Question

What is the net area between `f(x) = x^2+1/x` and the x-axis over `x in [2, 4 ]`?

Answer 1

`A=56/3+ln2` or `19.36` units.

Explanation:

Over the interval `x in [2,4]`, `x^2+1/x` is always positive, so the area we will be computing is between this graph and the positive `x`-axis.

To actually find the area, integrate `x^2+1/x` from `2` to `4`:
`A=int_2^4x^2+1/xdx`

Using the sum rule,
`A=int_2^4x^2dx+int_2^4 1/xdx`

And evaluating the integrals,
`A=[x^3/3]_2^4+[lnx]_2^4`

Doing some substitutions and algebra:
`A=((4)^3/3-(2)^3/3)+(ln4-ln2)`
`A=(64/3-8/3)+ln(4/2)`
`A=56/3+ln2`

If we wanted a precise answer, we could leave it like this. An approximation to 2 decimal places is `A=19.36` units.