Calculate the following integrals
$$\Large \int_0^a dx x^{2}e^{x}$$

Question

anonymous · Answer

In this case i need complete solution guys..

anonymous · Answer

$$ \int_0^a  x^{2}e^{x}dx$$

anonymous · Answer

tomorrow i can get same question on my exam paper

anonymous · Answer

honestly i dont know how to start with it,..

anonymous · Answer

you're in calculus 1 correct?

anonymous · Answer

here in germany we name it analysis
and if analysis = calculus you are right, 
its analysis I  thema..

anonymous · Answer

but a little bit light one analysis I for computer scientist

anonymous · Answer

@UnkleRhaukus Is there anyway to solve this integral without using  by parts?

anonymous · Answer

Have you heard of something called " By parts method?"

anonymous · Answer

unfortunately, not, i am complete new to integral, i have first looked at before two days ago for half hour youtube videos.. and i was not at universty when prof. lectured it =)

anonymous · Answer

Well there is certain ways of solving these types of equations that you learn later on i  just don't know whether they teach it or not in analysis 1.

anonymous · Answer

ok, i am going to invest more time on integrals ;) thanks anyways guys..

unklerhaukus · Answer

Looking at my standard table of integrals,
$$ [38] \qquad \int x^ne^x\text dx=x^nx^x-n\int x^{n-1}e^x \text dx$$

anonymous · Answer

Is it complete solution UnkleRhaukus ?

anonymous · Answer

this is by parts what unkle has said

$$\int udv=uv-\int vdu=x^2e^x-\int e^x2xdx\$$

anonymous · Answer

$$x^2e^x-2\int e^xxdx=x^2e^x-2[uv-\int vdu]=x^2e^x-2[xe^x-\int e^x2dx]$$

anonymous · Answer

$$x^2e^x-2xe^x+4\int e^xdx=x^2e^x-2xe^x+4e^x=e^x(x^2-2x+4)$$

anonymous · Answer

actually the last part is incorrect i took out th 2 so the du=1 and v = e^x so just get rid of one of the twos to get

\

$$e^x(x^2-2x+2)$$