Question

Find the area of the region. Use a graphing utility to verify your result. integral 0 to 7 9x · cubed root 3x + 1 dx

Answer 1

`\bf \displaystyle \int\limits_(0)^(7)\ 9x\sqrt[3]{3x+1}\cdot dx\\\\ -----------------------------\\\\ u=3x+1\implies \cfrac{du}{dx}=3\implies \cfrac{du}{3}=dx\\\\ -----------------------------\\\\ \displaystyle \int\limits_(0)^(7)\ 9x\sqrt[3]{u}\cdot \cfrac{du}{3}\implies \int_(0)^(7)\ 3x\sqrt[3]{u}\cdot dx\\\\ -----------------------------\\\\ now\qquad u=3x+1\implies u-1=3x\\\\ -----------------------------\\\\`


`\bf \displaystyle \int\limits_(0)^(7)\ (u-1)u^{(1)/(3)}\cdot dx\implies \int\limits_(0)^(7)\ \left( u^{(4)/(3)}-u^{(1)/(3)} \right) dx\\\\ -----------------------------\\\\ \textit{now, let's change the bounds, using u(x)} \\\\\\ u(x)=3x+1\qquad thus\qquad u(0)=1\qquad u(7)=22\\\\ -----------------------------\\\\`


`\bf \displaystyle \int\limits_(1)^(22)\ \left( u^{(4)/(3)}-u^{(1)/(3)} \right) dx\implies \cfrac{u^{(7)/(3)}}{(7)/(3)}-\cfrac{u^{(4)/(3)}}{(4)/(3)}\implies \left. \cfrac{3\sqrt[3]{u^7}}{7}-\cfrac{3\sqrt[3]{u^4}}{4}\right]_(1)^(22)`

upper-bound part


`\bf \left[ \cfrac{3\cdot 1356.187}{7} \right]-\left[ \cfrac{3\cdot 61.645}{4} \right] \\\\\\ 581.22-46.234\approx 534.98936648870`

and lower-bound part


`\bf \left[ \cfrac{3}{7} \right]-\left[ \cfrac{3}{4} \right]\implies -\cfrac{9}{28} \\\\\\ thus \\\\\\ 534.98936648870355525281-\left( -\cfrac{9}{28} \right) \\\\\\ 534.98936648870355525281+\cfrac{9}{28} \approx 535.31079506013212668138`