Question

can't understand why integral of xe^((x^2)/2) is e^((x^2)/2)
can someone explain how to calculate this integral please?

Answer 1

\[\LARGE xe^{\frac{x^2}{2}}\] this one?

Answer 2

do you know u-substitution?

Answer 3

The u-substitution is an integration method that essentially involves using the chain rule in reverse.

Answer 4

hmm let's try rewriting that fraction
\[\LARGE xe^{\frac{1}{2}x^2}\] so we can see the exponent clearly.

Answer 5

we need to pick a u so that when we take that derivative of u we can substitute..
so what will be our u?
Well we do see an x in front of the e.. Maybe we can have
\[\LARGE u=\frac{1}{2}x^2 \]

Answer 6

so what is the derivative of \[\large \frac{1}{2}x^2 ? \]

Answer 7

we can use product rule to make this a bit easier. Let f(x) = 1/2 and g(x) = x^2
recall that the product rule is f(x)g'(x) +f'(x)g(x)

Answer 8

so what is the derivative of f(x) = 1/2 ?

Answer 9

similarly what is the derivative of \[\large g(x) = x^2 \]

Answer 10

so if \[u= (1/2)x^2 -> u' = x\]

Answer 11

yeah but we have to write this differently
it will be
du = x dx

Answer 12

so now we have \[xe^u dx\]

Answer 13

hmm how do I explain this.... we just have \[e^u \]
and substitute the u back .. don't forget to add +C

Answer 14

remember du = x dx ?
since there's nothing on the left hand side.... oh man .. I'm having trouble explaining this part. . . usually when there's nothing ... it's just blank like 1 sorry I'm in panic mode I can't figure out how to type this well

Answer 15

but we just have \[\LARGE e^u +C \] now substitute \[\LARGE u=\frac{1}{2}x^2 \]
back into
\[\LARGE e^u +C \]

Answer 16

ok, so \[xe^u dx -> xe^u (1/x) dy -> e^u du\] (+c...)

Answer 17

du

Answer 18

you don't have to write dx
you don't have to write du

Answer 19

du = x dx

it's just .... @ganeshie8 can you help me explain this?
I know what's going on just can't put it into words

Answer 20

Ok, I think I understand, thank you very much!

Answer 21

but in the end you should have \[\LARGE e^u +C \] now plug in u back into this equation

Answer 22

and once we have our antiderivative, we can simply check by taking the derivative.

Answer 23

you should have something like \[\LARGE e^{\frac{1}{2}x^2} +C\]

Answer 24

now take the derivative of that equation...

Answer 25

that's how to check :)

Answer 26

so we get back to \[e^((x^2)/2) \]

Answer 27

yes, +c

Answer 28

mhm

Answer 29

if you take the derivative of your antiderivative, we should be getting that x in front of that e

Answer 30

Yes, I wrote the integral

Answer 31

\[\LARGE f(x) = e^{\frac{1}{2}x^2} +C \]
\[\LARGE f'(x) = xe^{\frac{1}{2}x^2} +C \]

because the derivative is \[\frac{1}{2}x^2 \rightarrow \frac{1}{2} (2x) + 0(x^2) \rightarrow x\] and that's placed in front of the e.

Answer 32

the derivative of any e^x is always in the front and the exponent part of e^x just stays there :)

Answer 33

Excellent. Thanks a lot!