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Calculus 1 : Regions

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Example Questions

Example Question #71 : Area

Find the indefinite integral

$\int10sin(s)+7cos(s)ds$

Possible Answers:

-20cos(s)+14sin(s)+C

-20cos(s)+14cos(s)+C

20cos(s)-14sin(s)+C

42cos(s)

-10cos(s)+7sin(s)+C

Correct answer:

-10cos(s)+7sin(s)+C

Explanation:

Know the integrals and derivatives of trigonometric functions from memory, they occur very often.

$\frac{\mathrm{d}}{\mathrm{d} x}sin(x)=cos(x)$ and $\frac{\mathrm{d} }{\mathrm{d} x}cos(x)=-sin(x)$

Now notice the pattern of 'going backwards' from taking derivatives to integrating:

$\int sin(x)dx=-cos(x)$ and $\int cos(x)dx=sin(x)$

We get,

$\int10sin(s)+7cos(s)ds=-10cos(s)+7sin(s)+C$

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Example Question #71 : Area

Find the indefinite integral

$\int (1+4t)^3dt$

Possible Answers:

$\frac{(1+42t)^4}{16}+C$

$\frac{(1+4t)^4}{16}+C$

$\frac{(1+4t)^2}{16}+C$

$\frac{(1+4t)^4}{4}+C$

$\frac{(1+16t)^4}{4}+Ct$

Correct answer:

$\frac{(1+4t)^4}{16}+C$

Explanation:

We have to use what is called direct substitution. We let some quantity be represented by a new variable, and take the derivative of that. Then we replace the quantity which was in the original equation, with this variable. Commonly, mathematicians use the letter for this substitution; nicknaming it a 'u-sub'.

Using direct substitution, Let u=1+4t

Recall the power rule for differentiation:

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=n\cdot x^{n-1}$

then take the derivative of both sides of the equation,

$du=\frac{\mathrm{d} }{\mathrm{d} t}(1+4t)=4dt$

$du=4dt\rightarrow \frac{du}{4}=dt$ .

You can replace (1+4t) with and now you can replace with $\frac{du}{4}$ .

Now recall the power rule for integration:

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

Finally, take the $\frac{1}{4}$ fraction outside the integral and replace the old expression with the new variable. And then integrate and when you are done, replace the new variable back with the old quantity, as if you never even did it.

$\frac{1}{4}\int u^3du=\frac{1}{4}\cdot \frac{u^4}{4}=\frac{(1+4t)^4}{16}+C$

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Example Question #4062 : Calculus

Find the indefinite integral

$\int z^3\sqrt{2+z^4}dz$

Possible Answers:

$\frac{(2+z^4)^{1/2}}{6}+C$

$\frac{(2+z^4)^{-1/2}}{6}+C$

$\frac{(42+z^4)^{3/2}}{6}+C$

$\frac{(2+z^4)^{-3/2}}{6}+C$

$\frac{(2+z^4)^{3/2}}{6}+C$

Correct answer:

$\frac{(2+z^4)^{3/2}}{6}+C$

Explanation:

Begin by making the substitution and taking the derivative, and recall the power rule for differentiation.

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=n\cdot x^{n-1}$

And so,

Let u=2+z^4

$du=4z^3dz\rightarrow \frac{du}{4}=z^3dz$ .

Now we replace 2+4z with and z^3dz with $\frac{du}{4}$

Recall the power rule for integration,

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

Finally, pull out the $\frac{1}{4}$ from the integrand and write the new integral and then solve. After you solve the integral, put the original quantity back in. That is replace with what it originally equalled.

$=\frac{1}{4}\int \sqrt{u}du=\frac{1}{4}\int u^{\frac{1}{2}}du=\frac{1}{6}u^{3/2}=\frac{1}{6}(2+z^4)^{3/2}+C$

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Example Question #3031 : Functions

y_1=x and $y_2=\sqrt{x}$

Find the area of the region created by the two functions.

Possible Answers:

$\frac{1}{2}$

$\frac{1}{6}$

$\frac{2}{3}$

$\frac{5}{6}$

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 would be $\sqrt{x}-x$ .

This means your integral should look like

$\int_0^1\sqrt{x}-x dx$ .

Then solve using the power rule

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

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

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Example Question #75 : How To Find Area Of A Region

y_1=cos(x) and y_2=sin(x)

Find the region (to the nearest thousandth) created by the functions and bound by the y-axis and $x=\frac{\pi}{4}$ .

Possible Answers:

$\sqrt2-1$

0.414

0.707

Correct answer:

0.414

Explanation:

First, set up your integral, using the given bounds, by subtracting the lower function from the upper function.

In this case, it would be cos(x)-sin(x) .

Therefore the integral would look like

$\int_0^\frac{\pi}{4}cos(x)-sin(x)dx$ .

Then solve using the rules for integrals of the trig functions:

$\\ \int_0^\frac{\pi}{4}cos(x)-sin(x)dx=[sin(x)-(-cos(x))]_0^\frac{\pi}{4}\\ =\left[sin\left(\frac{\pi}{4}\right)+cos\left(\frac{\pi}{4}\right)\right]-[sin(0)+cos(0)]\\=\left[\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\right]-[0+1]\\=\left[\frac{2\sqrt2}{2}\right]-[1]\\ =\sqrt2-1\approx 0.414$ .

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Example Question #4063 : Calculus

y_1=x and y_2=3x

Find the region created by the two functions bounded by y-axis and x=3 .

Possible Answers:

$\frac{18}{3}$

$\frac{27}{2}$

Correct answer:

Explanation:

First, set up your integral, using the given bounds, by subtracting the lower function from the upper function.

In this case, in the given bounds, the integral will be

$\int_0^33x-xdx$ .

Then solve using the power rule

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

$\\ \int_0^33x-xdx=\left[\frac{3x^2}{2}-\frac{x^2}{2}\right]_0^3\\ \\=\left[\frac{3(3)^2}{2}-\frac{(3)^2}{2}\right]-\left[\frac{3(0)^2}{2}-\frac{(0)^2}{2}\right]\\ \\=\left[\frac{3(9)}{2}-\frac{9}{2}\right]-[0]\\ \\=\frac{27}{2}-\frac{9}{2}\\ \\=\frac{18}{2}=9$

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Example Question #3031 : Functions

y_1=x and y_2=x^2-2

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

Possible Answers:

-4.5

4.5

Correct answer:

4.5

Explanation:

First, set up your integral using the bounds found either by graphing or setting the two functions up to equal one another and solving.

Then subtract the lower function from the upper function.

In this case, in the given bounds, the integral will be

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

Then solve using the power rule

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

$\\ \int_{-1}^2 x-(x^2-2)dx=\int_{-1}^2 x-x^2+2 dx\\ \\=\left[\frac{x^2}{2}-\frac{x^3}{3}+2x\right]_{-1}^2\\ \\=\frac{(2)^2}{2}-\frac{(2)^3}{3}+2(2)-\left[\frac{(-1)^2}{2}-\frac{(-1)^3}{3}+2(-1)\right]\\ \\=\frac{4}{2}-\frac{8}{3}+4-\frac{1}{2}-\frac{1}{3}+2\\ \\=8-3-\frac{1}{2}=4.5$

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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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Calculus 1 : Regions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #71 : Area

Example Question #71 : Area

Example Question #4062 : Calculus

Example Question #3031 : Functions

Example Question #75 : How To Find Area Of A Region

Example Question #4063 : Calculus

Example Question #3031 : Functions

Example Question #4064 : Calculus

Example Question #151 : Regions

Example Question #74 : Area

All Calculus 1 Resources

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