Question

the integral of 4/(x^2 +4) from -10 to 10

Answer 1

Have you learned Trig Sub yet?

Answer 2

I don't know...what is it? I am using integration and limits in this section

Answer 3

you know, limits of integration and such

Answer 4

What are some of the method of integration you have learned?

Answer 5

rule, algerbra, u-sub, integration by parts, trigonometricl integrals, improper integrals

Answer 6

We may rewrite you integrand as following :\[4\over 4({1 \over 4}x^2+1)\]

Answer 7

Also I have learned indeterminate forms

Answer 8

canceling out 4s we get this \[1 \over {1 \over4}x^2+1\]

Answer 9

There is a general form:
\[\int\limits \frac{du}{u^2+a^2}=\frac{1}{a}\arctan(\frac{u}{a})+C\]
So the integral is just:
\[4*\frac{1}{2}\arctan(\frac{x}{2})+C=2\arctan(\frac{x}{2})+C\]

Answer 10

Yeas, I have used that one, but the answer I received from the problem, (1/2)arctan(5) is incorrect

Answer 11

Then you have your bounds. -10 to 10
You have:
\[2(\arctan(-5)-\arctan(5))\]

Answer 12

Then you can get an approximation if you want or leave it as exact.

Answer 13

So I cannot use the bounds 0 to 10, and then multiply by two?

Answer 14

Not necessarily. That only works if it is an even function.

Answer 15

and arctan is odd I believe.

Answer 16

It is. :P So you can't multiply it by 2.

Answer 17

oh....

Answer 18

So you have 2(-arctan(5)-arctan(5))=-4arctan(5). Since arctan is odd arctan(-x)=arctan(x)

Answer 19

but 2(arctan(-5)-arctan(5)) doesn't work either. I put in -4arctan(5) also before coming to this site and that doesn't work either.

Answer 20

neither does 4arctan(5)

Answer 21

http://www.wolframalpha.com/input/?i=integral+of+4%2F%28x^2%2B4%29%2Cx%2C-10%2C10

Answer 22

Before I came here I also put that answer in with no such luck. Here is a clip of what I am doing.

Answer 23

You were right, in this case you can do 0 to 10 and multiply by 2. Because 1/x^2+4 is even. (I jumped ahead of myself) Either way, you should still get the same answer. Do they want an approximation?

Answer 24

It doesn't say, but when I used wolfram's approximation, it's labeled incorrect

Answer 25

Then I really don't know what to tell you man :/ We did the problem correct.

Answer 26

That's what I think too. Guess it is a mistake in the system. I will ask my teacher about it. Thanks for helping!

Answer 27

No problem. Sorry for the misthought on the even/odd. You don't look at the integral you look at the original function xP