Get the answer to "Evaluate the integral. ∫−22ez2ez2+1dz"

naivlingr

Answered question

2021-10-28

Evaluate the integral.

\int_{- 2}^{2} \frac{e^{\frac{z}{2}}}{e^{\frac{z}{2} + 1}} d z

aprovard

Skilled2021-10-29Added 94 answers

Step 1
To Determine: evaluate the following integral .include absolute values only when needed.
Given: we have an integral

\int_{- 2}^{2} \frac{e^{\frac{z}{2}}}{e^{\frac{z}{2} + 1}} d z

Explanation: we have

\int_{- 2}^{2} \frac{e^{\frac{z}{2}}}{e^{\frac{z}{2} + 1}} d z

let us consider

u = e^{\frac{z}{2}} + 1

then

d u = \frac{1}{2} e^{\frac{z}{2}} d z

2 d u = e^{\frac{z}{2}} d z

the limits also get changed

z = - 2 t h e n u = e^{\frac{- 2}{2}} + 1 = e^{- 1} + 1 = \frac{1}{e} + 1

z = 2 t h e n u = e^{\frac{2}{2}} + 1 = e^{1} + 1

Step 2
so our integral becomes

\int_{- 2}^{2} \frac{e^{\frac{z}{2}}}{e^{\frac{z}{2} + 1}} d z = \int_{\frac{1}{e} + 1}^{e + 1} \frac{2}{u} d u

= 2 {[\ln | u |]}_{\frac{1}{e} + 1}^{e + 1}

= 2 [\ln (e + 1) - \ln (\frac{1}{e} + 1)]

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