I'm trying to find the area between:
$y = 2x^2 - 1$ and $y = x^2$
I have found that the intersection points are at $(-1,1)$ and $(1,1)$.
But the part that confuses me is that $y = 2x^2 - 1$ goes below the x-axis and because of that I don't know how to calculate the integral.
$\endgroup$ 24 Answers
$\begingroup$As several others have pointed out, if $f(x)\geq g(x)$ over the interval $a\leq x \leq b$, then the area between the graphs of $f$ and $g$ is $$\int_a^b\left(f(x)-g(x)\right)dx.$$ It does not matter if both graphs are above the $x$-axis or not; what matters is that the inequality $f(x)\geq g(x)$ is maintained.
Assuming you believe that the area computation works when $0\leq g(x) \leq f(x)$, here's an explanation of why it still works without the zero on the left. Simply pick a number $c$ such that $c<g(x)$ for all $a\leq x \leq b$. (This can certainly be done, if $g$ is continuous on $[a,b]$.) Then, $0\leq g(x)+c \leq f(x)+c$. Thus, the area between $f(x)+c$ and $g(x)+c$ can certainly be computed via the integral formula, because both graphs lie completely above the $x$-axis. On the other hand, this area is certainly the same as the area between just $f$ and $g$, as one is just a vertical shift of the other. For your functions, the picture looks like so with $c=2$.
Finally, note that $$\int_a^b \left((f(x)+c)-(g(x)+c)\right)dx = \int_a^b \left(f(x)-g(x)\right)dx,$$ which is why the simple formula on the right (without the $+c$s) works.
$\endgroup$ 1 $\begingroup$It wouldn't matter as long the difference is between the two function and then integrated within bounds.
Basically, when you integrating a single function with bounds. You are just taking the area between function and x-axis as (y)−(0), y=0 is the x-axis.
But now, you are finding area with respect to another function below it. So, it doesn't matter whether graph of second function is below or above x-axis.
If you want to verify this consider the value of
$C_1:y=x^2$
$C_2:y=2x^2-1$
at $x=0$,
$C_1(0):y_0=0$
$C_2(0):y_0=-1$
And the difference? Well that's $C_1(0)-C_2(0)=1$ .
So, it wouldn't matter whether one of the graph goes below x-axis! As long as given that, you are integrating in form of $\int_a^b y.dx$ where bounds a,b corresponds to point on x-axis!
Here's the graph.
And the area you require, can be easily calculated by:
$$\large\int_{-1}^{1}x^2-(2x^2-1).dx$$
Without caring for whether $C_2:y=2x^2-1$ goes below x-axis. However be careful when only a single function is given not involving any difference between two function which demands you explicitly finding TOTAL AREA.
$\endgroup$ $\begingroup$it is $$I=2\int_{0}^{1} x^2-(2x^2-1)dx$$
$\endgroup$ $\begingroup$Area can easily found by using,
$\displaystyle\int_{-1}^{1} x^2-(2x^2-1)dx$
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