[SOLVED] Online Definite Integral Calculators faulty?

**dtb** · April 19th 2009, 04:19 AM

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

**Moo** · April 19th 2009, 04:56 AM

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

**dtb** · April 19th 2009, 05:16 AM

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?

**Plato** · April 19th 2009, 06:12 AM

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.

**dtb** · April 19th 2009, 06:32 AM

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

**Plato** · April 19th 2009, 07:10 AM

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

Read my reply. The use if absolute value will give area.

**dtb** · April 19th 2009, 08:21 AM

Nah I was just wondering when/why in real world situations one might one evaluate the integral vs. find the area, that's all

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