Find the area of the region given the boundaries

[beefiestcrib55]

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

Homework Equations

The Attempt at a Solution

 

[member 587159]

Hello beefiestcrib55!



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?

 

[Eclair_de_XII]

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

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 $\lim_{x \rightarrow -\infty} {tan^{-1}(x)}=-\frac{\pi}{2}$.