- #1

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## Homework Statement

Find the area bounded by y = 3x^2 + 1, x = 0, x = 2, y = 0

## Homework Equations

## The Attempt at a Solution

Not sure how to do this? Is this like finding the upper and lower sums?

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- Thread starter BuBbLeS01
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- #1

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Find the area bounded by y = 3x^2 + 1, x = 0, x = 2, y = 0

Not sure how to do this? Is this like finding the upper and lower sums?

- #2

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>Not sure how to do this? Is this like finding the upper and lower sums?

If you draw a picture, it is quickly seen that the area is the integral of f(x) = 3x^2 from 0 to 2: \int_0^2 3x^2 dx.

Using upper and lower sums:

You would have to calculate the limit of the upper sum as the interval is divided to smaller and smaller subintervals. Then you should do the same for the lower sum and conclude that because the two limits are equal, the integral (=area) must also be equal to this number.

- #3

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- #4

HallsofIvy

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One of the reasons you learn about "upper and lower sums" (more generally Riemann sums) is to help you set up integrals. If you were to divide that area into rectangles you could approximate the area as the sum of those rectangles. If you divide the interval from x= 0 to x=2 into "n" equal parts so that each part has length 2/n, then

[itex]A= \sum_{i=1}^{n} (3(2i/n)^2+ 1)(2/n)= \frac{24}{n}\sum_{i=1}^{n}i^2+ (2/n)\sum_{i=1}^{n}1= \frac{24}{n^3}\sum_{i=1}^{n}i^2+ 2[/tex]. If you are very clever and can recall a formula for [itex]\sum i^2[/itex] you can get that as a function of n and then take the limit as n goes to infinity.

Of course, you can see that that is very tedious to do in general! Do you know about "anti-derivatives"?

- #5

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I don't know how to integrate 0.

Working:

1 bounded area +ve

A= 2[int]0 ydx

A=2[int]0 3x^2 +1 dx

A= [x^3 + x]2/0

A=[2^3+2^1]-[0^3+0^1]

A=10 Squareunits

Problem with this is it covers the -negative area aswell when the y=0 should stop it.

Can't find anything in my notes.Can't integrate zero.

- #6

rock.freak667

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and that the area is simply given by

[tex]\int_0 ^{2} (3x^2+1)dx[/tex]

- #7

HallsofIvy

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?? What happened to " BuBbLeS01". I don't know what you mean by "without the y= 0"! Without some lower boundary you don't have a region to find the area of. The upper boundary is given y= 3x

I don't know how to integrate 0.

Working:

1 bounded area +ve

A= 2[int]0 ydx

A=2[int]0 3x^2 +1 dx

A= [x^3 + x]2/0

A=[2^3+2^1]-[0^3+0^1]

A=10 Squareunits

Problem with this is it covers the -negative area aswell when the y=0 should stop it.

Can't find anything in my notes.Can't integrate zero.

I can "integrate 0": the integral of any constant is that constant time x: [itex]\int 0 dx= 0+ C[/itex].

- #8

- 602

- 0

Anti-derivative of 3x^2 + 1 = x^3 + x

Plug in 2 and 0 =(2^3 + 2) - (0^3 + 0) = 10

- #9

HallsofIvy

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Yes, that's right. Although it's a lot more fun to set up the Riemann sums, don't you think!

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