Answer to Question #132263 in Calculus for Navya Sharma

Question #132263

Evaluate limit 0to ln3 ∫ e^x(1+e^x)^1\2dx .


1
Expert's answer
2020-09-14T17:58:25-0400
0ln3ex(1+ex)12dx\displaystyle\int_{0}^{\ln3}e^x(1+e^x)^{{1 \over 2}}dx

ex(1+ex)12dx\int e^x(1+e^x)^{{1 \over 2}}dx

u=1+ex,du=exdxu=1+e^x, du=e^xdx


ex(1+ex)12dx=u12du=23u32+C=\int e^x(1+e^x)^{{1 \over 2}}dx=\int u^{{1 \over 2}}du=\dfrac{2}{3}u^{{3 \over 2}}+C=

=23(1+ex)32+C==\dfrac{2}{3}(1+e^x)^{{3 \over 2}}+C=

0ln3ex(1+ex)12dx=[23(1+ex)32]ln30=\displaystyle\int_{0}^{\ln3}e^x(1+e^x)^{{1 \over 2}}dx=\big[\dfrac{2}{3}(1+e^x)^{{3 \over 2}}\big]\begin{matrix} \ln3 \\ 0 \end{matrix}=

=23(1+3)3223(1+1)32==\dfrac{2}{3}(1+3)^{{3 \over 2}}-\dfrac{2}{3}(1+1)^{{3 \over 2}}=


=163423=\dfrac{16}{3}-\dfrac{4\sqrt{2}}{3}




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