Question

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

Answer 1

Show below

Explanation:

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

`y_2=10-1/(2x)`

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

Answer 2

`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