Thread: Simple integration by parts problem.

step by step, how do you work the integration of (t^3(e^t) dt)

I know the answer is (t^3-3t^2+6t-6)(exp(t))

Originally Posted by mathman66

step by step, how do you work the integration of (t^3(e^t) dt)

I know the answer is (t^3-3t^2+6t-6)(exp(t))

here's the method of integration by parts. if we have a product of the form u*v', then

int{uv'} = uv - int{u'v}

in other words, you chose one function to differentiate, and another one to integrate then the integral of their product is the (integral of function1) times function2 minus the integral of (function1 times the derivative of function2).

how do you know which function to choose to do what? well, the first rule is simple. do what you can! there are some functions that you don't know the integral of but you know the derivative of, so obviously that would be the function you choose to differentiate. otherwise, choose the functions in such a way that the integral becomes simpler.

here we have t^3 * e^t, now we can integrate and differentiate either of these easily, so we have the can do part down. however, if we choose to integrate the t^3, what happens? it becomes t^4 and that makes the problem worst. so obviously we will choose the t^3 to differentiate and we'll end up with t^2, then do it again and end up with t and do it yet again and end up with 1. e^t stays the same throughout, so we don't worry about that. let's see how it works. we will have to do integration by parts three times as i mentioned, the first time i'll do it with explanation then the last two times i'll do it straight and hope you follow.

int{(t^3)(e^t)}dt
let u = t^3 => u' = 3t^2
let v' = e^t => v = e^t

so int{uv'} = uv - int{u'v}
=> int{(t^3)(e^t)}dt = (t^3)(e^t) - 3*int{(t^2)(e^t)}dt ...now do by parts again on this integral.

int{(t^2)(e^t)}dt = (t^2)(e^t) - 2*int{t(e^t)}dt ........by parts again on this integral

int{t(e^t)}dt = t(e^t) - int{e^t}dt = t(e^t) - e^t

Now back substitute to find our integral.

=> int{(t^2)(e^t)}dt = (t^2)(e^t) - 2*(t(e^t) - e^t) = (t^2)(e^t) - 2t(e^t) + 2e^t = (e^t)(t^2 - 2t + 2)

So finally our integral will be:

int{(t^3)(e^t)}dt = (t^3)(e^t) - 3*((e^t)(t^2 - 2t + 2)) + C
.........................= (t^3)(e^t) - 3(e^t)(t^2 - 2t + 2) + C .....factor out e^t
.........................= (e^t)(t^3 - 3t^2 + 6t - 6) + C

The colors aren't as explanatory as i thought they'd be. if you're confused by them don't worry about it, just follow the steps

i could write out the full expression with all the colors so you can see what gets plugged into what if you want

Originally Posted by mathman66

step by step, how do you work the integration of (t^3(e^t) dt)

I know the answer is (t^3-3t^2+6t-6)(exp(t))

It is quite obviously a cubic polynomial times e^t, so you can find the
integral by differentiation (At^3+Bt^2+Ct+D)e^t then equating coefficients
between the derivative and t^3 e^t.

RonL

Thanks, that helped a lot. I think my problem is that it gets confusing sometimes, especially with an engineers handwritting.

Originally Posted by CaptainBlack

It is quite obviously a cubic polynomial times e^t, so you can find the
integral by differentiation (At^3+Bt^2+Ct+D)e^t then equating coefficients
between the derivative and t^3 e^t.

RonL

I liked that suggestion so much, that i'm going to follow it. i think mathman66 specifically wanted the by parts way though. this is for fun

int{t^3 * e^t}dt will be a polynomial of the form (At^3+Bt^2+Ct+D)e^t + E, so we can equate d/dt [(At^3+Bt^2+Ct+D)e^t] to t^3 * e^t

Now d/dt [(At^3+Bt^2+Ct+D)e^t] = (At^3+Bt^2+Ct+D)e^t + (3At^2 + 2Bt + C)e^t
................................................= (At^3+Bt^2+Ct+D + 3At^2 + 2Bt + C)(e^t)
................................................= [At^3 + (3A + B)t^2 + (2B + C)t + (C + D)](e^t)



Now equating the coefficients to our original expression we get:
A = 1
3A + B = 0 => B = -3
2B + C = 0 => C = 6
C + D = 0 => D = -6



So our integral will be: [t^3 - 3t^2 + 6t - 6](e^t) + C