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

2 Answers
Apr 29, 2018

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Explanation:

y_1=-3/(8x(x-8))

y_2=10-1/(2x)

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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(y_2-y_1)*dx

A=int_2^8(10-1/(2x))-(-3/(8x(x-8)))*dx

A=int_2^8(10-1/(2x))+(3/(8x(x-8)))*dx

A=int_2^8(10-1/(2x))*dx+int_2^8(3/(8x(x-8)))*dx

int(10-1/(2x))*dx=10*x-ln(abs(x))/2

int_2^8(10-1/(2x))*dx=(ln(2)-40)/2-(ln(8)-160)/2=-(ln(8)-ln(2)-120)/2=59.31

int(3/(8x(x-8)))=-(3*(ln(abs(x))-ln(abs(x-8))))/64=-3((ln(abs(x))-(((lnx/ln8))))/64)

int_2^8(3/(8x(x-8)))=

then complete the steps normally

Apr 30, 2018

18 units^2

Explanation:

x=2 and x=8 are two vertical lines and y=10-1/2x 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=1/2(9+6)6=45 units^2

y=-3/8x(x-8) => y=-3/8x^2+3x

This is an nn shaped parabola that passes through x=2 and x=8 . If we find the area under the curve by intergration

int_2^8 -3/8x^2+3x dx

[-1/8x^3+3/2x^2]_2^8

[-512/8 +192/2] - [-8/8 +12/2]

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