How do you solve the following questions?

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How do you solve the following questions?

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Answer 1 · 2018-04-20T16:52:42

i) You should notice that

$A_{shaded} = A_{rectangle} - A_{under root graph}$ $A_"shaded"= A_"rectangle" - A_"under root graph"$

$A_{shaded} = 2 - \int_{0}^{1} \sqrt{3 x + 1} d x$ $A_"shaded" = 2 - int_0^1 sqrt(3x+ 1) dx$

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

$A_{shaded} = 2 - \frac{1}{3} \int_{1}^{4} \sqrt{u} d u$ $A_"shaded" = 2- 1/3int_1^4 sqrt(u) du$

$A_{shaded} = 2 - \frac{1}{3} {[\frac{2}{3} u^{\frac{3}{2}}]}_{1}^{\frac{3}{2}}$ $A_"shaded" = 2 - 1/3[2/3u^(3/2)]_1^4$

$A_{shaded} = 2 - \frac{1}{3} [\frac{2}{3} {(4)}^{\frac{3}{2}} - \frac{2}{3} {(1)}^{\frac{3}{2}}]$ $A_"shaded" = 2 - 1/3[2/3(4)^(3/2) - 2/3(1)^(3/2)]$

$A_{shaded} = 2 - \frac{1}{3} (\frac{16}{3} - \frac{2}{3})$ $A_"shaded" = 2 - 1/3(16/3 - 2/3)$

$A_{shaded} = 2 - \frac{14}{9}$ $A_"shaded" = 2 - 14/9$

$A_{shaded} = \frac{4}{9}$ $A_"shaded" = 4/9$

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

$V = π \int_{0}^{1} 2^{2} d x - π \int_{0}^{1} {(\sqrt{3 x + 1})}^{2} d x$ $V = piint_0^1 2^2dx - pi int_0^1 (sqrt(3x + 1))^2 dx$

$V = π {[4 x]}_{0}^{1} - π {[\frac{3}{2} x^{2} + x]}_{0}^{2} d x$ $V = pi[4x]_0^1 - pi[3/2x^2 + x]_0^1 dx$

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

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

$V = \frac{3}{2} π$ $V = 3/2 pi$

$V = 0.477$ $V = 0.477$

iii) I will let another contributor do this one.

Hopefully this helps!

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