the curve is the integral of the function evaluated at the interval given.

Write the integral to be evaluated.

$\int_{1}^{3}x^3:dx = \left[\frac{1}{4}x^4\right]_{1}^{3} = \frac{1}{4}(3)^4-\frac{1}{4}(1)^4$

$\frac{81}{4}-\frac{1}{4} =\frac{80}{4}=20$

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Question

Find the area of the region bounded by the following two curves:

$y=\sqrt{x}$

Answer

In order to find the area of the region bounded by the two curves, we must first find the bounds of the region by determning where the curves intersect:

$x^2=\sqrt{x}\rightarrow x^2-\sqrt{x}=0\rightarrow x^{\frac{1}{2}}(x^{\frac{3}{2}}-1)=0$

$x^{\frac{1}{2}}=0\rightarrow x=0$

$x^{\frac{3}{2}}-1=0\rightarrow x^{\frac{3}{2}}=1\rightarrow x=1$

The curves intersect at x=0 and x=1, so these are the bounds of the region for which we want to determine the area. Our next step is to set up an integral with these bounds, where the bottom curve is subtracted from the upper curve, which can then be evaluated to find the area of the region:

$A=\int_{0}^{1}(\sqrt{x}-x^2)dx=\left[\frac{2}{3}x^{\frac{3}{2}}-\frac{1}{3}x^3\right]_0^1=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$

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Question

Find the area of the region bounded by the following two functions:

Answer

Before we can find the area of the region bounded by the two curves, we must first determine the bounds of that region by finding where the two functions intersect. We do this by setting them equal to each other and solving for x:

$4x+16=2x^2+10\rightarrow 2x^2-4x-6=0\rightarrow 2(x^2-2x-3)=0$

$2(x-3)(x+1)=0\rightarrow x=-1,x=3$

The functions intersect at x=-1 and x=3, so these are the bounds of the area. Now we set up an integral with these bounds, where the lower function is subtracted from the upper function:

$A=\int_{-1}^{3}[(4x+16)-(2x^2+10)]dx=\int_{-1}^{3}(4x-2x^2+6)dx$

$=\left[2x^2-\frac{2}{3}x^3+6x\right]_{-1}^{3}=(18-18+18)-\left(2+\frac{2}{3}-6\right)=\frac{64}{3}$

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