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.

1 Answer
Aug 22, 2017

Answer:

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.