Question

What is the net area between `f(x)=(x^2-5)/(x^3-5x` in `x in[1,2]` and the x-axis?

Answer 1

Area `=ln 2=0.693147" "`square unit

Explanation:

Given `y=(x^2-5)/(x^3-5x)` from `x=1` to `x=2` and the x-axis

`y=(x^2-5)/(x^3-5x)=1/x`

Area `int_1^2 (1/x)dx=[ln x]_1^2=ln 2- ln 1=ln 2`

Area `=ln 2=0.693147" "`square unit

graph{(y-(x^2-5)/(x^3-5x))(y+10000x-10000)(y+10000x-20000)(y-0)=0[-0.1,4,-2,2]}

God bless...I hope the explanation is useful