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Calculus 1 : Regions

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Example Questions

Example Question #161 : Regions

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 #162 : Regions

$y=\sqrt x$

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

Possible Answers:

$\frac{16}{3}$

$\frac{4}{3}$

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

$\frac{8}{3}$

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 #163 : Regions

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

Possible Answers:

$\frac{32}{3}$

$\frac{64}{3}$

$\frac{48}{3}$

$\frac{16}{3}$

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 #164 : Regions

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

Possible Answers:

$\frac{21}{2}$

$\frac{-3}{2}$

$\frac{-17}{2}$

$\frac{3}{2}$

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 #165 : Regions

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 #166 : Regions

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:

$\frac{34}{3}$

$\frac{44}{3}$

$\frac{43}{6}$

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 #167 : Regions

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

Possible Answers:

$\pi/6$

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

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

$\frac{1}{2}$

$\pi/3$

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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Example Question #168 : Regions

Find the area under the curve f(x)=20-3x^2 in the region bounded by the -axis, the lower bound x=1 and the upper bound x=2

Possible Answers:

Correct answer:

Explanation:

To find the area under the curve

f(x)=20-3x^2

Integrate it from the specified bounds:

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

A=(40-8)-(20-1)=32-19

A=13

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

Find the area enclosed by the lines $f(x)=\frac{1}{3}x$ , g(x)=2x-10 , and the x-axis.

Possible Answers:

102

Correct answer:

Explanation:

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:

g(5)=2(5)-10=0

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:

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

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

x=6

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=A_1+A_2

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

A=5

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Example Question #85 : Area

$f(x)=e^{2x}{}$

Find the area under the curve drawn by the function f(x) on the interval of x=0 to x=4 .

Possible Answers:

$=\frac{1}{2}e^{4}-e$

$=\frac{1}{2}e^{8}-1$

$=\frac{1}{2}e^{8}-\frac{1}{2}$

$=\frac{1}{2}e^{2}+\frac{1}{2}$

Correct answer:

$=\frac{1}{2}e^{8}-\frac{1}{2}$

Explanation:

In order to find the area under f(x) on the interval of x=0 to x=4 , you must evaluate the definite integral

$\int_{}0}^{4}f(x)dx{}$

$=\int_{}0}^{4}e^{2x}dx$

First, antidifferentiate the function.

$=-\frac{1}{2}e^{2x}\mid_{0}^{4}{}$

Then, substitute values for .

$=\frac{1}{2}e^{2*4}-\frac{1}{2}e^{2*0}$

Finally, evaluate in terms of

$=\frac{1}{2}e^{2*4}-\frac{1}{2}*(1)$

$=\frac{1}{2}e^{8}-\frac{1}{2}$

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Calculus 1 : Regions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #161 : Regions

Example Question #162 : Regions

Example Question #163 : Regions

Example Question #164 : Regions

Example Question #165 : Regions

Example Question #166 : Regions

Example Question #167 : Regions

Example Question #168 : Regions

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

Example Question #85 : Area

All Calculus 1 Resources

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