- #1

- 23

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I evaluated the integral from 0 to 2 of x^2 - -x

I got 14/3

Is this correct?

- Thread starter kenny87
- Start date

- #1

- 23

- 0

I evaluated the integral from 0 to 2 of x^2 - -x

I got 14/3

Is this correct?

- #2

- 68

- 0

are you sure the function is y=-x?

- #3

- 23

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yeah, i just double checked them. y=x^2 and y= -x

- #4

- 68

- 0

I setup the problem, and evaluated the integral with respect to y

and got 1/6.

i said, integral from y=1 to y = 0 and y^(.5)-y dy

- #5

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I know 1/6 isnt the correct answer because its not one of the choices...

the choices are

2/3, 8/3, 4, 14/3, and 16/3

these are ap problems and i know they try to trick you though, so im really not sure if my answer is correct or not... i might have made up some math

- #6

- 68

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and when you put the limits in the answer should be 2/3.

(2)^3/3-2^2/2 = 2/3

- #7

- 23

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what i did:

1/3*x^3 - 1/2*x^2 and then evaluated it at 2 and 0.... what did i do wrong?

- #8

- 68

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lets call f(x)=x^2 and g(x)=-x

then integral from f(x)-g(x) = x^2+x

integrate that function

and we get x^3/+x^2/2

plug in the limits

and we get 14/3..

sorry for the wrong answer sheesh.. i feel dumb.

- #9

- 23

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no problem. thank you so much!

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