# Definite Integrals

#### DougD720

Hello everyone, i have been teaching myself some basic calculus from a coupld textbooks (high-school level) my mom brought home. I'm currently in an Analysis class, next year i will be taking a BC Calc. course.

My question is this, (it comes from the fact that i've kind of been skipping around in the books).

The definite integral of a curve from x1 to x2 gives you the Exact area under the curve right?

I only ask because the books both cover a lot of riemann sums and stuff, like breaking the curve up into smaller portions (rectangles) and such. Why would one do that when you can calculate the Exact area from the definite integral?

Or am i wrong, haha, sorry for the extremely basic question, i've only learned the basics of limits, derivatives and integrals.

Thanks for the help =)

#### mathman

Riemann sums are used in the proof that the definite integral gives the area.

#### HallsofIvy

Homework Helper
Hello everyone, i have been teaching myself some basic calculus from a coupld textbooks (high-school level) my mom brought home. I'm currently in an Analysis class, next year i will be taking a BC Calc. course.

My question is this, (it comes from the fact that i've kind of been skipping around in the books).

The definite integral of a curve from x1 to x2 gives you the Exact area under the curve right?
That's one specific use: if y= f(x) is positive for x between x1 and x2, then $\int_{x_0}^{x_1}f(x)dx$ is the exact area under the curve.

I only ask because the books both cover a lot of riemann sums and stuff, like breaking the curve up into smaller portions (rectangles) and such. Why would one do that when you can calculate the Exact area from the definite integral?

Or am i wrong, haha, sorry for the extremely basic question, i've only learned the basics of limits, derivatives and integrals.

Thanks for the help =)
Well, except when some really nasty teacher gives you a homework assignment where you are required to compute a limit of Riemann sums in order to learn exactly how they give the same thing as the anti-derivative, there is no good reason not to evaluate the anti-derivative at the limits of integration.

However, it is a really good idea to understand the concept of Riemann sum because it so often is a good way setting up the definite integral in a specific application. For example, "work" done by a varying force can be calculated by treating the force as constant over short intervals so that the work over each interval is a product of force times distance: $f(x_i)\Delta x$ and then summing over the intervals:
$$\Sigma f(x_i)\Delta x$$, a Riemann sum. Taking the limit as the number of intervals goes to infinity (and the length of each interval to 0), this becomes the integral
[tex]\int_{x_0}^{x_1}f(x)dx[/itex]

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#### DougD720

I see, thank you, very helpful, i'll go back to that part, thanks!

By the way, when you guys are writing the mathmatical notation (such as above) on the computer, what program are you using? I know how to do it in word, it's just annoying at times.

Thanks again!

#### dextercioby

Homework Helper
We use LaTex. This forum has a compiler enabled, so one can easily type/copy-paste the code and put it btw [ tex ] [ /tex ] (without the spaces) to create formulas, or [ itex ] [ /itex ] (without the spaces) to put the same formulas inside the written text.

The are numerous tutorials on LaTex on the internet. Just google for "LaTex tutorial"

Daniel.

#### Gib Z

Homework Helper
I've always believed that The Fundamental Theorem Of Calculus, which provided us the Ability to work out the exact area under a continuous smooth function was the original way of working out the Definite Integral, and the Riemann Sums were merely an attempt at a rigorous definition of the Definite Integral.

Yes, it helps us understand what the definite integral actually is, and useful for teaching students who haven't learned Calculus yet how to find the area under some curves (even if the method takes a while), but it is one one of the many different definitions of the Definite Integral.