Fnd ∫x^2e^x
Intergration by parts (IBP). - This accours every time you have a mulitplication in an intergral. This quesion wants you to find the area under the graph x2ex. For any intergration by parts question simpally learn by heart uv - ∫vdu/dx dx. Then learn what the terms mean and where to find them in the question (shown below) - also it doesn't matter which way arround you decide to use the terms becuase 2x6 is the same as 6x2 - but if it is easier to answer the question one way rather than the other you should - for example a hint for this question would be that it is much easier to have ex as a the second term. Unlike when you had the product rule in differnetiation, where u and v were the two differnt terms in the question, in intergration the two terms being mulipied are u and dv/dx. u=the first term in this case = x2. So you differentiate this to get du/dx = 2x. Then the dv/dx= ex and remember this is easy as ex allways intergates to iteself, so v = ex. Then you can just nicely put these values into the IBP formula, to give: (1) x2ex - ∫ 2xex - this is not a perfect answer tho, we need to do IBP again! As the intergral in (1) cannot be solved it has another mulitpication inside an intergral. So ignoring the x2ex part of (1) and just concentrating on the intergral ∫ 2xex: u=2x du/dx= 2 dv/dx and v = ex so putting this in to the IBP formula gives: (2) 2xex - ∫ 2ex = ∫ 2xex now we can put this back into (1). So (1) = x2ex - 2xex + ∫ 2ex and we know that the intergral of ex is itself + c and this is the same for 2ex so (1) = x2ex - 2xex + 2ex +c. ex allways integrates and differentiates to itself so this means that if we had used the products the other way arround in the IBP equation uv - ∫vdu/dx dx the intergral is going to be still unsolvable. This is shown: ex = u and x2 = dv/dx ∫ 1/3x3ex doesn't work!
