Question

integrate 4x^2/e^-2x

Answer 1

What ideas do you have?

Answer 2

hold on its 4x^2/e^-2x

Answer 3

@bloop it's \(e^{-2x}\)

Answer 4

integration by parts, only im confused on the concept

Answer 5

ok good

Answer 6

so first we want to make our lives as easy as possible let's state what the definition/formula for integration by parts is

Answer 7

integral udv=uv-integral vdu

Answer 8

good, so we have this. Do you understand how I brought up the denominator?\[\int(4x^2e^{2x})dx\]

Answer 9

yes i see how you got that

Answer 10

ok so now let's pick our u and dv

Answer 11

u=4x^2 and dv=e^2x

Answer 12

alright so now we need du and v

Answer 13

du=8x*dx

Answer 14

im kinda confused on how to get the integral of e^2x

Answer 15

what's the deriv of e^x?

Answer 16

its just e^x right?

Answer 17

yep so now we have a constant in front of our function, just like 4x^2, if you were to integrate that what would you do?

Answer 18

well if i were integrating 4x^2 i would get 4x^3/3
would the same rule apply to e^2x?

Answer 19

yup (minus the exponent rule)tell me what you get

Answer 20

e^2x+1/2x+1

Answer 21

not quite ok so let me explain it this way let u be a function of x e^u=u'e^u

Answer 22

(Derivative)

Answer 23

so if we let k be a constant. e^(kx)=ke^(kx) is the derivative, so what is the integral?

Answer 24

2e^2x?

Answer 25

close, but that would be the derivative

Answer 26

e^2x/2

Answer 27

yup so now let's put that into our eq and what do we get?

Answer 28

=uv-int(vdu)
=4x^2*e^2x/2-int(e^2x/2*8x

Answer 29

wait why do you have division?

Answer 30

thats v right?

Answer 31

in the integral

Answer 32

but yea that's v

Answer 33

v is e^2x/2
thats the division inside the integral

Answer 34

you've written \[=4x^2*e^2x/2-\int(e^2x/2*8x)dx
\] ohk sorry hard to read like that

Answer 35

if you throw a \[ and a \)<--that should be a bracket around it, it will write it all fancy

Answer 36

so now let's look at that integral

Answer 37

Somplify and pull out the constants to start

Answer 38

simplify*

Answer 39

\[2x^2*e^2x-int(4x*e^2x)\)

Answer 40

lmao the bracket thing didnt work

Answer 41

\[2x^2*e^2x-int(4x*e^2x)\]

Answer 42

that parenthesis needed to be a bracket

Answer 43

now pull out the constant

Answer 44

ok final answer
2x^2*e^2x-4*x^2/2*e^2x/2

Answer 45

not quite but close you have one more thing

Answer 46

\[2x^2*e^{2x}-4*\int xe^{2x}/2 dx\] this is actually what you have, you can't integrate that the way you did guess what you get to do again???!!??

Answer 47

and oops that /2 shouldnt be there

Answer 48

wait im confused again..

Answer 49

ok after removing that 4, you have \(2x^2e^{2x}−4\int xe^{2x} dx\)

Answer 50

do you see why?

Answer 51

yea the integral constant rule

Answer 52

ok now that integral, you have to use integration by parts... again :/

Answer 53

the constant gets kicked to the outside of the integral

Answer 54

yup the constant does

Answer 55

ewww whyyyyyyyyy

Answer 56

because it is a product haha

Answer 57

so what is your u? dv?

Answer 58

so okay
u=x dv=e^2x

du=1 v=e^2x/2

Answer 59

You have to use int. by parts again because it's still just as un-integrateable as before. Often in these cases you need to integrate by parts twice, sometimes more :S

Answer 60

Good so apply, le int by parts and then solve just remember that -4 will multiply the whole thing

Answer 61

So true @agent0smith

Answer 62

btw I can't get to my questions for some reason right now they won't load so if i'm not here that's why