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$y = x^2 + 2x$, and $y = -x + 4$ are two curves. How to find the bounded area of these two curves and first Moment of area using integration. I got an answer of $71/3$ for the bounded area. But I am not sure about my answer. Can you please help me to find the correct answer and the first moment of area. We have to find the area between -4 and 2.

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    What is a first moment of an area? – 2017-02-21

2 Answers 2

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Here's the figure, including the shaded area underneath both curves:

enter image description here

Integrate to get the area:

$\int\limits_{x=0}^4 \min[x^2 + 2 x, -x + 4]\ dx$

or

$\int\limits_{x=0}^1 x^2 + 2 x\ dx + \int\limits_{x=1}^4 -x + 4\ dx = {35 \over 6}$,

where the transition point $x=1$ was found by solving $x^2 + 2 x = -x + 4$ for $x$.

The first moment is:

$\int\limits_{x=0}^4 x \min [x^2 + 2 x, -x + 4]\ dx = {119 \over 12}$.

For those interested, all the above was done in Mathematica:

 Plot[{x^2 + 2 x, -x + 4, Min[x^2 + 2 x, -x + 4]}, {x, 0, 4},
 Filling -> {3 -> Axis}]

 Integrate[x Min[x^2 + 2 x, -x + 4], {x, 0, 4}]
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    Which software you drew the plot in? Just out of curiosity. – 2017-02-21
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    I used *Mathematica* for both the graphics and the symbolic integration. – 2017-02-21
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    Thank you. I wouldn't have thought it's Mathematica. I guess I should learn more beyond the Plot[] and Integrate[]. – 2017-02-21
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I am not sure about your $\frac{71}{3}$. I think the are should be $$\int_{-4}^{1}(-x+4)dx-\int_{-4}^{-2}(x^2+2x)dx-\int_{0}^{1}(x^2+2x)dx+\left|\int_{-2}^{0}(x^2+2x)dx\right|$$ which is $\frac{125}{6}$? Having seen @David G. Stork answer, I am calculating are between $-4$ and $1$. Your question leaves out which area you have in mind.

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    What about the first moment of area? – 2017-02-22
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    The limit is -4 to 2 – 2017-02-22