# How do you solve the following questions?

## Apr 20, 2018

i) You should notice that

${A}_{\text{shaded"= A_"rectangle" - A_"under root graph}}$

${A}_{\text{shaded}} = 2 - {\int}_{0}^{1} \sqrt{3 x + 1} \mathrm{dx}$

The integral can be evaluated thru a simple u-substitution. Let $u = 3 x + 1$. Then $\mathrm{du} = 3 \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{3}$.

${A}_{\text{shaded}} = 2 - \frac{1}{3} {\int}_{1}^{4} \sqrt{u} \mathrm{du}$

${A}_{\text{shaded}} = 2 - \frac{1}{3} {\left[\frac{2}{3} {u}^{\frac{3}{2}}\right]}_{1}^{4}$

${A}_{\text{shaded}} = 2 - \frac{1}{3} \left[\frac{2}{3} {\left(4\right)}^{\frac{3}{2}} - \frac{2}{3} {\left(1\right)}^{\frac{3}{2}}\right]$

${A}_{\text{shaded}} = 2 - \frac{1}{3} \left(\frac{16}{3} - \frac{2}{3}\right)$

${A}_{\text{shaded}} = 2 - \frac{14}{9}$

${A}_{\text{shaded}} = \frac{4}{9}$

ii) What we essentially have to do is find the volume if the area under the line $y = 2$ on $\left[0 , 1\right]$ is rotated around the x-axis and then subtract the volume if the area under $y = \sqrt{3 x + 1}$ on $\left[0 , 1\right]$ is rotated around the x-axis.

$V = \pi {\int}_{0}^{1} {2}^{2} \mathrm{dx} - \pi {\int}_{0}^{1} {\left(\sqrt{3 x + 1}\right)}^{2} \mathrm{dx}$

$V = \pi {\left[4 x\right]}_{0}^{1} - \pi {\left[\frac{3}{2} {x}^{2} + x\right]}_{0}^{1} \mathrm{dx}$

$V = 4 \pi - \pi \left(\frac{3}{2} + 1\right)$

$V = 4 \pi - \frac{3}{2} \pi - \pi$

$V = \frac{3}{2} \pi$

$V = 0.477$

iii) I will let another contributor do this one.

Hopefully this helps!