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Calculus 1 : Functions

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Example Question #198 : Regions

To find area under a curve, you must:

Possible Answers:

integrate the function twice.

differentiate the function twice.

integrate the function once.

differentiate the function once.

Correct answer:

integrate the function once.

Explanation:

To find the area of a curve, you are not finding the slope (thus not differentiating) but rather integrating. Specifically, you only differentiate once to find 2-dimensional area. Thus, the answer is "integrate the function once."

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Example Question #199 : Regions

Find the area under the curve $y=45x^{2}+10x+14$ between x=0 and x=1 .

Possible Answers:

449

435

Correct answer:

Explanation:

In order to find the area underneath a curve for a specific range of values, we need to set up a definite integral expression. To do this, we plug our specific information into the following skeleton:

$\int_{a}^{b}f(x) dx$

We plug in the equation of the curve in for f(x) , the smaller of our x-values for , and the larger of our x-values for . For this problem, our set-up looks like this:

$\int_{0}^{1}(45x^{2}+10x+14)dx$

Next, we integrate our expression, ignoring our and values for now. We get:

$F(x)= 15x^{3}+5x^{2}+14x$

Now, to find the area under the part of the curve we're looking for, we plug our and values into our integrated expression and find the difference, using the following skeleton:

area = F(b)-F(a)

In this problem, this looks like:

area = F(1)-F(0)

Which plugged into our integrated expression is:

$area = [15(1)^{3}+5(1)^{2}+14(1)] - [15(0)^{3}+5(0)^{2}+14(0)]$

area = [15+5+14] - [0+0+0]

area = [34] - [0]

With our final answer being:

area = 34

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Example Question #200 : Regions

Find the area under the curve $y=42x^{6}+\frac{4}{7}x^{3}+76x$ between x=0 and x=3 .

Possible Answers:

$\frac{94329}{2}$

$-\frac{212880}{7}$

$\frac{94329}{7}$

$\frac{212880}{7}$

$-\frac{94329}{7}$

Correct answer:

$\frac{94329}{7}$

Explanation:

$\int_{a}^{b}f(x) dx$

We plug in the equation of the curve in for f(x) , the smaller of our x-values for , and the larger of our x-values for . For this problem, our set-up looks like this:

$\int_{0}^{3}(42x^{6}+\frac{4}{7}x^{3}+76x)dx$

Next, we integrate our expression, ignoring our and values for now. We get:

$F(x)= 6x^{7}+\frac{1}{7}x^{4}+38x^{2}$

Now, to find the area under the part of the curve we're looking for, we plug our and values into our integrated expression and find the difference, using the following skeleton:

area = F(b)-F(a)

In this problem, this looks like:

area = F(3)-F(0)

Which plugged into our integrated expression is:

$area = [6(3)^{7}+\frac{1}{7}(3)^{4}+38(3)^{2}] - [6(0)^{7}+\frac{1}{7}(0)^{4}+38(0)^{2}]$

$area = [6(2187)+\frac{1}{7}(81)+38(9)] - [6(0)+\frac{1}{7}(0)+38(0)]$

$area = [13122+\frac{81}{7}+342)] - [0+0+0]$

$area = [\frac{91854}{7}+\frac{81}{7}+\frac{2394}{7})] - [0]$

$area = \frac{94329}{7}$

With our final answer being:

$area = \frac{94329}{7}$

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Example Question #121 : How To Find Area Of A Region

Find the area under the curve $y=12x^{2}+6x+12$ between x=1 and x=3 .

Possible Answers:

152

108

190

Correct answer:

152

Explanation:

$\int_{a}^{b}f(x) dx$

We plug in the equation of the curve in for f(x) , the smaller of our x-values for , and the larger of our x-values for . For this problem, our set-up looks like this:

$\int_{1}^{3}(12x^{2}+6x+12)dx$

Next, we integrate our expression, ignoring our and values for now. We get:

$F(x)= 4x^{3}+3x^{2}+12x$

Now, to find the area under the part of the curve we're looking for, we plug our and values into our integrated expression and find the difference, using the following skeleton:

area = F(b)-F(a)

In this problem, this looks like:

area = F(3)-F(1)

Which plugged into our integrated expression is:

$area = [4(3)^{3}+3(3)^{2}+12(3)] - [4(1)^{3}+3(1)^{2}+12(1)]$

area = [4(27)+3(9)+12(3)] - [4(1)+3(1)+12(1)]

area = [108+27+36] - [4+3+12]

area = [171] - [19]

With our final answer being:

area = 152

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Example Question #122 : How To Find Area Of A Region

Find the area under the curve $y=12x^{3}+12x^{2}+ 12x+12$ between x=-1 and x=2 .

Possible Answers:

180

135

121

Correct answer:

135

Explanation:

$\int_{a}^{b}f(x) dx$

We plug in the equation of the curve in for f(x) , the smaller of our x-values for , and the larger of our x-values for . For this problem, our set-up looks like this:

$\int_{-1}^{2}(12x^{3}+12x^{2}+ 12x+12 )dx$

Next, we integrate our expression, ignoring our and values for now. We get:

$F(x)= 3x^{4}+4x^{3}+6x^{2}+12x$

Now, to find the area under the part of the curve we're looking for, we plug our and values into our integrated expression and find the difference, using the following skeleton:

area = F(b)-F(a)

In this problem, this looks like:

area = F(2)-F(-1)

Which plugged into our integrated expression is:

$area = [3(2)^{4}+4(2)^{3}+6(2)^{2}+12(2)] - [3(-1)^{4}+4(-1)^{3}+6(-1)^{2}+12(-1)]$

area = [3(16)+4(8)+6(4)+12(2)] - [3(1)+4(-1)+6(1)+12(-1)]

area = [48+32+24+24] - [3-4+6-12]

area = [128] - [-7]

With our final answer being:

area = 135

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Example Question #123 : How To Find Area Of A Region

Find the area under the curve $y=9x^{2}+14x+5$ between x=-2 and x=2 .

Possible Answers:

Correct answer:

Explanation:

$\int_{a}^{b}f(x) dx$

We plug in the equation of the curve in for f(x) , the smaller of our x-values for , and the larger of our x-values for . For this problem, our set-up looks like this:

$\int_{-2}^{2}(9x^{2}+14x+5)dx$

Next, we integrate our expression, ignoring our and values for now. We get:

$F(x)= 3x^{3}+7x^{2}+5x$

Now, to find the area under the part of the curve we're looking for, we plug our and values into our integrated expression and find the difference, using the following skeleton:

area = F(b)-F(a)

In this problem, this looks like:

area = F(2)-F(-2)

Which plugged into our integrated expression is:

$area = [3(2)^{3}+7(2)^{2}+5(2)] - [3(-2)^{3}+7(-2)^{2}+5(-2)]$

area = [3(8)+7(4)+5(2)] - [3(-8)+7(4)+5(-2)]

area = [24+28+10] - [-24+28-10]

area = [62] - [-6]

With our final answer being:

area = 68

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Example Question #124 : How To Find Area Of A Region

Find the area under the curve y=7x+3 between x=3 and x=10 .

Possible Answers:

$\frac{1931}{2}$

$\frac{1589}{2}$

$-\frac{1589}{2}$

Correct answer:

$\frac{1589}{2}$

Explanation:

$\int_{a}^{b}f(x) dx$

We plug in the equation of the curve in for f(x) , the smaller of our x-values for , and the larger of our x-values for . For this problem, our set-up looks like this:

$\int_{3}^{10}(7x+3)dx$

Next, we integrate our expression, ignoring our and values for now. We get:

$F(x)= \frac{17}{2}x^{2}+3x$

Now, to find the area under the part of the curve we're looking for, we plug our and values into our integrated expression and find the difference, using the following skeleton:

area = F(b)-F(a)

In this problem, this looks like:

area = F(10)-F(3)

Which plugged into our integrated expression is:

$area = [\frac{17}{2}(10)^{2}+3(10)] - [\frac{17}{2}(3)^{2}+3(3)]$

$area = [\frac{17}{2}(100)+3(10)] - [\frac{17}{2}(9)+3(3)]$

$area = [\frac{1700}{2}+30] - [\frac{153}{2}+9]$

$area = [\frac{1700}{2}+\frac{60}{2}] - [\frac{153}{2}+\frac{18}{2}]$

$area = [\frac{1760}{2}] - [\frac{171}{2}]$

With our final answer being:

$area = \frac{1589}{2}$

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Example Question #125 : How To Find Area Of A Region

Find the area under the curve $y=100x^{3}+6x$ between x=1 and x=2 .

Possible Answers:

440

541

918

384

706

Correct answer:

384

Explanation:

$\int_{a}^{b}f(x) dx$

We plug in the equation of the curve in for f(x) , the smaller of our x-values for , and the larger of our x-values for . For this problem, our set-up looks like this:

$\int_{1}^{2}(100x^{3}+6x)dx$

Next, we integrate our expression, ignoring our and values for now. We get:

$F(x)= 25x^{4}+3x^{2}$

Now, to find the area under the part of the curve we're looking for, we plug our and values into our integrated expression and find the difference, using the following skeleton:

area = F(b)-F(a)

In this problem, this looks like:

area = F(1)-F(2)

Which plugged into our integrated expression is:

$area = [25(2)^{4}+3(2)^{2}] - [25(1)^{4}+3(1)^{2})]$

area = [25(16)+3(4)] - [25(1)+3(1)]

area = [400+12] - [25+3]

area = [412] - [28]

With our final answer being:

area = 384

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Example Question #121 : Area

Find the area of the region bounded on top by 2x+2 and on bottom by x^2 from x=0 to x=2

Possible Answers:

$\frac{16}{3}$

$-\frac{8}{3}$

Correct answer:

$\frac{16}{3}$

Explanation:

To find the area bounded by the two curves, we must integrate:

$\int_{a}^{b}[f(x)-g(x)]dx$ , where f(x) is the top function and g(x) is the bottom function.

Using the above formula, we get

$\int_{0}^{2}(2x+2-x^2)dx=2^2+2(2)-\frac{2^3}{3}-(0)=\frac{16}{3}$

The integration was perfomed using the following rule:

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

The definite integration was performed by taking the result of the integral and evaluating it at the upper limit, and then subtracting the evaluation at the lower limit.

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Example Question #121 : How To Find Area Of A Region

Write the function describing the area below $y=\sqrt{x}$ and above y=(x-5)^2-20

Possible Answers:

$\frac{(x-5)^3}{3}-20x-\frac{2x^{\frac{3}{2}}}{3}+C$

$-\frac{(x-5)^3}{3}+20x+\frac{2x^{\frac{3}{2}}}{3}+C$

$\frac{(x-5)^3}{3}-20x-x^{\frac{3}{2}}+C$

$(x-5)^3-20x-\frac{2x^{\frac{3}{2}}}{3}+C$

Correct answer:

$\frac{(x-5)^3}{3}-20x-\frac{2x^{\frac{3}{2}}}{3}+C$

Explanation:

To determine the function that describes the area between two functions, we must integrate over the difference between the upper and lower functions:

$\int ((x-5)^2-20-x^{\frac{1}{2}})dx$

For ease, we can split the integral into three integrals:

$\int (x-5)^2dx -\int 20 dx-\int x^{\frac{1}{2}}dx$

For the first integral, we must make the following substitution:

u=x-5, du=dx

Rewriting the integral and integrating, we get

$\int u^2=\frac{e^3}{3}+C$

which was found using the following rule:

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

The second integral is equal to

$\int 20 dx=20x+C$

and was found using the following rule:

$\int dx=x +C$

The third integral, found using the same rule as the first integral, is equal to

$\int x^{\frac{1}2{}} dx=\frac{2x^{\frac{3}{2}}}{3}+C$

Combining the results - and adding all the constants of integration to make a single C - we get

$\frac{(x-5)^3}{3}-20x-\frac{2x^{\frac{3}{2}}}{3}+C$ .

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #198 : Regions

Example Question #199 : Regions

Example Question #200 : Regions

Example Question #121 : How To Find Area Of A Region

Example Question #122 : How To Find Area Of A Region

Example Question #123 : How To Find Area Of A Region

Example Question #124 : How To Find Area Of A Region

Example Question #125 : How To Find Area Of A Region

Example Question #121 : Area

Example Question #121 : How To Find Area Of A Region

All Calculus 1 Resources

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