Answer
$\displaystyle\int\limits_0^1(1+r)^{3}dr=\dfrac{15}{4}$
Work Step by Step
$\displaystyle\int\limits_0^1(1+r)^{3}dr$
Let's apply the cube of a binomial rule to the integral. This rule is $(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{2}$
$\displaystyle\int\limits_0^1(1+r)^{3}dr=\int\limits_0^1(1+3r^{2}+3r+r^{3})dr=...$
Now, integrate each term separately and apply the second part of the fundamental theorem of calculus:
$...=\displaystyle\int\limits_0^1dr+3\int\limits_0^1r^{2}dr+3\int\limits_0^1rdr+\int\limits_0^1r^{3}dr=...$
$...=r+(3)\Big(\dfrac{1}{3}\Big)r^{3}+(3)\Big(\dfrac{1}{2}\Big)r^{2}+\dfrac{1}{4}r^{4}\Big|_0^1=r+r^{3}+\dfrac{3}{2}r^{2}+\dfrac{1}{4}r^{4}\Big|_0^1$
$...=\Big[1+(1)^{3}+\dfrac{3}{2}(1)^{2}+\dfrac{1}{4}(1)^{4}\Big]-\Big[0+(0)^{3}+\dfrac{3}{2}(0)^{2}+\dfrac{1}{4}(0)^{4}\Big]$
$...=1+1+\dfrac{3}{2}+\dfrac{1}{4}=\dfrac{15}{4}$
