Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 328: 33

$\frac{15}{4}$

Evaluate the integral: $\int^1_0(1+r)^3dr$ Recall the second part of The Fundamental Theorem of Calculus: $\int^b_af(x)dx = F(b) - F(a)$ Find $F(x)$: $F(x) = \int(1+x)^3dx$ $F(x) = \frac{(1+x)^4}{4}$ Now Evaluate $F(b)-F(a)$: $F(1)-F(0)$ $\frac{(1+1)^4}{4} - \frac{(1+0)^4}{4}$ $\frac{16}{4} - \frac{1}{4} = \frac{15}{4}$