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Calculus 1 : Functions

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Example Questions

Example Question #4064 : Calculus

y_1=x and y_2=x^2

What is the area of the region created by the two functions?

Possible Answers:

$\frac{1}{6}$

$\frac{5}{6}$

$\frac{-1}{6}$

$\frac{-5}{6}$

Correct answer:

$\frac{1}{6}$

Explanation:

First, graph the two functions in order to identify the boundaries of the region. You will find that they are [0,1] .

Therefore, when you set up your integral, it will be from zero to one. Next you set up your integral by subtracting the lower function from the upper function.In this case it will be x-x^2 .

This means your integral should look like

$\int_0^1 x-x^2 dx$ .
Then solve using the power rule

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

$\\ \int_0^1 x-x^2 dx=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1\\ \\=\frac{(1)^2}{2}-\frac{(1)^3}{3}-\left[\frac{(0)^2}{2}-\frac{(0)^3}{3}\right]\\ \\=\frac{1}{2}-\frac{1}{3}\\ \\=\frac{1}{6}$

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Example Question #151 : Regions

y_1=x and y_2=-x^2

What is the area of the region created by the two functions?

Possible Answers:

$\frac{5}{6}$

$\frac{-1}{6}$

$\frac{-5}{6}$

$\frac{1}{6}$

Correct answer:

$\frac{1}{6}$

Explanation:

First, graph the two functions in order to identify the boundaries of the region. You will find that they are [-1,0] . Therefore, when you set up your integral, it will be from zero to one.

Next you set up your integral by subtracting the lower function from the upper function. In this case, it would be x^2-x .

This means your integral should look like

$\int_{-1}^0x^2-x dx$ .

Then solve using the power rule

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

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

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Example Question #74 : Area

y=sin(x)

What is the area of the region created by the function and the given bounds y-axis and $\pi$ ?

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

Set up your integral using the given bounds, then solve:

Remember the rules of trigonometric functions,

$\int sin(x)=-cos(x)$ .

Therefore our equation becomes,

$\\ \int_0^\pi sin(x) dx=-cos(x)]_0^\pi\\ \\=-cos(\pi)-(-cos(0))\\ \\=-(-1)-(-(1))\\ \\=1+1=2$ .

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Example Question #81 : How To Find Area Of A Region

y=sinx

What is the area of the region created by the function and the given bounds $-\pi$ and $\pi$ ?

Possible Answers:

$\pi$

Correct answer:

Explanation:

Set up your integral using the given bounds, then solve.

Remember the rules of trigonometric functions,

$\int sin(x)=-cos(x)$ .

Therefore our equation becomes,

$\\ \int_{-\pi}^\pi sin(x) dx=-cos(x)]_{-\pi}^\pi\\ \\=-cos(\pi)-(-cos(-\pi))\\ \\=-(-1)-(-(-1))\\ \\=1-(1)=0$ .

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Example Question #82 : How To Find Area Of A Region

$y=\sqrt x$

What is the area of the region created by the bounds y-axis and x=4 ?

Possible Answers:

$\frac{16}{3}$

$\frac{4\sqrt2}{3}$

$\frac{4}{3}$

$\frac{8}{3}$

Correct answer:

$\frac{16}{3}$

Explanation:

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

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

$\\ \int_{0}^4 \sqrt x dx=\int_0^4x^{\frac{1}{2}} dx \\ \\=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\left|_0^4=\frac{2x^{\frac{3}{2}}}{3}\left|_0^4\\ \\=\frac{2(4)^{\frac{3}{2}}}{3}=\frac{2\cdot8}{3}\\ \\=\frac{16}{3}$ .

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Example Question #83 : How To Find Area Of A Region

What is the area of the region created by the function y=-x^2+4 and the -axis?

Possible Answers:

$\frac{64}{3}$

$\frac{32}{3}$

$\frac{16}{3}$

$\frac{48}{3}$

Correct answer:

$\frac{32}{3}$

Explanation:

First, graph the two functions in order to identify the boundaries of the region. You will find that they are [-2,2] .

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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Example Question #84 : How To Find Area Of A Region

What is the area of the region created by y=x+2 and the bounds y-axis and x=3 ?

Possible Answers:

$\frac{-3}{2}$

$\frac{3}{2}$

$\frac{-17}{2}$

$\frac{21}{2}$

Correct answer:

$\frac{21}{2}$

Explanation:

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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Example Question #85 : How To Find Area Of A Region

Find the area created by y=2x+3 with the boudaries y-axis and x=6 .

Possible Answers:

Correct answer:

Explanation:

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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Example Question #86 : How To Find Area Of A Region

Find the area of the region encompassed by the curves f(x)=10-x^3 and $g(x)=\frac{1}{2}x^2$ , and the y-axis.

Possible Answers:

$\frac{43}{6}$

$\frac{44}{3}$

$\frac{34}{3}$

Correct answer:

$\frac{44}{3}$

Explanation:

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

f(x_u)=g(x_u)

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

This equation holds for the value of x_u=2

Therefore the lower and upper bounds are x_l=0,x_u=2 . 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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Example Question #87 : How To Find Area Of A Region

What is the area of $y=\sin(x)$ on the interval $[0,\pi/3]$ ?

Possible Answers:

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

$\pi/6$

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

$\frac{1}{2}$

$\pi/3$

Correct answer:

$\frac{1}{2}$

Explanation:

We slice the region into thin vertical strips of thickness $\Delta x$ and height $\sin(x_{k})$ and then sum up all strips, each of area $\sin(x_{k})\Delta x$ .

This gives us an approximate expression for the area:

$Area\approx \sum_{k=1}^{n}\sin(x_{k})\Delta x$ where $x_{k}=0+k\Delta x$ and $\Delta x= \frac{\pi/3}{n}$ .

We take the limit as the number of slices approaches infinity over this interval and we get the definite integral:

$\\ Area=\lim_{n\to\infty}\sum_{k=1}^{n}\sin{x_{k}}\Delta x\\ \\=\int_{0}^{\pi/3}\sin(x)dx\\ \\=-\cos(\pi/3)-[-\cos(0)])\\ \\=-\frac{1}{2}+1=\frac{1}{2}$

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #4064 : Calculus

Example Question #151 : Regions

Example Question #74 : Area

Example Question #81 : How To Find Area Of A Region

Example Question #82 : How To Find Area Of A Region

Example Question #83 : How To Find Area Of A Region

Example Question #84 : How To Find Area Of A Region

Example Question #85 : How To Find Area Of A Region

Example Question #86 : How To Find Area Of A Region

Example Question #87 : How To Find Area Of A Region

All Calculus 1 Resources

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