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Calculus 2 : Integral Applications

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Example Questions

Example Question #33 : Area Under A Curve

Find the area under the curve for f(x)=|-3x+2x^2+x^3| from x=-1 to x=1 , rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{-1}^{1}|-3x+2x^2+x^3|dx$

Solution:

This function is negative from x=[0,1] , and positve everywhere else. Split this integral up into 2 pieces, multiplying x=[0,1] region by , and sum everything up.

First piece:

$\int_{-1}^{0}|-3x+2x^2+x^3|dx$

$-\frac{3}{2}x^2+\frac{2}{3}x^3+\frac{1}{4}x^4\Big|_{-1}^{0}$

$0-(-\frac{3}{2}-\frac{2}{3}+\frac{1}{4}) = \frac{3}{2}+\frac{2}{3}-\frac{1}{4} = \frac{23}{12}$

Second piece:

$\int_{0}^{1}|-3x+2x^2+x^3|dx$

$\frac{3}{2}x^2-\frac{2}{3}x^3-\frac{1}{4}x^4\Big|_{0}^{1}$

$\frac{3}{2}-\frac{2}{3}-\frac{1}{4} = \frac{7}{12}$

Sum:

$\frac{23}{12}+\frac{7}{12} = \frac{30}{12} = \frac{5}{2}=2.5$

When rounded to the nearest integer, the area under the curve is

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Example Question #261 : Integrals

Find the area under the curve for f(x)=sin(x)+1 from $x=\frac{\pi}{4}$ to $x=\frac{\pi}{2}$

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}sin(x)+1dx$

Solution:

$\int sin(x)+1dx$

$x-cos(x)\Big|_{\frac{\pi}{4}}^{\frac{\pi}{2}}$

$({\frac{\pi}{2}}-cos({\frac{\pi}{2}}))-({\frac{\pi}{4}} - cos({\frac{\pi}{4}}))$

$\frac{\pi}{4} + \frac{1}{\sqrt2} = 1.4925$

Rounded to the nearest integer, the area under the curve is

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Example Question #91 : Integral Applications

Find the area under the curve for f(x)=2sin(x)+x from x=0 to $x=\pi$

Possible Answers:

$\frac{1}{2}\pi^2 + 6$

$\pi^2 + 2$

$\frac{1}{2}\pi + 2$

$\frac{1}{2}\pi^2 + 4$

Correct answer:

$\frac{1}{2}\pi^2 + 4$

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{0}^{\pi}2sin(x)+xdx$

Solution:

$\int2sin(x)+xdx$

$\frac{1}{2}x^2-2cos(x)\Big|_{0}^{\pi}$

$\frac{1}{2}\pi^2-2\cdot cos(\pi)-(-2cos(0)) = \frac{1}{2}\pi^2-2\cdot cos(\pi) + 2cos(0)$

$\frac{1}{2}\pi^2-2\cdot-1 + 2 = \frac{1}{2}\pi^2 + 4$

The area under the curve is $\frac{1}{2}\pi^2 + 4$

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Example Question #36 : Area Under A Curve

Find the area under the curve for f(x)=|3x^2-4x| from x=-2 to x=2

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{-2}^{2}|3x^2-4x|dx$

Solution:

This function is negative from $x=[0,\frac{4}{3}],$ and positve everywhere else. Split this integral up into 3 pieces, multiplying x=[0,\frac{4}{3}] region by -1, and sum everything up.

1st piece:

$\int_{-2}^{0}3x^2-4xdx$

$3\cdot\frac{1}{3}x^3-\frac{4}{2}x^2\Big|_{-2}^{0}$

$x^3-2x^2\Big|_{-2}^{0}$

$0-((-2)^3-2\cdot4) = 8+8 = 16$

2nd piece:

$\int_{0}^{\frac{4}{3}}|3x^2-4x|dx$

$2x^2 - x^3\Big|_{0}^{\frac{4}{3}}$

$2\cdot(\frac{4}{3})^2 - (\frac{4}{3})^3 = \frac{32}{9}-\frac{64}{27} = \frac{32}{27}$

3rd piece:

$\int_{\frac{4}{3}}^{2}|3x^2-4x|dx$

$x^3 - 2x^2\Big|_{\frac{4}{3}}^{2}$

$(8-2\cdot4)-((\frac{4}{3})^3-2\cdot(\frac{4}{3})^2) = \frac{32}{9}-\frac{64}{27} = \frac{32}{27}$

Sum:

$16+\frac{32}{27} + \frac{32}{27} = \frac{496}{27} = 18.37$

When rounded to the nearest integer, the area under the curve is

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Example Question #37 : Area Under A Curve

Find the area under the curve for f(x)=(x+6)^2-36 from x=1 to x=4

Possible Answers:

111

123

107

127

Correct answer:

111

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{1}^{4} (x+6)^2-36dx$

Solution:

First, simplify the function and then evaluate the integral.

1. Simplify the function

(x+6)^2-36

x^2+36+12x-36

x^2+12x

2. Evaluate the integral

$\int_{1}^{4} x^2+12xdx$

$\frac{1}{3}x^3+6x^2\Big|_{1}^{4}$

$\frac{64}{3}+6\cdot16-(\frac{1}{3}+6) = 111$

The area under the curve is 111

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Example Question #38 : Area Under A Curve

Find the area under the curve for f(x)=(x-4)^2+4 from x=-1 to x=4 , when rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{-1}^{4} (x-4)^2+4dx$

Solution:

First, simplify the function and then evaluate the integral.

1. Simplify

(x-4)^2+4

x^2+16-8x+4

x^2-8x+20

2. Evaluate the integral

$\int_{-1}^{4} x^2-8x+20dx$

$\frac{1}{3}x^3-4x^2+20x\Big|_{-1}^{4}$

$\frac{1}{3}\cdot64-4\cdot16+80-(-\frac{1}{3}-4-20)$

$\frac{64}{3}-64+80+\frac{1}{3}+24$

$\frac{65}{3}+40 = \frac{185}{3} = 61.667$

When rounded tot he nearest integer, the area under the curve is

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Example Question #271 : Integrals

Find the area under the curve of y=tan(x) from to $2 \pi$

Possible Answers:

$\frac{1}{2}$

$\infty$

Correct answer:

Explanation:

We can represent the area as:

$\int_{0}^{2\pi}tan(x) dx= \int_{0}^{2\pi}\frac{sin(x)}{cos(x)} dx$

u=cos(x) , $\frac{du}{dx}=-sin(x)$

$\int \frac{sin(x)}{cos(x)} dx= \int -\frac{1}{u}du=-ln(u)+C$

F(x)= -ln(cos(x)) +C

By the fundamental theorem of calculus:

$\int_{0}^{2\pi} \frac{sin(x)}{cos(x)} dx= F(2\pi)-F(0)= -ln(cos(2\pi))+ln(cos(0))$

=0+0=0

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Example Question #31 : Area Under A Curve

Determine:

$\int_{0}^{5} \int_{0}^{5}1 dx dy$ .

Hint: Do the inside integral first and then the outside integral second.

Possible Answers:

Correct answer:

Explanation:

Looking at the inside integral:

$\int_{0}^{5}1 dx= 5-0=5$

Having done the inside integral, we can do the outside integral

$\int_{0}^{5}5 dy= 5*5-5*0=25$

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Example Question #2022 : Calculus Ii

Find the area between the two curves given by the functions f(x)=x^3 and $g(x)=\sqrt[3]{x}$

Possible Answers:

$\frac{1}{3}$

$\frac{11}{3}$

$\frac{1}{2}$

$\frac{13}{12}$

None of the other answers

Correct answer:

$\frac{1}{2}$

Explanation:

The area between two curves y=f_1(x) and y=f_2(x) is given by the formula $A=\int^b_a(f_1(x)-f_2(x))dx$ , where f_1(x) is the upper bound curve, f_2(x) is the lower bound curve, and and are the solutions to the equation f_1(x)=f_2(x) . The graphs of f(x)=x^3 and $g(x)=\sqrt[3]{x}$ are given in the figure below:

Cube graphs

As we can see, the graph of g(x) is the upper bound for the area and the graph of f(x) is the lower bound for the area. It is also apparent that the limits of integration are given by a=0 and b=1 . To see this algebraically, for graphs often do not give us clear answers for limits of integration, we would solve the equation f(x)=g(x) . Plugging in f(x) and g(x) , we obtain:

$x^3=\sqrt[3]{x}$

$x^3=x^{1/3}$

$x^3-x^{1/3}=0$

$x^{1/3}(x^{2/3}-1)=0$

Setting each factor equal to gives us $x^{1/3}=0\rightarrow x=0$ , and $x^{2/3}-1=0\rightarrow x^{2/3}=1\rightarrow x=1$ . With these limits of integration, our integral for the area becomes:

$A=\int_0^1(x^{1/3}-x^3)dx=\frac{3}{4}x^{4/3}-\frac{x^4}{4}\biggr|_0^1$

$=\biggr(\frac{3}{4}(1^{4/3})-\frac{1}{4}\biggr(\frac{1^4}{4}\biggr)\biggr)-\biggr(\frac{3}{4}(0^{4/3})-\frac{1}{4}\biggr(\frac{0^4}{4}\biggr)\biggr)$

$=\biggr(\frac{3}{4}-\frac{1}{4}\biggr)-0=\frac{1}{2}$

Therefore the area between the curves must be $\frac{1}{2}$ .

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Example Question #2022 : Calculus Ii

Find the area in between the parabola 9-x^2 and the x-axis from x=0 to x=3 .

Possible Answers:

Correct answer:

Explanation:

To calculate the area between two functions, take the integral of the function on top minus the function on bottom. The intersection of the functions will be the bound on the integral. In this particular case the top function is the parabola described as 9-x^2 and the bottom function is the x-axis, or in other words zero.

Therefore the basic integral looks as follows.

$\int (9-x^2)-0dx$

The question indicates the upper and lower bound as x=3 and x=0 , applying these bounds to the integral results in,

$\int_{0}^{3}(9-x^2)dx$ .

To calculate the integral recall the following rule of integration.

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

where C represents a constant.

Applying this rule to the integral in question results in,

$(9x-\frac{x^3}{3})|_{0}^{3}$

Substituting in the bounds results in the solution for area.

$9(3)-\frac{3^3}{3}=18$

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Calculus 2 : Integral Applications

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #33 : Area Under A Curve

Example Question #261 : Integrals

Example Question #91 : Integral Applications

Example Question #36 : Area Under A Curve

Example Question #37 : Area Under A Curve

Example Question #38 : Area Under A Curve

Example Question #271 : Integrals

Example Question #31 : Area Under A Curve

Example Question #2022 : Calculus Ii

Example Question #2022 : Calculus Ii

All Calculus 2 Resources

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