How to Use Integral to Find Area Under a Curve: A Comprehensive Guide

[ryanuser]

We can find an area under a curve with integral, but how if the curve is under the x axis? Do e still use integral?

In general where can we use integral?

 

[dkotschessaa]

We can find an area under a curve with integral, but how if the curve is under the x axis? Do e still use integral?

Yes, and you will get a negative number. This may seem confusing (how can we have negative area?), but here is a good thread about it: https://www.physicsforums.com/showthread.php?t=292737

Check out HallsofIvy's excellent response. I'll quote him if that's ok

...There is no such thing as negative area (and "area" is the correct word here, not "space"). The integral as area is one simple application. When you integrate
f(x)=x3
from -1 to 1, you are NOT calculating an area. Strictly speaking when you integrate any
∫baf(t)dt
you are NOT calculating an area unless that is the particular application you are using the integral for. The integral of a function, from a to b, say, is a number. What that number means depends on the application.

In general where can we use integral?

If a function is continuous on an interval, it will be integrable on that interval. (That's a mouthful!)

Here's a neat wolfram demonstration as to why that's true: http://demonstrations.wolfram.com/ContinuousFunctionsAreIntegrable/

-Dave K

 

[EJChoi]

If the curve is under the x-axis the integral has a negative value.

Basically this gives you the net area under the curve.

If you want to find the total area both under and above the x-axis, you'll need to integrate the modulus of the function.

i.e \displaystyle \int^{2\pi}_{0} sin(x) = 0

But..

\displaystyle \int^{2\pi}_{0} |sin(x)| = 4

 

[Goa'uld]

If the curve is under the x-axis the integral has a negative value.

Basically this gives you the net area under the curve.

If you want to find the total area both under and above the x-axis, you'll need to integrate the modulus of the function.

i.e \displaystyle \int^{2\pi}_{0} sin(x) = 0

But..

\displaystyle \int^{2\pi}_{0} |sin(x)| = 4

I don't mean to be rude, but you need dx in the integral for that to be true. You could confuse some people, otherwise.

 

[EJChoi]

I don't mean to be rude, but you need dx in the integral for that to be true. You could confuse some people, otherwise.

Oops I was being sloppy again...

 

[Goa'uld]

...

In general where can we use integral?

In addition to area it can be used to find volume and surface area of solids, arc lengths, mass, average value of a function, the centroid, just to name a few simple applications.

 

[HallsofIvy]

Notice that the integral gives signed area- negative if the graph goes below the x- axis. Strictly speaking, area is always positive so if the problem actually asks for the area between a curve and the x-axis, where the curve goes both above and below the x-axis, is \int |f(x)| dx

 

[WannabeNewton]

If a function is continuous on an interval, it will be integrable on that interval. (That's a mouthful!)

This is only a sufficient condition; it is not a necessary condition. The condition of Riemann integrable is much more general. The indicator function associated with the Cantor set has uncountably many discontinuities but it is Riemann integrable.

 

[micromass]

If a function is continuous on an interval, it will be integrable on that interval. (That's a mouthful!)

A bit nitpicking, but you of course want your interval to be closed and bounded! I'm sure you mean this.

 

[jbunniii]

This is only a sufficient condition; it is not a necessary condition. The condition of Riemann integrable is much more general. The indicator function associated with the Cantor set has uncountably many discontinuities but it is Riemann integrable.

Indeed, we can go a bit further and state a necessary and sufficient condition for Riemann integrability of a bounded function $f$ defined on a closed, bounded interval: $f$ is Riemann integrable if and only if the set of points where it is discontinuous has Lebesgue measure zero. The indicator function of the Cantor set qualifies because its set of discontinuities is precisely the Cantor set, which has measure zero.