Question

How do I find a definite integral by computing an area?

Answer 1

I don't have a one sentence answer

Explanation:

Above the x-axis
If `f(x)` is non-negative over the interval `[a,b]`, then
`int_a^b f(x) dx` is the area of the region between the graph and the `x`-axis between `x=a` and `x=b`

Below the x-axis
If `f(x)` is non-positive over the interval `[a,b]`, then
`int_a^b f(x) dx` is the negative of area of the region between the graph and the `x`-axis between `x=a` and `x=b`

Example:

`int_0^3 sqrt(9-x^2) dx`.

`a=0` and `b=3`.
`f(x) = sqrt(9-x^2)` is never negative, so it is not negative on `[0,3]`

Therefore the integral is equal to the area under the curve and above the `x`-axis between `x=0` and `x=3`.

The graph of `y = sqrt(9-x^2)` is the part of `y^2 = 9-x^2` that has non-negative `y`-values. It is the upper semicircle for `x^2+y^2 = 9`
The part between `x=0` and `x=3` is a quarter of a circle with radius 3.

graph{y = sqrt(9-x^2)*(sqrt(1.5^2-(x-1.5)^2))/(sqrt(1.5^2-(x-1.5)^2)) [-2.63, 6.137, -0.812, 3.572]}

So
`int_0^3 sqrt(9-x^2) dx` is `1/4` of the area of the circle with radius `3`
`int_0^3 sqrt(9-x^2) dx = (9 pi)/4`