1. Quick integral question

Does the integral$\displaystyle \int^3_0(x^2-1)dx$ represent the total area bounded by $\displaystyle f ( x ) = x^2 - 1$ and the x-axis , between x = 0 and x = 3? or is the area larger?

2. Originally Posted by wren17
Does the integral$\displaystyle \int^3_0(x^2-1)dx$ represent the total area bounded by $\displaystyle f ( x ) = x^2 - 1$ and the x-axis , between x = 0 and x = 3? or is the area larger?
Well I will give you a hint...the area between the x-axis and your two curves is given by

$\displaystyle A_{region}=\int_0^{3}{\color{red}{{\bigg|}}}x^2-1{\color{red}{{\bigg|}}}{dx}$

3. I think i figured out that the area is 6 but im not sure if its larger or not

4. Originally Posted by wren17
I think i figured out that the area is 6 but im not sure if its larger or not
$\displaystyle x^2-1=0\Rightarrow{x=\pm{1}}$

1 being the only relevant result since -1 is outside of our domain

$\displaystyle \therefore\int_0^{3}|x^2-1|dx=\int_1^{3}x^2-1dx-\int_0^{1}x^2-1dx=\frac{22}{3}\approx{7.3333}$

Now

$\displaystyle \int_0^{3}x^2-1dx=\bigg[\frac{x^3}{3}-x\bigg]\bigg|_{x=0}^{3}=6$

So now what can you conclude?