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

Chandasouk

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.

Mark44

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

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

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?

Redbelly98

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

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

No, no, no! The two [itex]x^2[/itex] terms cancel they don't add!

Chandasouk

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

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

Looks like you have set the integral up correctly. Can you show the rest of your work? I am getting 343/3 as well.