C4 doubts

I had a go at one C4 question paper today and and I came across this question no 6:

http://www.school-portal.co.uk/GroupDownloadFile.asp?GroupId=1152243&ResourceId=3992062

Pls can you check if my working out is correct?
*integral sign* t(t+1) (-2t) dt (ON TOP OF *integral sign* I HAD 0 AND ON BOTTOM ROOT 2)
*integral sign* (-2t^3 -2t^2) dt
[-2t^4/ 4 - 2t^3/ 3 ]

Then subbing in 0 gave 0. so 0- ({6+4sqrt2}/ 3)

I am asking if my method is OK because the MS is doing something different , they are removing the - sign and when they do this they are swapping the limits around? :frown:

Reply 1

Smaug123

15

It would be great if you could use LaTeX to write out maths.
You need a positive area, because it's an area. The reason you swap the limits is:

\int_a^b f = -\int_b^a f

, because it's a useful convention that works.

(edited 10 years ago)

Reply 2

ztibor

10

Original post by laurawoods

I had a go at one C4 question paper today and and I came across this question no 6:

http://www.school-portal.co.uk/GroupDownloadFile.asp?GroupId=1152243&ResourceId=3992062

Pls can you check if my working out is correct?
*integral sign* t(t+1) (-2t) dt (ON TOP OF *integral sign* I HAD 0 AND ON BOTTOM ROOT 2)
*integral sign* (-2t^3 -2t^2) dt
[-2t^4/ 4 - 2t^3/ 3 ]

Then subbing in 0 gave 0. so 0- ({6+4sqrt2}/ 3)
[ :frown:

Your correct substitution would be

0-((-6-4sqrt2)/3)=

\frac{6+4\sqrt{2}}{3}

C4 doubts

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