Question

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

Answer 1

The net area will be `-1.17` square units.

Explanation:

Start by finding the antiderivative of `f(x)`, call it `g(x)`.

`int f(x) = -int ln(x^2 + 1)`

Use integration by parts. Let `u = ln(x^2 + 1)` and `dv = 1dx`. Then `du = (2x)/(x^2 + 1)dx` and `v = x`.

`int u dv = uv - int v du`

`int ln(x^2 + 1) = ln(x^2 + 1)x - int x(2x)/(x^2 + 1)dx`

`int ln(x^2 + 1) = xln(x^2 + 1) - 2int x^2/(x^2 + 1) dx`

`int ln(x^2+ 1) = xln(x^2 + 1)- 2int 1 dx+ 2int 1/(x^2 + 1)dx`

`int ln(x^2 + 1) = -xln(x^2 + 1) + 2x + 2arctanx`

`-int ln(x^2 + 1) = 2x - 2arctanx - xln(x^2 + 1) + C`

Now let's consider the region whose area we must determine.

We now set up a definite integral.

`int_1^2 ln(x^2 + 1)`. This will make:

`A = 2(2) - 2arctan(2) - 2ln(5) - (2 - 2arctan(1) - ln(2))`

`A ~~ -1.17`

Hopefully this helps!