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.

Aug 22, 2017

Veronica is correct.

Explanation:

We have:

$f \left(x\right) = {x}^{2} - 1$

And

$I = {\int}_{0}^{3} \setminus f \left(x\right) \setminus \mathrm{dx}$

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

For the roots of $f \left(x\right) = 0$ we have:

${x}^{2} - 1 = 0 \implies x = \pm 1$

And, as we can see, for $x \in \left[0 , 3\right]$ 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:

$A = - {\int}_{0}^{1} f \left(x\right) \mathrm{dx} + {\int}_{1}^{3} f \left(x\right) \mathrm{dx}$

Which will of course be larger than the net area.

Hence, Veronica is correct.