# How do you find the area between y=-3/8x(x-8), y=10-1/2x, x=2, x=8?

Apr 29, 2018

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#### Explanation:

${y}_{1} = - \frac{3}{8 x \left(x - 8\right)}$

${y}_{2} = 10 - \frac{1}{2 x}$

this is a sketch for your functions you can use this website to sketch them[www.desmos.com]

the area between the two curve equal

$A = {\int}_{2}^{8} \left({y}_{2} - {y}_{1}\right) \cdot \mathrm{dx}$

$A = {\int}_{2}^{8} \left(10 - \frac{1}{2 x}\right) - \left(- \frac{3}{8 x \left(x - 8\right)}\right) \cdot \mathrm{dx}$

$A = {\int}_{2}^{8} \left(10 - \frac{1}{2 x}\right) + \left(\frac{3}{8 x \left(x - 8\right)}\right) \cdot \mathrm{dx}$

$A = {\int}_{2}^{8} \left(10 - \frac{1}{2 x}\right) \cdot \mathrm{dx} + {\int}_{2}^{8} \left(\frac{3}{8 x \left(x - 8\right)}\right) \cdot \mathrm{dx}$

$\int \left(10 - \frac{1}{2 x}\right) \cdot \mathrm{dx} = 10 \cdot x - \ln \frac{\left\mid x \right\mid}{2}$

${\int}_{2}^{8} \left(10 - \frac{1}{2 x}\right) \cdot \mathrm{dx} = \frac{\ln \left(2\right) - 40}{2} - \frac{\ln \left(8\right) - 160}{2} = - \frac{\ln \left(8\right) - \ln \left(2\right) - 120}{2} = 59.31$

$\int \left(\frac{3}{8 x \left(x - 8\right)}\right) = - \frac{3 \cdot \left(\ln \left(\left\mid x \right\mid\right) - \ln \left(\left\mid x - 8 \right\mid\right)\right)}{64} = - 3 \left(\frac{\ln \left(\left\mid x \right\mid\right) - \left(\left(\left(\ln \frac{x}{\ln} 8\right)\right)\right)}{64}\right)$

${\int}_{2}^{8} \left(\frac{3}{8 x \left(x - 8\right)}\right) =$

then complete the steps normally

Apr 30, 2018

$18 u n i t {s}^{2}$

#### Explanation:

$x = 2$ and $x = 8$ are two vertical lines and $y = 10 - \frac{1}{2} x$ is a diagonal line that passes through $x = 2$ at (2,9) and $x = 8$ at (8,6). This gives us a trapezium that has two parallel sides of length 9 and 6 and a height of 6 so the area would be

$A = \frac{1}{2} \left(9 + 6\right) 6 = 45 u n i t {s}^{2}$

$y = - \frac{3}{8} x \left(x - 8\right)$ $\implies y = - \frac{3}{8} {x}^{2} + 3 x$

This is an $\cap$ shaped parabola that passes through $x = 2 \mathmr{and} x = 8$ . If we find the area under the curve by intergration

${\int}_{2}^{8}$ $- \frac{3}{8} {x}^{2} + 3 x$ $\mathrm{dx}$

${\left[- \frac{1}{8} {x}^{3} + \frac{3}{2} {x}^{2}\right]}_{2}^{8}$

$\left[- \frac{512}{8} + \frac{192}{2}\right] - \left[- \frac{8}{8} + \frac{12}{2}\right]$

$32 - 5 = 27$

So the area enclosed by all four graphs is the area of the trapezium less the area under the curve.

45 - 27 = 18