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Calculus 2 : Integral Applications

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Example Questions

Example Question #41 : Area Under A Curve

Find the area bounded by the functions,

g(x)=x+6

f(x)=2(x+3)^2-1

Set up the integral and simplify the integrand.

Plot for area between curves problem

Possible Answers:

$A=\int_{-\frac{11}{4}-\frac{\sqrt{33}}{4}}^{\frac{11}{4}+\frac{\sqrt{33}}{4}}(2x^2+11x+11)dx$

$A=\int_{\frac{11}{4}-\frac{\sqrt{33}}{4}}^{\frac{11}{4}+\frac{\sqrt{33}}{4}}(2x^2+11x+11)dx$

$A=\int_{-\frac{11}{4}-\frac{\sqrt{33}}{4}}^{-\frac{11}{4}+\frac{\sqrt{33}}{4}}(-2x^2-11x-11)dx$

$A=\int_{\frac{7}{4}-\frac{\sqrt{209}}{4}}^{\frac{7}{4}+\frac{\sqrt{209}}{4}}(2x^2+11x+11)dx$

$A=\int_{-\frac{7}{2}-\frac{\sqrt{209}}{2}}^{-\frac{7}{2}+\frac{\sqrt{209}}{2}}(x^2+7x+7)dx$

Correct answer:

$A=\int_{-\frac{11}{4}-\frac{\sqrt{33}}{4}}^{-\frac{11}{4}+\frac{\sqrt{33}}{4}}(-2x^2-11x-11)dx$

Explanation:

Find the area bounded by the functions:

g(x)=x+6

f(x)=2(x+3)^2-1

Looking at the plot of the function we can see that $g(x)\geq f(x)$ for all in the region. The area formula is therefore,

$A=\int_{a}^{b}|g(x)-f(x)|dx$

, for g(a)=f(a) and g(b)=f(b)

To find the intersection points, set the functions equal to each other and solve for

g(x)=f(x)

x+6=2(x+3)^2-1

x+6=2(x^2+6x+9)-1

x+6=2x^2+12x+17

2x^2+11x+11=0

$x =\frac{-11\pm\sqrt{11^2-4(11)(2))}}{2(2)}$

$=-\frac{11}{4}\pm \frac{\sqrt{33}}{4}$

Therefore, the two lines intersect for the following values of :

$x = \left \{ -\frac{11}{4}-\frac{\sqrt{33}}{4},- \frac{11}{4}+\frac{\sqrt{33}}{4}\right \}$ .

So we will integrate over this interval,

$A=\int_{-\frac{11}{4}-\frac{\sqrt{33}}{4}}^{-\frac{11}{4}+\frac{\sqrt{33}}{4}}|(x+6)-[2(x+3)^2-1])dx$

$=\int_{-\frac{11}{4}-\frac{\sqrt{33}}{4}}^{-\frac{11}{4}+\frac{\sqrt{33}}{4}}[(x+6)-(2x^2+12x+17)]dx$

Finally we simplify the integrand to arrive at:

$=\int_{-\frac{11}{4}-\frac{\sqrt{33}}{4}}^{-\frac{11}{4}+\frac{\sqrt{33}}{4}}(-2x^2-11x-11)dx$

If you were to evaluate the integration is simple, but evaluating the limits would be quite tedious. The outcome is:

$=\left [ -\frac{2}{3}x^3-\frac{11}{2}x^2-11x\right ]_{\frac{11}{4}-\frac{\sqrt{33}}{4}}^{\frac{11}{4}+\frac{\sqrt{33}}{4}}$

$=\frac{11\sqrt{33}}{8}$

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Example Question #41 : Area Under A Curve

What is the area under the curve f(x)=x^2-1 bounded by the x-axis from x=2 to x=4?

Possible Answers:

$\frac{50}{3}$

$\frac{51}{3}$

$\frac{49}{3}$

$-\frac{50}{3}$

Correct answer:

$\frac{50}{3}$

Explanation:

When setting up this problem, it should look like this: $\int_{2}^{4}x^2-1dx$ . Then, integrate. Remember that when integrating, raise the exponent by 1 and then also put that result on the denominator: $\frac{x^3}{3}-x$ . Then, evaluate first at 4 and then at 2. Subtract the results. $(\frac{64}{3}-4)-(\frac{8}{3}-2)=\frac{56}{3}-2=\frac{50}{3}$ .

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Example Question #41 : Area Under A Curve

What is the area under the curve f(x)=x^2-3x-4 bounded by the x-axis from x=4 to x=5?

Possible Answers:

$\frac{19}{6}$

$\frac{17}{6}$

$\frac{21}{2}$

$\frac{17}{3}$

Correct answer:

$\frac{17}{6}$

Explanation:

First, set up the integral expression: $\int_{4}^{5}x^2-3x-4dx$ . Then, integrate. Remember, when integrating, raise the exponent by 1 and then put that result on the denominator: $\frac{x^3}{3}-\frac{3x^2}{2}-4x$ . Then, evaluate at 5 and then 4. Subtract those results: $(\frac{125}{3}-\frac{75}{2}-20)-(\frac{64}{3}-\frac{48}{2}-16)$ . Simplify to get your final answer of $\frac{17}{6}$ .

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Example Question #102 : Integral Applications

What is the area under the curve f(x)=x^3-2x+1 bounded by the x-axis from x=2 to x=3?

Possible Answers:

$\frac{51}{4}$

$\frac{53}{4}$

$\frac{49}{2}$

$\frac{49}{4}$

Correct answer:

$\frac{49}{4}$

Explanation:

First, set up the integral expression: $\int_{2}^{3}x^3-2x+1dx$ . Then, integrate. Remember to raise the exponent by 1 and then put that result on the denominator: $\frac{x^4}{4}-\frac{2x^2}{2}+x$ . Evaluate at 3 and then 2. Subtract those 2 results: $(\frac{81}{4}-9+3)-(\frac{16}{4}-4+2)=\frac{65}{4}-4=\frac{49}{4}$ .

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Example Question #271 : Integrals

What is the area under the curve f(x)=x^2-3x+1 bounded by the x-axis and from x=4 to x=5?

Possible Answers:

$\frac{47}{6}$

$\frac{7}{6}$

$\frac{49}{6}$

$\frac{52}{6}$

$\frac{53}{6}$

Correct answer:

$\frac{47}{6}$

Explanation:

First, set up the integral expression:
$\int_{4}^{5}x^2-3x+1dx$

Now, integrate:

$\frac{x^3}{3}-\frac{3x^2}{2}+x$

Evaluate at 5 and 4. Subtract the results:

$(\frac{125}{3}-\frac{75}{2}+5)-(\frac{64}{3}-\frac{48}{2}+4)=(\frac{125}{3}-\frac{75}{2}+5)$

Simplify to get:

$\frac{47}{6}$

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Example Question #281 : Integrals

What is the area under the curve f(x)=2x^3-2x+1 bounded by the x-axis and from x=1 to x=2?

Possible Answers:

6.5

5.5

5.75

5.25

Correct answer:

5.5

Explanation:

First, set up the integral expression:

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

Now, integrate. Remember to raise the exponent by 1 and then also put that result on the denominator:

$\frac{2x^4}{4}-\frac{2x^2}{2}+x=\frac{x^4}{2}-x^2+x$

Evaluate at 2 and then 1. Subtract the results:

$(8-4+2)-(\frac{1}{2}-1+1)=6-\frac{1}{2}=5.5$

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Example Question #44 : Area Under A Curve

What is the area under the curve f(x)=x^2-5x+4 bounded by the x-axis from x=4 to x=5?

Possible Answers:

$\frac{5}{3}$

$\frac{19}{6}$

$\frac{11}{6}$

$\frac{13}{6}$

$\frac{7}{6}$

Correct answer:

$\frac{11}{6}$

Explanation:

First, set up the integral expression:

$\int_{4}^{5}x^2-5x+4dx$

Now, integrate. Remember to raise the exponent by 1 and then also put that result on the denominator:

$\frac{x^3}{3}-\frac{5x^2}{2}+4x$

Evaluate at 5 and then 4. Subtract the results:

$(\frac{125}{3}-\frac{125}{2}+20)-(\frac{64}{3}-\frac{80}{2}+16)$

Simplify to get your answer of: $\frac{11}{6}$

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Example Question #108 : Integral Applications

What is the area under the curve f(x)=x^2+4x+4 from x=4 to x=5 , bounded by the x-axis?

Possible Answers:

$\frac{8}{3}$

$\frac{1}{3}$

$\frac{11}{3}$

$\frac{2}{3}$

$\frac{5}{3}$

Correct answer:

$\frac{5}{3}$

Explanation:

First, set up the integral expression:

$\int_{4}^{5}x^2+4x+4dx$

Next, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{x^3}{3}+\frac{4x^2}{2}+4x=\frac{x^3}{3}+2x^2+4x$

Now, evaluate at 5 and then 4. Subtract the results:

$(\frac{125}{3}+50+20)-(\frac{64}{3}+32+16)=22-\frac{61}{3}=\frac{5}{3}$

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Example Question #281 : Integrals

What is the area under the curve f(x)=x^2-7x+9 bounded by the x-axis from x=0 to x=1?

Possible Answers:

$\frac{17}{6}$

$\frac{35}{2}$

$\frac{23}{6}$

$\frac{35}{6}$

$\frac{37}{6}$

Correct answer:

$\frac{35}{6}$

Explanation:

First, write out the integral expression:

$\int_{0}^{1}x^2-7x+9dx$

Next, integrate. Remember to add one to the exponent and also put that result on the denominator

$\frac{x^3}{3}-\frac{7x^2}{2}+9x$

Next, evaluate at 1 and then 0. Subtract the results:

$(\frac{1}{3}-\frac{7}{2}+9)-(0)=\frac{35}{6}$

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Example Question #282 : Integrals

What is the area under the curve f(x)=x^3-2x^2+1 bounded by the x-axis from x=3 to x=4?

Possible Answers:

$\frac{133}{12}$

$\frac{233}{12}$

$\frac{233}{2}$

$\frac{233}{3}$

$\frac{233}{4}$

Correct answer:

$\frac{233}{12}$

Explanation:

First, write out the integral expression for this problem:

$\int_{3}^{4}x^3-2x^2+1dx$

Next, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{x^4}{4}-\frac{2x^3}{3}+x$

Next, evaluate at 4 and then 3. Subtract the results:

$(\frac{256}{4}-\frac{128}{3}+4)-(\frac{81}{4}-\frac{54}{3}+3)$

Simplify to get your answer:

$\frac{175}{4}-\frac{74}{3}+1=\frac{233}{12}$

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Calculus 2 : Integral Applications

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #41 : Area Under A Curve

Example Question #41 : Area Under A Curve

Example Question #41 : Area Under A Curve

Example Question #102 : Integral Applications

Example Question #271 : Integrals

Example Question #281 : Integrals

Example Question #44 : Area Under A Curve

Example Question #108 : Integral Applications

Example Question #281 : Integrals

Example Question #282 : Integrals

All Calculus 2 Resources

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