Given that, the integral is ∫ex5dx. ∫ex5dx=∫ex15dx=∫ex·15dx=∫ex5dx Let u=x5 then, dudx=15du5=dx

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College Algebra

1st Edition

ISBN:9781938168383

Author:Jay Abramson

Publisher:Jay Abramson

Chapter6: Exponential And Logarithmic Functions

Section6.7: Exponential And Logarithmic Models

Problem 27SE: Prove that bx=exln(b) for positive b1 .

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Given that, the integral is $\int \sqrt[5]{e^{x}} d x$ .

$\begin{array}{rcl} \int \sqrt[5]{e^{x}} d x & = & \int {(e^{x})}^{\frac{1}{5}} d x \\ = & \int {(e)}^{x \cdot \frac{1}{5}} d x \\ = & \int {(e)}^{\frac{x}{5}} d x \end{array}$

Let $u = \frac{x}{5}$ then,

$\begin{array}{rcl} \frac{d u}{d x} & = & \frac{1}{5} \\ \frac{d u}{5} & = & d x \end{array}$

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