Question

Given that `int_0^3f(x)dx=6` and `int_0^2f(x)dx=5`. Show that `int_2^3f(x)dx`. How to do this?

Answer 1

`int_2^3 f(x)\ dx=1`

Explanation:

If we consider an integral as the area under a curve, we can say that the area under the curve on the interval `[2,3]` can be expressed as the area under the curve on the interval `[0,3]` minus the area under the curve on the interval `[0,2]`.

You can think of it as taking the area under `[0,3]` and then removing the area `[0,2]`, which would leave you with the area under `[2,3]`.

This knowledge lets us put up the following equation of the integrals:
`int_2^3 f(x)\ dx=int_0^3 f(x)\ dx-int_0^2 f(x)\ dx`

Since we know the values for the integrals on the right hand side, we get:
`int_2^3 f(x)\ dx=6-5`

This means that the integral is equal to:
`int_2^3 f(x)\ dx=1`