Question

Heeeelp Pllllleasease

Answer 1

Do I have to use Partial Fraction Decomposition?

Answer 2

\(\large u= x^2\)

Answer 3

Oh yea you are right. oops. lol

Answer 4

@agent0smith I know the answer is zero. But i don't remember why...
\[-\frac{ 13 }{ 2 }+\frac{ 13 }{ 2 } cancels\]

Answer 5

Does the - Infinity squared turn to positive? Because NegxNeg=Pos

Answer 6

or use
\(u=e^{-x^2}\)

Answer 7

Wouldn't that make is more messy though? I mean i got the correct answer, I just dont really understand how the infinity exponents work with e.

Answer 8

@iTz_Sid
`I know the answer is zero. But i don't remember why...`
Whenever you integrate an odd function between -k to +k, the result is zero, because the function is anti-symmetric.
This also explains why this integral can be answered as zero based on the above.

Answer 9

How do you know that it is odd? because of the -x^2?

Answer 10

let u=e^(-x^2), du=-2xe^(-x^2)
so
\(I=-13/2\int_{-\infty}^{\infty}du=-(13/2)u=-13/2e^{-x^2}+C\)

Answer 11

The function is odd because it is the product of x (odd) and e^(-x^2) (even).
An odd function multiplied by an even function is even.

Answer 12

Oh, how is e^-x^2 odd though? Because of the exponent?

Answer 13

e^(-x^2) is even, because of the exponent 2.

Answer 14

`An odd function multiplied by an even function is even.`
correction:
An odd function multiplied by an even function is odd.

Answer 15

It is even but the product of x and e^(-x^2) is odd
Try the following. Let f(x) be even and g(x) be odd.
Then f(-x) = f(x) and g(-x) = -g(x)
Then let h(x) = f(x).g(x)
h(-x) = f(-x).g(-x) = f(x).-g(x) = -f(x)(g(x) = -h(x)
So h(x) the product of an odd and even function is odd.

So the answer will be 0.

Answer 16

The integral could be evaluated as that of -1/2 e^(-x^2) if you wished but it is faster to use your knowledge of odd functions.

Answer 17

Okay so, Even functions are always zero. and Odd functions always go to infinity.
So... Even functions are always Convergent and Odd functions are always Convergent.

Answer 18

Right?

Answer 19

no not quite.

\[\int\limits_{-a}^{a} f(x) dx = 0 ~ if ~ f~ is ~odd\]
\[\int\limits_{-a}^{a} f(x) = 2 \int\limits_{0}^{a} f(x)~ dx ~ if~f~is~even\]

Answer 20

Do not confuse odd and even functions with convergence.
We are referring to the special case of integrating an odd function over limits symmetric with the y-axis.
Yes, if you actually do the integration, this will eliminate the case of divergent functions.
And yes, your calculation of the limit of the function at inf is relevant!

Answer 21

if you draw the situation it will be clear why


Remember we are dealing with definite integrals, not necessarily areas.