Question

For number 6, evaluate the definite integral.

Answer 1

$\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{(8+2x)^2}}\cdot dx\impliedby \textit{now, let's do some substitution}\\\\ -------------------------------\\\\ u=8+2x\implies \cfrac{du}{dx}=2\implies \cfrac{du}{2}=dx\\\\ -------------------------------$

$\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{u^2}}\cdot \cfrac{du}{2}\implies \cfrac{1}{2}\int\limits_{0}^{28}\ u^{-\frac{2}{3}}\cdot du\impliedby \begin{array}{llll} \textit{now let's change the bounds}\\ \textit{by using } u(x) \end{array}\\\\ -------------------------------\\\\ u(0)=8+2(0)\implies u(0)=8 \\\\\\ u(28)=8+2(28)\implies u(28)=64$

$\bf \\\\ -------------------------------\\\\ \displaystyle \cfrac{1}{2}\int\limits_{8}^{64}\ u^{-\frac{2}{3}}\cdot du\implies \cfrac{1}{2}\cdot \cfrac{u^{\frac{1}{3}}}{\frac{1}{3}}\implies \left. \cfrac{3\sqrt[3]{u}}{2} \right]_8^{64} \\\\\\ \left[ \cfrac{3\sqrt[3]{(2^2)^3}}{2} \right]-\left[ \cfrac{3\sqrt[3]{2^3}}{2} \right]\implies \cfrac{12}{2}-\cfrac{6}{2}\implies 6-3\implies 3$