Evaluate the following integrals. int_{-2}^2frac{e^{z/2}}{e^{z/2}+1}dz

Evaluate the following integrals. int_{-2}^2frac{e^{z/2}}{e^{z/2}+1}dz

Question
Integrals
asked 2021-01-08
Evaluate the following integrals.
\(\int_{-2}^2\frac{e^{z/2}}{e^{z/2}+1}dz\)

Answers (1)

2021-01-09
To Determine: evaluate the following integral .include absolute values only when needed.
\(\text{Given: we have an integral }\int_{-2}^2\frac{e^{z/2}}{e^{z/2}+1}dz\)
\(\text{Explanation: we have}\int_{-2}^2\frac{e^{z/2}}{e^{z/2}+1}dz\)
\(\text{let us consider }\)
\(u=e^{z/2}+1\)
\(\text{then}\)
\(du=\frac{1}{2}e^{z/2}dz\)
\(2du=e^{z/2}dz\)
\(\text{the limits also get changed }\)
\(z=-2\ then\ u=e^{-2/2}+1\ =e^{-1}+1\ =\frac{1}{e}+1\)
\(z=2\ then\ u=e^{2/2}+1\ =e^1+1\)
\(\text{so our integral becomes}\)
\(\int_{-2}^2\frac{e^{z/2}}{e^{z/2}+1}dz=\int_{\frac{1}{e}+1}^{e+1}\frac{2}{u}du\)
\(=2[\ln|u|]_{\frac{1}{e}+1}^{e+1}\)
\(=2[\ln(e+1)-\ln(\frac{1}{e}+1)]\)
0

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