\lim_{n\to\infty }n.\int_{-1}^{0}\left(x+e^x\right)^ndx
S stuart clark New member Joined Mar 3, 2011 Messages 25 Mar 14, 2011 #1 limn→∞n.∫−10(x+ex)ndx\displaystyle \lim_{n\to\infty }n.\int_{-1}^{0}\left(x+e^x\right)^ndxn→∞limn.∫−10(x+ex)ndx
limn→∞n.∫−10(x+ex)ndx\displaystyle \lim_{n\to\infty }n.\int_{-1}^{0}\left(x+e^x\right)^ndxn→∞limn.∫−10(x+ex)ndx
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Mar 14, 2011 #2 Wow, where are you getting these problems?. Try starting with parts: ∫−10n(x+ex)ndx=∫−10x+ex1+ex[(x+ex)n]′dx\displaystyle \int_{-1}^{0}n(x+e^{x})^{n}dx=\int_{-1}^{0}\frac{x+e^{x}}{1+e^{x}}[(x+e^{x})^{n}]'dx∫−10n(x+ex)ndx=∫−101+exx+ex[(x+ex)n]′dx Using parts, we get: 12−(−1+e−1)n+11+e−1−∫−10(x+ex)n[2ex−x+1e2x+2ex+1]dx\displaystyle \frac{1}{2}-\frac{(-1+e^{-1})^{n+1}}{1+e^{-1}}-\int_{-1}^{0}(x+e^{x})^{n}\left[\frac{2e^{x}-x+1}{e^{2x}+2e^{x}+1}\right]dx21−1+e−1(−1+e−1)n+1−∫−10(x+ex)n[e2x+2ex+12ex−x+1]dx Now, can you finish and see what the result is?.
Wow, where are you getting these problems?. Try starting with parts: ∫−10n(x+ex)ndx=∫−10x+ex1+ex[(x+ex)n]′dx\displaystyle \int_{-1}^{0}n(x+e^{x})^{n}dx=\int_{-1}^{0}\frac{x+e^{x}}{1+e^{x}}[(x+e^{x})^{n}]'dx∫−10n(x+ex)ndx=∫−101+exx+ex[(x+ex)n]′dx Using parts, we get: 12−(−1+e−1)n+11+e−1−∫−10(x+ex)n[2ex−x+1e2x+2ex+1]dx\displaystyle \frac{1}{2}-\frac{(-1+e^{-1})^{n+1}}{1+e^{-1}}-\int_{-1}^{0}(x+e^{x})^{n}\left[\frac{2e^{x}-x+1}{e^{2x}+2e^{x}+1}\right]dx21−1+e−1(−1+e−1)n+1−∫−10(x+ex)n[e2x+2ex+12ex−x+1]dx Now, can you finish and see what the result is?.
