Solved: "Evaluate the integrals ∫0ln⁡9eθ(eθ−1)12dθ"

arenceabigns

Answered question

2021-10-22

Evaluate the integrals

\int_{0}^{\ln 9} e^{θ} {(e^{θ} - 1)}^{\frac{1}{2}} d θ

wheezym

Skilled2021-10-23Added 103 answers

We have to evaluate the integral

\int_{0}^{\ln 9} e^{θ} {(e^{θ} - 1)}^{\frac{1}{2}} d θ

let

e^{θ} - 1 == t

on differentiating with respect to t we get

e^{θ} d θ = d t

d θ = \frac{d t}{e^{θ}}

now the limits

t = e^{θ} - 1

t = e^{\ln 9} - 1

t = 9 - 1

t = 8

also

t = e^{0} - 1

t = 1 - 1

t = 0

now the definite integral

\int_{0}^{8} e^{θ} t^{\frac{1}{2}} \frac{d t}{e^{θ}}

\int_{0}^{8} t^{\frac{1}{2}} d t

\Rightarrow {[\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}]}_{0}^{8}

\Rightarrow \frac{2}{3} {[t^{\frac{3}{2}}]}_{0}^{8}

\Rightarrow \frac{2}{3} [8^{\frac{3}{2} - 0}]

\Rightarrow \frac{2}{3} [2^{3 \times \frac{3}{2}}]

\Rightarrow \frac{2^{\frac{11}{2}}}{3} = \frac{32 \times 2^{\frac{1}{2}}}{3}

therefore

\int_{0}^{\ln 9} e^{θ} {(e^{θ} - 1)}^{\frac{1}{2}} d θ = \frac{32 \times 2^{\frac{1}{2}}}{3}

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