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Evaluate this definite integral 20x + 3x^2 - (2/3)x^3 from [-2, 5] ?

by Chandasouk
Tags: 2 or 3x3, definite, evaluate, integral
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Chandasouk
#1
Jun16-10, 02:05 PM
P: 165
1. The problem statement, all variables and given/known data

I get the answer 19/3 but the real answer is 343/3. Can you show work to how it is that? I did this many times and still came up with 19/3 instead

I got up to evaluating the integral part of finding the area between two curves but can never get the correct answer. I have a feeling I'm messing up signs somewhere.
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Mark44
#2
Jun16-10, 02:19 PM
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Are you sure you copied the problem correctly? I don't get either of your answers. It would be helpful for you to show what you did.
Chandasouk
#3
Jun16-10, 05:44 PM
P: 165
Yeah.

Let f(x) = 20 + X + X2

Let g(x) = X2-5X

Find the area of the region enclosed by the two graphs.

The graphs intersect at X = -2 and 5, so those are the bounds. F(x) is the upper curve.

So I take the definite integral of

[(20 + X - X^2) - (X^2-5X)dx] from [-2, 5]

The answer is 343/3, but I do not get that answer when I compute the integral at all

Mark44
#4
Jun16-10, 06:27 PM
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P: 21,311
Evaluate this definite integral 20x + 3x^2 - (2/3)x^3 from [-2, 5] ?

Quote Quote by Chandasouk View Post
Yeah.

Let f(x) = 20 + X + X2

Let g(x) = X2-5X

Find the area of the region enclosed by the two graphs.

The graphs intersect at X = -2 and 5, so those are the bounds.
No, they intersect at only a single point, and it's not at either of those values of x.
f(-2) = 20 -2 + 4 = 22
g(-2) = 4 + 10 = 14

f(5) = 20 + 5 + 25 = 50
g(5) = 25 - 25 = 0

If you set 20 + x + x2 = x2 - 5x, notice that you can subtract x2 from both sides, leaving you with a linear equation - an equation that has only one solution.

I don't see that the two graphs enclose any region, so I don't know how you can do this problem. Are you sure you have all the information straight?
Quote Quote by Chandasouk View Post
F(x) is the upper curve.

So I take the definite integral of

[(20 + X - X^2) - (X^2-5X)dx] from [-2, 5]

The answer is 343/3, but I do not get that answer when I compute the integral at all
Redbelly98
#5
Jun16-10, 06:34 PM
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Quote Quote by Chandasouk View Post
Yeah.

Let f(x) = 20 + X + X2

Let g(x) = X2-5X

Find the area of the region enclosed by the two graphs.
The expression posted in the thread title contains an x3 term, so this isn't even the same problem.

If you post the actual problem statement, exactly as it is written, that would be a big help.
Chandasouk
#6
Jun16-10, 06:59 PM
P: 165
The problem in my Calc book just stated


Let f(x) = 20 + X - X^2

Let g(x) = X^2-5X

Find the area of the region enclosed by the two graphs.

I set these two equal to each other

20 + X + X^2 = X^2-5X

Through Algebra, I found that

0 = 2X^2 - 6X -20

which factors to

2(X-5)(X+2)

Thus the curves intersect at X = -2 and X = 5. I made a mistake earlier saying that the graphs intersected at those two points; sorry.

f(x) is the upper curve. From there they just used used the formula for calculating the area between curves and obtained the answer 343/3 but i cannot work out the math to get that
HallsofIvy
#7
Jun16-10, 07:39 PM
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Quote Quote by Chandasouk View Post
The problem in my Calc book just stated


Let f(x) = 20 + X + X^2

Let g(x) = X^2-5X

Find the area of the region enclosed by the two graphs.

I set these two equal to each other

20 + X + X^2 = X^2-5X

Through Algebra, I found that

0 = 2X^2 - 6X -20
No, no, no! The two [itex]x^2[/itex] terms cancel they don't add!

which factors to

2(X-5)(X+2)

Thus the curves intersect at X = -2 and X = 5. I made a mistake earlier saying that the graphs intersected at those two points; sorry.

f(x) is the upper curve. From there they just used used the formula for calculating the area between curves and obtained the answer 343/3 but i cannot work out the math to get that
Chandasouk
#8
Jun16-10, 08:05 PM
P: 165
Aw damn, now I know why. Good catch. I wrote the equation for f(x) incorrectly. It was supposed to be

20 + X - X^2 not 2+ + X + X^2


Now what I wrote before makes sense
Susanne217
#9
Jun17-10, 06:11 AM
P: 317
Quote Quote by Chandasouk View Post
1. The problem statement, all variables and given/known data

I get the answer 19/3 but the real answer is 343/3. Can you show work to how it is that? I did this many times and still came up with 19/3 instead

I got up to evaluating the integral part of finding the area between two curves but can never get the correct answer. I have a feeling I'm messing up signs somewhere.
If you have trouble getting the right anti-derivative then you can always try to compare your result with one from http://integrals.wolfram.com/index.j...3&random=false
Redbelly98
#10
Jun17-10, 07:28 AM
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Quote Quote by Chandasouk View Post
f(x) is the upper curve. From there they just used used the formula for calculating the area between curves and obtained the answer 343/3 but i cannot work out the math to get that
Looks like you have set the integral up correctly. Can you show the rest of your work? I am getting 343/3 as well.


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