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Calculus 2 : Area Under a Curve

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Example Questions

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:

$\frac{23}{6}$

$\frac{35}{6}$

$\frac{37}{6}$

$\frac{17}{6}$

$\frac{35}{2}$

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 #51 : Area Under A Curve

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:

$\frac{233}{3}$

$\frac{233}{2}$

$\frac{233}{12}$

$\frac{233}{4}$

$\frac{133}{12}$

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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Example Question #282 : Integrals

Find the area under the given curve, o the interval $0\leq x\leq2$ :

$y(x)=e^{2x}+x^2+2$

Possible Answers:

$\frac{e^4-1}{2}+20/3$

$\frac{e^4}{2}+20/3$

$\frac{e^4}{2}$

Correct answer:

$\frac{e^4-1}{2}+20/3$

Explanation:

The area under the given curve is found using the following integral:

$\int_{0}^{2}(e^{2x}+x^2+2)dx=\frac{e^{2*2}}{2}+\frac{2^3}{3}+2*2-\frac{e^0}{2}=\frac{e^4-1}{2}+20/3$

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Example Question #112 : Integral Applications

Find the area of the region bounded by the curves $\small x = -y^2$ and $\small y = -x-2$

Area under curve problem 11

Possible Answers:

$\frac{13}{6}$

$\small \frac{9}{2}$

$\frac{7}{3}$

Correct answer:

$\small \frac{9}{2}$

Explanation:

Find the area of the region bounded by the curves $\small x = -y^2$ and $\small y = -x-2$

_______________________________________________________________

Definition

The area between two curves over the interval $a\leq y\leq b$ is defined by the integral:

$Area =\int_a^b [f(y)_{upper }-g(y)_{lower}]dy$

where $f(y)_{upper}\geq g(y)_{lower}$ for all such that $a\leq y\leq b$

_______________________________________________________________

Looking at the graphs of the curves we were given, we quickly see that integrating with respect to would be easiest. Attempting to integrate with respect to , we would not be able to assert $f(x)_{upper}\geq g(x)_{lower}$ for all values of between the intersection points. In fact, $\small x = -y^2$ would not even pass the vertical line test to be considered a function.

First let's find where the curves intersect. The coordinate of both points will be our limits of integration. Substitute the equation for the parabola into the equation of the line we were given:

y=(-y^2)-2

y^2 - y-2=0

(y+1)(y-2)=0

The solutions are therefore $y = \left \{ -1,2\right \}$

Area under curve problem 11

$Area =\int_a^b [f(y)_{upper }-g(y)_{lower}]dy$

$\small Area = \int_{-1}^2[-y^2-(-y-2)]dy$

$\small \small Area = \int_{-1}^2[-y^2+y+2)]dy$

$\small \small \small \small Area =\left[-\frac{1}{3}y^3+\frac{1}{2}y^2+2y \right ]_{-1}^{2}$

$\small Area = \left(-\frac{8}{3}+2+4 \right)-\left(\frac{1}{3}+\frac{1}{2}-2 \right )$

$\small \small Area = \left(\frac{10}{3} \right)-\left(-\frac{7}{6} \right )$

$\small Area = \frac{9}{2}$

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Calculus 2 : Area Under a Curve

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #281 : Integrals

Example Question #51 : Area Under A Curve

Example Question #282 : Integrals

Example Question #112 : Integral Applications

All Calculus 2 Resources

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