Find the Indefinite integral using integration by parts: S(x^2)(e^x) dxusing the equation Sudv = uv - Svdu I have figured out that u=x^2 and dv= e^x dx and du = dx and v=e^x. Assuming that is...
Find the Indefinite integral using integration by parts: S(x^2)(e^x) dx
using the equation Sudv = uv - Svdu I have figured out that u=x^2 and dv= e^x dx and du = dx and v=e^x. Assuming that is right, I have calculated the following:
(x^2)(e^x) - Se^x dx
(x^2)(e^x) - e^x +C
(e^x)((x^2) -1) +C
2 Answers
You need to use the following formula to integrate by parts such that:
`int udv = uv - int vdu`
You have made a right selection considering `u = x^2` , since differentiating it, its degree will minimize, but you did not differentiate it right. You should follow the next formula when differentiate `x^2` such that:
`(x^n)' = n*x^(n-1)`
Reasoning by analogy yields:
`(x^2)' = 2*x^(2-1) = 2x`
Hence, considering `u = x^2` and `dv = e^x dx` yieldS:
`u = x^2 => du = 2xdx`
`dv = e^x dx => v = e^x`
`int x^2 e^x dx = x^2*e^x - int 2x*e^x dx`
`int x^2 e^x dx = x^2*e^x - 2 int x* e^x dx`
You need to use parts againto sol,ve the integral `int x*e^x dx` such that:
`u = x => du = dx`
`dv = e^xdx => v = e^x`
`int x* e^x dx = x*e^x - int e^x dx`
`int x* e^x dx = x*e^x - e^x + c`
Substituting `x*e^x - e^x` for `int x* e^x dx` yields:
`int x^2 e^x dx = x^2*e^x - 2 (x*e^x - e^x) + c`
`int x^2 e^x dx = x^2*e^x - 2e^x*(x - 1) +` c
Factoring out `e^x` yields:
`int x^2 e^x dx = e^x*(x^2 - 2x + 2) + c`
Hence, evaluating the given indefinite integral using parts yields `int x^2 e^x dx = e^x*(x^2 - 2x + 2) + c` .
You are correct to take u=x^2 and dv = e^x. However, du = 2x dx (i.e. the derivative of x^2 is 2x not 1).
So you should get
x^2 e^x - S 2x e^x dx
S 2x e^x dx itself needs to be integrated by parts
Now u=2x dv=e^x, meaning du = 2, v = e^x
giving S 2x e^x dx = 2x e^x - 2 S e^x dx = 2x e^x - 2e^x + C
Putting this result into the first equation gives
x^2 e^x - 2x e^x + 2e^x + C