# Thread: [SOLVED] Online Definite Integral Calculators faulty?

1. ## [SOLVED] Online Definite Integral Calculators faulty?

Hi all,

I've been looking at two online calculators of Definite Integrals and I'm wondering if they can't handle limits that cross the x-axis

eg.
$\int_{-3}^{1} \! sin(x) \, dx.$

My solution is that the area is 2.4486 (being 1.989 + 0.459 area below axis added to area above axis)

But if you use either of these two popular online Integral calculators
Find the Numerical Answer to a Definite Integral - WebMath
Definite Integral Calculator at SolveMyMath.com

they give a result of -1.53 (negative area subtract smaller positive area)

So are they wrong? or is there a different purpose for this negative result?

Cheers

2. Hello,
Originally Posted by dtb
Hi all,

I've been looking at two online calculators of Definite Integrals and I'm wondering if they can't handle limits that cross the x-axis

eg.
$\int_{-3}^{1} \! sin(x) \, dx.$

My solution is that the area is 2.4486 (being 1.989 + 0.459 area below axis added to area above axis)

But if you use either of these two popular online Integral calculators
Find the Numerical Answer to a Definite Integral - WebMath
Definite Integral Calculator at SolveMyMath.com

they give a result of -1.53 (negative area subtract smaller positive area)

So are they wrong? or is there a different purpose for this negative result?

Cheers
When calculating an area, you have to subtract any area that is below the x-axis. And add the areas that are above the x-axis.
That's integrals, they can be negative, though areas in geometry are always positive.

If you work with the antiderivatives... :
An antiderivative of sin(x) is -cos(x) (just check by taking the derivative of -cos(x))
By definition, your integral is thus $-\cos(1)-[-\cos(-3)]=-\cos(1)+\cos(-3)=-\cos(1)+\cos(3)$
You can check this result with a calculator, it'll give -1.53

3. Hmm, I'm still not sure...

I can work it out that way if I leave all +/- symbols as they are:

I get
-1.9899 (area below x-axis) + 0.4597 (area above x-axis)
= -1.5302

Unfortunately, having looked at 5 calculus books, there isn't much mention of graphs that cross the x-axis....

My teacher was saying that you need to take each area separately - make a negative area positive - then add them together
ie. magnitude |-1.9899| + |0.4597|
= 1.9899 + 0.4597
= 2.4496

So the area is positive, and a sum of the part below the axis and the part above.

Any thoughts?

4. If $f$ is nonnegative and integrable on $[a,b]$ then $\int_a^b {f(x)dx}$ is a measure of the area bound by the graph of $f$ and the x-axis.

Therefore, $\int_0^{\frac{\pi }{2}} {\cos (x)dx} = 1$ means that there is 1 square unit in the region bound by $\cos(x)$ and the x-axis from $x=0$ to $x=\frac{\pi}{2}$.

However, $\int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {\cos (x)dx} = 0$ this is not area because the function is not nonnegative on that entire interval.

But $\int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {\left| {\cos (x)} \right|dx} = 2$ which is the correct bounded area.

5. I figured out why I was getting confused...

There's a difference between the "integral" and "the area between the graph and the x-axis"

or at least there is where the graph crosses the x-axis.

So, those online calculators will evaluate an integral, but they won't tell you the area.

Now I just need to find out what each type of result is used for

6. Originally Posted by dtb
There's a difference between the "integral" and "the area between the graph and the x-axis"
Now I just need to find out what each type of result is used for