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Calculus 1 : Functions

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Example Questions

Example Question #58 : Area

Find the area of the region bound by f(x) , the -axis, and the lines x=5,x=6 .

$f(x)=\frac{16}{x}$

Possible Answers:

2.917

0.182

18.23

29.17

Correct answer:

2.917

Explanation:

Find the area of the region bound by h(t), the x-axis, and the lines x=5,x=6

$f(x)=\frac{16}{x}$

To find area, set up an integral:

$\int_{5}^{6}\frac{16}{x}dx$

Recall that:

$\int\frac{1}{x}dx=ln(x)$

So we get:

$\int_{5}^{6}\frac{16}{x}dx=16ln(x)^6_5$

16ln(6)-16ln(5)=2.917

So our answer is:

2.917

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Example Question #3021 : Functions

A company produces printers that cost $\$100$ to make and sells them for $\$200$ . If the number of printers that a company sells in a year can be modeled by the equation y=-13x^2+770718x+450432 , how many printers should the company make every year in order to maximize profits?

Possible Answers:

26,753

36,326

None of the above.

31,623

29,643

Correct answer:

29,643

Explanation:

To solve this problem we must be able to find the point on the graph of the equation that is the maximum.

In order to achieve this we must take the derivative of the equation, set it equal to zero and solve for .

Taking the derivative of the equation using the power rule

$\frac{d}{dx}x^n=nx^{n-1}}$ ,

we find that the equation becomes

y'=-26x+770718 .

Setting the equation equal to 0 and solving for , we find that in order to maximize profits, the company should make 29,643 printers a year.

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Example Question #52 : How To Find Area Of A Region

Integrate

y=2xcos(4x)

Possible Answers:

$\frac{xsin(4x)}{2}+\frac{cos(4x)}{8}+C$

$\frac{xsin(4x)}{2}-\frac{cos(4x)}{8}$

$\frac{sin(4x)}{2}(x-1) +C$

$\frac{xsin(4x)}{2}+\frac{cos(4x)}{8}$

$\frac{xsin(4x)}{2}-\frac{cos(4x)}{8}+C$

Correct answer:

$\frac{xsin(4x)}{2}+\frac{cos(4x)}{8}+C$

Explanation:

This can be integrated through integration by parts.

$u=2x\rightarrow u'=2dx$

$v'=cos(4x)\rightarrow v=\frac{sin(4x)}{4}$

$\\\int 2xcos(4x)dx\\ \\=uv-\int u'v\\ \\=\frac{xsin(4x)}{2}-\int \frac{sin(4x)dx}{2}$

$=\frac{xsin(4x)}{2}+\frac{cos(4x)}{8}+C$

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Example Question #61 : How To Find Area Of A Region

Integrate 2xcos(x^2)dx

Possible Answers:

sin(2x)+C

sin(x^2)+C

-sin(x^2)

sin(x^2)

-sin(x^2)+C

Correct answer:

sin(x^2)+C

Explanation:

Let $u=x^2\rightarrow du=2xdx$

Therefore by u substitution we are able to integrate thus finding the area under the curve. Once you have integrated remember to plug your original variables back into the equation:

$\\ \int 2xcos(x^2)dx\\ \\=\int cos(u)du\\ \\=sin(u)+C\\ \\=sin(x^2)+C$

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Example Question #4054 : Calculus

Find the area between the curves $y=\sqrt{x}$ and y=x^3 .

Possible Answers:

$\frac{-5}{12}$

$\frac{5}{12}$

Correct answer:

$\frac{5}{12}$

Explanation:

Graph1

Looking at the graph, we see that the graphs intersect at the point (1,1) with $y=\sqrt{x}$ on top.

Therefore, our integral to find the area by partitioning the x-axis looks like

$\int_{0}^{1} [ \sqrt{x}-x^3]dx = \frac{5}{12}$ .

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Example Question #3021 : Functions

Find the area between x=-y^2+1 and the -axis.

Possible Answers:

$\frac{4}{3}$

$\frac{1}{3}$

$\frac{2}{3}$

Correct answer:

$\frac{4}{3}$

Explanation:

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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Example Question #4056 : Calculus

Calculate the area between the sine curve and the cosine curve for

$0 \leq x \leq \frac{\pi}{2}$ .

Possible Answers:

$\frac{-4}{\sqrt2}$

$2\sqrt2-2$

$2-4\sqrt2$

Correct answer:

$2\sqrt2-2$

Explanation:

To find the area between the 2 trigonometric waves, we have to look at two different regions between 0 and $\frac{\pi}{2}$ .

Cosine is above sine from 0 to $\frac{\pi}{4}$ , and sine is above cosine from $\frac{\pi}{4}$ to $\frac{\pi}{2}$ :

Calc area 1

Written as an integral, this area can be represented as follows:

$\int_{0}^{\frac{\pi}{4}}cosx - sinx dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}sinx - cosx dx$

Integrating yields:

sinx - (-cosx) from 0 to $\frac{\pi}{4}$ and -cosx - sinx from $\frac{\pi}{4}$ to $\frac{\pi}{2}$

Evaluate:

$sin\left(\frac{\pi}{4}\right)+cos\left(\frac{\pi}{4}\right)-[sin(0)+cos(0)] + \left[-cos\left(\frac{\pi}{2}\right)-sin\left(\frac{\pi}{2}\right)\right]-\left[-cos\left(\frac{\pi}{4}\right)-sin\left(\frac{\pi}{4}\right)\right]$

$\left[\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\right]-\left[(0+1) \right ] + [0-1]\left(-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}\right)$

$\frac{2}{\sqrt2}-1-1+\frac{2}{\sqrt2}$

$\frac{4}{\sqrt2}-2$ rationalizing the denominator:

$\frac{4\sqrt2}{2}-2=2\sqrt2 -2$

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Example Question #4057 : Calculus

Find the area between the functions f(x)=3x^2 + 1 and g(x)=5-x^2 .

Possible Answers:

$\frac{16}{3}$

$\frac{20}{3}$

Correct answer:

$\frac{16}{3}$

Explanation:

The area between the two curves goes between the two points of intersection:

Calc area 2

To figure out the left and right bounds of this region, solve the system of equations y=f(x), y=g(x) by setting the two equations equal to each other and solving for x:

-x^2 +5 = 3x^2 + 1 add x^2 to both sides

5=4x^2 + 1 subtract 1 from both sides

4=4x^2 divide by 4

1=x^2

The solutions are -1 and 1, so we're finding the area for $-1\leq x \leq1$ .

For this region, the function -x^2 + 5 is above 3x^2 + 1 , so our integral looks like this:

$\int_{-1}^{1} -x^2+5- (3x^2 + 1)dx$

We can simplify before solving to $\int_{-1}^{1}-4x^2 +4 dx$ and continue solving.

$-\frac{4}{3}x^3 +4x$ from -1 to 1

$-\frac{4}{3}(1)^3 + 4(1) - \left[-\frac{4}{3}(-1)^3 + 4(-1)\right]$

$-\frac{4}{3}+4 -\left[\frac{4}{3}-4\right]=-\frac{4}{3}-\frac{4}{3}+4+4 = \frac{16}{3}$

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Example Question #4058 : Calculus

Find the area between f(x)=5-x and $g(x)=\frac{6}{x}$ .

Possible Answers:

5.0672

1.6589

12.2506

0.9872

0.0672

Correct answer:

0.0672

Explanation:

First, figure out the boundaries of the region between the two curves. To do this, set the two functions equal to each other to determine where the curves intersect:

$5-x = \frac{6}{x}$ multiply both sides by x

5x-x^2 = 6

This can be re-written in more standard quadratic form:

x^2 -5x +6 = 0

This can be factored as (x-2)(x-3)=0 , so our two solutions are 2 and 3. We are then evaluating the definite integral between 2 and 3. In this region, the equation 5-x is greater. Evaluate:

$\int_{2}^{3}5-x-\frac{6}{x}dx = 5x - \frac{1}{2}x^2 -6lnx$ between 2 and 3

$5(3)-\frac{1}{2}(3)^2 - 6ln(3) - \left[5(2)-\frac{1}{2}(2)^2-6ln(2)\right]$

$15 - 4.5 - 6.5917 - [10-2-4.1589]\approx 0.0672$

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Example Question #4059 : Calculus

Find the dot product $\left<5,6,-1\right>$ and $\left<-2,3,9\right>$ .

Possible Answers:

None of these

Correct answer:

Explanation:

The way to find a dot product is to multiply the first number of one vector with the first number in the other. Then you do the same with the second and third numbers in each vector. Then you add up the three values and the answer is the dot product.

(5)(-2)+(6)(3)+(-1)(9)

=-10+18-9

=-1

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #58 : Area

Example Question #3021 : Functions

Example Question #52 : How To Find Area Of A Region

Example Question #61 : How To Find Area Of A Region

Example Question #4054 : Calculus

Example Question #3021 : Functions

Example Question #4056 : Calculus

Example Question #4057 : Calculus

Example Question #4058 : Calculus

Example Question #4059 : Calculus

All Calculus 1 Resources

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