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integration help needed!

hi :smile:
i have a C1 mock tomorrow and have being doing some questions and my weakest topic is integration. i'm doing this problem now from my textbook and every time i try i don't get the answer it says. it's probably me making a mistake but was wondering if anyone could check to see whether they got the textbook's answer?
basically its an area under a curve, and i know i cant draw it but you need to integrate -x^2 + x + 2. with the limits 2, and -1, then take them away from each other. the curve by the way has the equation y =(2-x)(1-x)
the answer is apparently 4.5 but i'm getting different answers each time :/
any help deeply appreciated!
ALSO quick question, when integrating an expression how do you know when to put +c on the end? some questions condone it when used so just wondering as i'm not sure.
thanks and sorry for long post!
Reply 1
Avatar for raheem94
raheem94
13
Original post by georgiaaaxo
hi :smile:

ALSO quick question, when integrating an expression how do you know when to put +c on the end? some questions condone it when used so just wondering as i'm not sure.


You have to always add +c except when evaluating a definite integral(i.e. in which the limits are given to you).
(edited 12 years ago)
Reply 2
Avatar for Ree69
Ree69
14
I thought I'd point out −x2+x+2≠(2−x)(1−x) -x^2 + x + 2 \neq (2-x)(1-x) so obviously something has gone wrong there.

You need a '+c' at the end of a solution to an integral if it's indefinite - i.e., if there's no limits.
Reply 3
Avatar for CJAW
CJAW
9
Firstly, you you only put a +c on the end if it is not between limits. So for the on you are doing, you don't need a + c because it is a definite integral (between limits)

The intergral should be -x^3/3 +x^2/2 + 2x
So if you stick the limits in you get:
(-8/3 + 2 + 4) - (-1/2 + 1/2 -2 )
= 16/3
Is that right?
Reply 4
Avatar for raheem94
raheem94
13
Original post by georgiaaaxo
hi :smile:

basically its an area under a curve, and i know i cant draw it but you need to integrate -x^2 + x + 2. with the limits 2, and -1, then take them away from each other. the curve by the way has the equation y =(2-x)(1-x)
the answer is apparently 4.5 but i'm getting different answers each time :/
any help deeply appreciated!


I get the answer as 4.5.
You have expanded the brackets wrongly,
(2−x)(1−x)=2−2x−x+x2=2−3x+x2 (2-x)(1-x) = 2 - 2x - x +x^2 = 2 - 3x + x^2
Reply 5
Avatar for georgiaaaxo
georgiaaaxo
OP
8
thankyou so much everyone! oops can't believe i expanded it wrongly :tongue: i'm so stupid haha. thankyou again!!!

Original post by raheem94
I get the answer as 4.5.
You have expanded the brackets wrongly,
(2−x)(1−x)=2−2x−x+x2=2−3x+x2 (2-x)(1-x) = 2 - 2x - x +x^2 = 2 - 3x + x^2




Original post by raheem94
You have to always add +c except when the evaluating a definite integral(i.e. in which the limits are given to you).




Original post by Ree69
I thought I'd point out −x2+x+2≠(2−x)(1−x) -x^2 + x + 2 \neq (2-x)(1-x) so obviously something has gone wrong there.

You need a '+c' at the end of a solution to an integral if it's indefinite - i.e., if there's no limits.




Original post by CJAW
Firstly, you you only put a +c on the end if it is not between limits. So for the on you are doing, you don't need a + c because it is a definite integral (between limits)

The intergral should be -x^3/3 +x^2/2 + 2x
So if you stick the limits in you get:
(-8/3 + 2 + 4) - (-1/2 + 1/2 -2 )
= 16/3
Is that right?




Original post by raheem94
I get the answer as 4.5.
You have expanded the brackets wrongly,
(2−x)(1−x)=2−2x−x+x2=2−3x+x2 (2-x)(1-x) = 2 - 2x - x +x^2 = 2 - 3x + x^2
(edited 12 years ago)
Reply 6
Avatar for georgiaaaxo
georgiaaaxo
OP
8
sorry to quote you all again but i just did it and i'm not getting 4.5 :frown:
when you expand the brackets i definitely get 2-3x + x^2.
i substitute in 2 and -1 separately but am obviously getting the wrong values. could anyone tell me what values they got when they subbed 2 and the -1 in?
thankyou!

Original post by raheem94
You have to always add +c except when the evaluating a definite integral(i.e. in which the limits are given to you).




Original post by Ree69
I thought I'd point out −x2+x+2≠(2−x)(1−x) -x^2 + x + 2 \neq (2-x)(1-x) so obviously something has gone wrong there.

You need a '+c' at the end of a solution to an integral if it's indefinite - i.e., if there's no limits.




Original post by CJAW
Firstly, you you only put a +c on the end if it is not between limits. So for the on you are doing, you don't need a + c because it is a definite integral (between limits)

The intergral should be -x^3/3 +x^2/2 + 2x
So if you stick the limits in you get:
(-8/3 + 2 + 4) - (-1/2 + 1/2 -2 )
= 16/3
Is that right?




Original post by raheem94
I get the answer as 4.5.
You have expanded the brackets wrongly,
(2−x)(1−x)=2−2x−x+x2=2−3x+x2 (2-x)(1-x) = 2 - 2x - x +x^2 = 2 - 3x + x^2
Reply 7
Avatar for CJAW
CJAW
9
Original post by georgiaaaxo
sorry to quote you all again but i just did it and i'm not getting 4.5 :frown:
when you expand the brackets i definitely get 2-3x + x^2.
i substitute in 2 and -1 separately but am obviously getting the wrong values. could anyone tell me what values they got when they subbed 2 and the -1 in?
thankyou!


Okay, so the integral is x^3/3 - 3x^2/2 + 2x

With the limits that gives (8/3 - 6 + 4) - (-1/3 - 3/2 -2) = 4.5 :smile:
Reply 8
Avatar for raheem94
raheem94
13
Original post by georgiaaaxo
sorry to quote you all again but i just did it and i'm not getting 4.5 :frown:
when you expand the brackets i definitely get 2-3x + x^2.
i substitute in 2 and -1 separately but am obviously getting the wrong values. could anyone tell me what values they got when they subbed 2 and the -1 in?
thankyou!


∫−12(2−3x+x2)dx=[2x−32x2+x33]−12 \displaystyle \int^2_{-1} (2-3x + x^2)dx = [2x -\frac32x^2 +\dfrac{x^3}{3}]^2_{-1}

I get the answer as 4.5.
Reply 9
Avatar for georgiaaaxo
georgiaaaxo
OP
8
Original post by CJAW
Okay, so the integral is x^3/3 - 3x^2/2 + 2x

With the limits that gives (8/3 - 6 + 4) - (-1/3 - 3/2 -2) = 4.5 :smile:




Original post by raheem94
∫−12(2−3x+x2)dx=[2x−32x2+x33]−12 \displaystyle \int^2_{-1} (2-3x + x^2)dx = [2x -\frac32x^2 +\dfrac{x^3}{3}]^2_{-1}

I get the answer as 4.5.


ah okay, think i have it now. thankyou sooooooo much! :biggrin:
Reply 10
Avatar for BigWilfo
BigWilfo
Original post by raheem94
You have to always add +c except when the evaluating a definite integral(i.e. in which the limits are given to you).


NOOOOOOOO

You only add +C when there are NO limits, ie/ an indefinite intergral!

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