Answer
$\frac{15}{4}$
Work Step by Step
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 $F(b)-F(a)$:
$F(1)-F(0)$
$\frac{(1+1)^4}{4} - \frac{(1+0)^4}{4}$
$\frac{16}{4} - \frac{1}{4} = \frac{15}{4}$