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**1. Homework Statement**

The curve of y= 4/(x^2+4), the x-axis, and the vertical lines x = -2 and x = 2

- Thread starter beefiestcrib55
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The curve of y= 4/(x^2+4), the x-axis, and the vertical lines x = -2 and x = 2

- #2

Math_QED

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To begin, every member here will appreciate if you type out your work. Most members won't even bother if you post images.

There are a couple of things wrong about your approach. You set up the integral correctly, but then a couple of things went wrong:

1) Reflect about your answer! Can an area be negative?

2) You need to perform a substitution, you cannot (without experience) evaluate the integral into a primitive function in one step!

3)##\int \frac{1}{x^2 + 1} dx = \arctan(x) + c ##, not ##\arctan(x^2)+c##

4) Although your bounds should be changed by the substitution, why would you fill in ##1## as upperbound if you have ##2## as upperbound?

- #3

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Specifically, ##\int \frac{dx}{x^2+a^2}=\frac{1}{a}tan^{-1}(\frac{x}{a})## or in his case, ##\int \frac{dx}{\frac{x^2}{a^2}+1}=\frac{1}{a}tan^{-1}(\frac{x}{a})##. Also, ##tan^{-1}(0)≠\frac{\pi}{2}##. For future reference, ##\lim_{x \rightarrow \infty} {tan^{-1}(x)}=\frac{\pi}{2}##, and ##\lim_{x \rightarrow -\infty} {tan^{-1}(x)}=-\frac{\pi}{2}##.##\int \frac{1}{x^2 + 1} dx = \arctan(x) + c##, not ##arctan(x^2)+c##

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