## Calculus 8th Edition

Published by Cengage

# Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises - Page 347: 84

#### Answer

$$\int_0^2(x-1)e^{(x-1)^2}dx=0$$

#### Work Step by Step

To evaluate the integral $$\int_0^2(x-1)e^{(x-1)^2}dx$$ we will use substitution $(x-1)^2=t$ which gives us $2(x-1)dx=dt\Rightarrow (x-1)dx=\frac{dt}{2}$. The integration bounds would be: for $x=0$ we have $t=1$ and for $x=2$ we have $t=1$. Putting this into the integral we get: $$\int_0^2(x-1)e^{(x-1)^2}dx=\int_{1}^1e^t\frac{dt}{2}=0$$

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