Question

two integrals questions

Answer 1

\[\int\limits_{4}^{9} \frac{ dx }{ 9+x ^{2} }\]

Answer 2

@Kainui

Answer 3

whoops the limits are 3 and -3

Answer 4

Do you know how to do trig substitution? Try it the best you can or explain what you know and I'll help fill in the gaps.

Answer 5

\[\int\limits_{4}^{9}\frac{ 2+x }{ 2\sqrt{x} }\]

Answer 6

oh derivative of arctan is 1/1+x^2 right?

Answer 7

True

Answer 8

how do i simplify so that i can get it into that form?

Answer 9

You could multiply the integral by \[\LARGE \frac{1/9}{1/9}\] that will give you \[\LARGE (\frac{x}{3})^2\] in your integral and the rest should look a lot more like the integral you want.

Answer 10

so then the integral equals arctan((x/3)^2)?

Answer 11

\[I=\int\limits_{-3}^{3}\frac{ dx }{ 9+x^2 }=2\int\limits_{0}^{3}\frac{ dx }{ 9+x^2 }=2\left[ \frac{ 1 }{ 3 }\tan^{-1} \frac{ x }{ 3 } \right] from~0\to 3\]

Answer 12

Not quite, you should do a substitution, like u=x/3. Then you'll get something that looks like arctan(u) not arctan(u^2), the squared is part of the derivative of arctan.

Answer 13

ok lemme try it

Answer 14

For your first question:\[\int\limits_{-3}^{3}\frac{ 1 }{ 9+x ^{2} }dx\]First rewrite the integral as\[2\int\limits_{0}^{3}\frac{ 1 }{ 9+x ^{2} }dx\]because f(x) is an even function. Then apply trigonometric substitutions.

Answer 15

what is an even function? I never learned about that

Answer 16

does it simply mean the function is always positive?

Answer 17

\[=\frac{ 2 }{ 3 }\left[ \tan^{-1} \frac{ 3 }{ 3 }-\tan^{-1} 0 \right]=\frac{ 2 }{ 3 }\left[ \frac{ \pi }{ 4 }-0 \right]=?\]

Answer 18

if f(-x)=f(x) ,then it is even.
if f(-x)=-f(x),then it is odd.

Answer 19

An even function just satisfies this condition: \[\LARGE f(x)=f(-x)\] so we can see: \[\LARGE \frac{1}{1+x^2}=\frac{1}{1+(-x)^2}\] So it's even. That just means it is reflected across the y-axis, which is convenient for simplifying.

Similarly odd functions satisfy: \[\LARGE -f(x)=f(-x)\] So they will cancel themselves out instead of doubling themselves. Imagine integrating sine from -pi to pi. You will just get 0 since both humps cancel each other out.

Answer 20

ok

Answer 21

An even function is a function symmetric about the y-axis. Thus if f(-x) = f(x) then f(x) is an even function and\[\int\limits_{-a}^{a}f(x)dx =2\int\limits_{0}^{a}f(x)dx\]Similarly, if f(-x) = -f(x) then f(x) is symmetric about the origin and f(x) is an odd function, in which case\[\int\limits_{-a}^{a}f(x)dx =0\]

Answer 22

but wait

Answer 23

when i multiply by 1/9/1/9, i get 1/9 on the top

Answer 24

You can pull constants outside the integral though.

Answer 25

ok

Answer 26

here \[f(x)=\frac{ 1 }{ 9+x^2 },f(-x)=\frac{ 1 }{ 9+(-x)^2 }=\frac{ 1 }{ 9+x^2 }=f(x)\]

Answer 27

so far i have
\[\frac{ 2 }{ 9 }\int\limits_{0}^{3}\frac{ 1 }{ 1+(\frac{ x }{ 3 })^{2} }dx\]

Answer 28

@surjithayer you're not supposed to be giving her the answers or doing the work for them. How are they supposed to learn?? Teach!!!

Answer 29

then the answer is pi/6

Answer 30

thank you guys!

Answer 31

it's annoying i can only give one best response

Answer 32

Good, now you know that: \[\LARGE \arctan(u)+C=\int\limits \frac{du}{1+u^2}\] so you need to just make a substitution to help you figure out what extra amount you would get from the chain rule.

Answer 33

yea i subbed u=x/3, then found the integral to be \[\frac{ 2 }{ 3 }\int\limits_{0}^{3}\frac{ 1 }{ 1+u ^{2} }du\]

Answer 34

then used the arctan derivative and evaluated

Answer 35

how about my second question?

Answer 36

Oh ok good. Alright, I have to go, maybe try closing this and asking it separately?

Answer 37

ok, thanks anyways