Area between two curves x^2+2x+1, 2x + 5

gluck

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Jul 30, 2007
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Find the area between the following two curves

∫(x)= x^2 + 2x+ 1 ∫(x)= 2x+5

I am completely lost on this one.

I set each to 0, so the limits are 1 and 5. Then integrate:

1. 1/3X^3 + x^2 + 1x =

[1/3(1)^3 + (1)^2 + 1(1)] - [1/3(5)^3 + 5^2 + 1(5)]

= 69.333

2. x^2 + 5x =

[1^2+5(1)]-[5^2+5(5)]

= 44

Add the 2 and the answer is 113.333
 
I'm sorry, but this post is a mass of confusion. Can you just write down the problem exactly as stated in your text or work papers? Perhaps your work will make some sense after that.
 
areaia0.gif

Always start with a graph.
That will tell us how to use the integrals.
 
gluck said:
I am completely lost on this one.
It would certainly appear so.... :shock:

I'm sorry that this topic and many of the prerequisite topics were not covered in class. You may need more help than we are able here to provide, simply because of the hours of missing instruction. But, for a start:

1) A function is not generally equal to its integral, general or otherwise. So "f(x) = int[f(x)]dx = (some number)" is going to be wrong at least 99.99999% of the time. Instead, the integral generally gives you an entirely different function, so you would have something like "int[f(x)]dx = F(x)". Numerically, you would have something like "int[a,b][f(x)]dx = F(b) - F(a) = (some number)". If this is all news to you, then you might want to hire a tutor and spend a few weeks learning about "anti-differentiation" and "integration".

2) I would be very surprised if the assignment actually named both functions as "integral of x". But even if the assignment is that astonishingly badly written, let's give the functions sensible names, at least for the purposes of this thread. :wink:

. . . . .f(x) = x<sup>2</sup> + 2x + 1

. . . . .g(x) = 2x + 5

For further information, you might want to review "functions" and "function notation". (An hour or two of study should suffice.)

3) To find the area between two heights, one generally subtracts, rather than adding. For instance, if you wanted to find the area of wall to cut out between what will be the bottom of a window you're going to install and the top of that window, you wouldn't find the area below the window, the area from the top down to the floor, and then add, would you? You'd find the height of the top, the height of the bottom, and then subtract, right?, only then multiplying by the width. You need to do something very similar for this exercise.

4) Before you integrate, you might want to try graphing the two functions, to find which is "on top" and which is "underneath". (Study "graphing" "quadratic" and "linear" functions; there are loads of lessons online.) Once you know which is above and which is below, you'll know which to subtract from the other.

5) Also, to find the limits of integration, you will need to find the intersection of the two graphs. Setting each equal to zero will only tell you where each is equal to zero, which is not what you want. Instead, you want to find where they are equal to each other, so set them equal to each other, and solve the resulting quadratic equation. (Do a search for "solving quadratic" equations; there are loads of lessons on this, too.)

6) Once you have the correct subtraction, do that subtraction, and then simplify. (Search for "subtracting" "polynomials" or similar search terms.)

7) Once you have the correctly-simplified polynomial and the correct intersection points, you then have the correct integrand and limits of integration. Do the integration, and evaluate at the limits.

If the above doesn't make sense, please hire a tutor to work you through the various points. (A qualified tutor, working with you face-to-face, should be able to get you through this in only a few weeks.) Otherwise, please try the exercise again, following the various bits of advice provided, and referencing the graphic you've been given.

Thank you! :D

Eliz.
 
gluck said:
∫(x)= x^2 + 2x+ 1 ∫(x)= 2x+5
I see. Next time, just use "f"

f(x) = stuff

and don't reuse it.

g(x) = other stuff
 
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