Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 330: 83

$$e^2-1$$

Given $$ \int_{-1}^{1} e^{u+1} d u $$ Since \begin{aligned} \int_{-1}^{1} e^{u+1} d u&=\int_{-1}^{1} e^{u}\cdot e d u\\ &=e\int_{-1}^{1} e^{u}du\\ &= e[e^u]\bigg|_{-1}^1\\ &= e[e-\frac{1}{e}]\\ &= e\cdot \frac{e^2-1}{e}\\ &= e^2-1 \end{aligned}