Integration

[hel]

I was having difficulties integrating and answering this question... If anyone could help, I would really appreciate it!

[dessskris]

was the equation of the graph of f(x) given? like in part (a) or somewhere in the question? I don't know how to solve without it but

k⁴-6k²+8=0

(k²-4)(k²-2)=0

k= -2, 2, -√2, √2

Since 2<k<1, k=-√2

Therefore

∫_-2^-√2 (f(x) -3x+2) dx = 2k+4

∫_-2^-√2 (f(x) -3x+2) dx = 4-2√2

Then... I don't know how to proceed. I'm working backwards btw.

[hel]

was the equation of the graph of f(x) given? like in part (a) or somewhere in the question? I don't know how to solve without it but

k⁴-6k²+8=0

(k²-4)(k²-2)=0

k= -2, 2, -√2, √2

Since 2<k<1, k=-√2

Therefore

∫_-2^-√2 (f(x) -3x+2) dx = 2k+4

∫_-2^-√2 (f(x) -3x+2) dx = 4-2√2

Then... I don't know how to proceed. I'm working backwards btw.

that's quite useful, thanks. the graph of f(x)=x^3.

You can see the whole question in N10 P1 TZ0, Q10.

https://mr-t-tes.wikispaces.com/file/view/Nov+2010+Mathematics+SL+paper+1+QP.pdf

Edited May 1, 2011 by hel

[jess1ca]

Same here, I'm getting every term correct except the term in k.

EDIT: I see what I did wrong, I accidentally added an extra negative sign somewhere. Did you figure it out?

I think the most foolproof way to solve this is to find the area under 3x-2, then find the area under x^3 separately, and then subtract them. Careless mistakes always come from combining too many steps!

Edited May 3, 2011 by SmilingAtLife:)

[pap6r]

if you still need the solution, here~

(sorry the pic's so big)

Edited May 4, 2011 by pap6r