Integrate this equation by parts

[Xetman]

Homework Statement

Integrate e^ttdt using integration by parts.2. The attempt at a solution

Seems like a very easy problem; however, I just started learning integration by parts-- didn't understand this.

So the book's method:

u=t dv=e^tdt
du=dt v= e^t

uv-∫vdu=te^t-∫e^tdt=te^t-e^t+C.My method:

u=e^t dv=tdt
du=e^tdt v=t²/2uv-∫vdu=e^tt²/2-∫(t²/2)e^tdt

My method seems wrong, so can anyone please answer me why my method won't work?
Is it just about assigning the correct values to the correct variables?

 

[lurflurf]

Your method is not wrong, but it is not helpful. Integration by parts offers several options. We want to replace our integral by a simpler one. Your integral is more complicated. ∫te^t dt can be traded for ∫t^2 e^t dt or ∫e^t dt, the former is worse the latter better. As you practice you will see what choice improves the situation and what does not.

 

[MrWarlock616]

∫te^t dt can be traded for ∫t^2 e^t dt or ∫e^t dt, the former is worse the latter better.

can you please explain how ? I'm confused..

 

[tiny-tim]

Hi MrWarlock616!

te^t is in two parts, t and e^t

you can push e^t up, and t down …

then e^t stays as e^t, and t becomes 1, then 0​

of you can push e^t down, and t up …

then e^t stays as e^t, and t becomes t²/2, then t³/6, then …​

the idea is to push in the direction that makes one of them disappear

 

[MrWarlock616]

Hi MrWarlock616!

te^t is in two parts, t and e^t

you can push e^t up, and t down …

then e^t stays as e^t, and t becomes 1, then 0​

of you can push e^t down, and t up …

then e^t stays as e^t, and t becomes t²/2, then t³/6, then …​

the idea is to push in the direction that makes one of them disappear

This must a new concept. I'm not familiar with the push up and down..but thanks!

 

[Curious3141]

The thing to keep in mind is that for elementary integration by parts problems, the tricky term is the integral one is left with. From the original integrand (which is the product of two terms), one term gets differentiated, while the other term gets integrated. Do that in your head, and if the product between the two resulting terms can now be integrated again more easily, then it's the right approach.

Writing it out makes it seem more complicated than it is!

 

[SithsNGiggles]

This must a new concept. I'm not familiar with the push up and down..but thanks!

The pushing "up and down" means "integrating and differentiating," respectively. I imagine tiny_tim was thinking of "pushing down" the degree of the "t" part of the original integrand, and the resulting integral (the v du part) would simply be e^t, as opposed to the one you got using your method.