Explained: "Evaluate the following definite integral. ∫1eln⁡2xdx"

ruigE

Answered question

2020-12-28

Evaluate the following definite integral.

\int_{1}^{e} \ln 2 x d x

Clelioo

Skilled2020-12-29Added 88 answers

\int_{1}^{e} \ln 2 x d x

Substitute u =2x

\frac{d u}{d x} = 2

d x = \frac{1}{2} d u

= \frac{1}{2} \int_{2}^{2 e} \ln (u) d u

Now solve

\int_{2}^{2 e} \ln u d u

Integrate by parts:

\int f g^{'} = f g - \int f^{'} g

f = \ln u, f^{'} = \frac{1}{u}

g'=1, g=u

\Rightarrow {[u \ln (u)]}_{2}^{2 e} - \int_{2}^{2 e} 1 d u

Now solve

\int_{2}^{2 e} 1 d u = {[u]}_{2}^{2 e}

Plug in the solved integrals:

{[u \ln (u)]}_{2}^{2 e} - \int_{2}^{2 e} 1 d u

= {[u \ln (u) - u]}_{2}^{2 e}

Plug in the solved integrals:

\frac{1}{2} \int_{2}^{2 e} \ln (u) d u

= {[\frac{u \ln (u)}{2} - \frac{u}{2}]}_{2}^{2 e}

Undo the substitution u=2x

= {[x \ln (2 x) - x]}_{1}^{e}

= [e \ln (2 e) - e] - [1 \ln (2) - 1]

= e \ln (2 e) - e - \ln (2) + 1

Jeffrey Jordon

Expert2021-11-05Added 2605 answers

Answer is given below (on video)

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