Question

Alex and Veronica were discussing the definite integral `int_0^3 (x^2 − 1)dx`. Alex said it represented the total area bounded by `f(x) = x^2 - 1` and the `x`-axis , between `x = 0` and `x = 3`. Veronica said the total area was larger?

Who was right and why? Also, calculate the area.

Answer 1

Veronica is correct.

Explanation:

We have:

`f(x) = x^2-1`

And

`I = int_0^3 \ f(x) \ dx`

If we look at the graph of the function we have:
graph{x^2-1 [-10, 10, -4, 10]}

For the roots of `f(x)=0` we have:

`x^2-1=0=> x=+-1`

And, as we can see, for `x in[0,3]` part of the bounded area is below the `x`-axis, and part of it above .

When we evaluate the Integral, I, we get the net area . where the area above is counted as positive, and the area below is counted as negative.

Hence, the actual area is as follows:

`A = -int_0^1 f(x) dx + int_1^3 f(x) dx`

Which will of course be larger than the net area.

Hence, Veronica is correct.