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

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Homework Help Overview

The problem involves evaluating a definite integral of the function 20x + 3x² - (2/3)x³ over the interval [-2, 5]. Participants are discussing the correct setup and evaluation of this integral, as well as the area between two curves defined by related functions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Some participants express confusion over the correct evaluation of the integral and the area between curves, with one participant stating they consistently arrive at an incorrect answer of 19/3 instead of the expected 343/3.
  • There are discussions about the correct formulation of the functions involved, with participants questioning the intersection points and the setup of the integral.
  • Participants also suggest checking the problem statement for accuracy and ensuring the correct functions are being used in the calculations.

Discussion Status

The discussion is ongoing, with participants providing insights into potential errors in setup and calculations. Some have offered to help clarify the integral evaluation process, while others are still trying to reconcile their results with the expected answer.

Contextual Notes

There is mention of a possible misunderstanding regarding the functions involved, as well as the need for clarity on the problem statement. Participants are also reflecting on the signs and terms used in their calculations, indicating a potential source of error.

Chandasouk
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Homework Statement



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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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.
 
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
 
Chandasouk said:
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?
Chandasouk said:
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
 
Chandasouk said:
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.
 
The problem in my Calc book just statedLet 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
 
Last edited:
Chandasouk said:
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
 
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^2Now what I wrote before makes sense
 
Chandasouk said:

Homework Statement



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.jsp?expr=20x+++3x^2+-+(2/3)x^3&random=false
 
  • #10
Chandasouk said:
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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