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C4 Question

Hey. I really need some help on this Q.

1)A curve has the gradient function k(2x-1)² and passes through the origin and the point (1,2). FInd the equation of the curve.

Thnx
Reply 1
Avatar for Srathmore
Srathmore
OP
11
Sorry, i have another question. It's attached. Basically just to find the area of the shaded parts. Thanks
graph.bmp259.8KB
Reply 2
Avatar for yusufu
yusufu
16
integrating gives y = k/6 (2x-1)3 + C

thereafter use simultaneous equations with the points given

--------------

Srathmore
Sorry, i have another question. It's attached. Basically just to find the area of the shaded parts. Thanks

what is the limit on the right hand side? between 3 and ??
Reply 3
Avatar for Srathmore
Srathmore
OP
11
Undry1
integrating gives y = k/6 (2x-1)3 + C

thereafter use simultaneous equations with the points given

--------------


what is the limit on the right hand side? between 3 and ??


4, sorry.
Reply 4
Avatar for Srathmore
Srathmore
OP
11
Undry1
integrating gives y = k/6 (2x-1)3 + C

thereafter use simultaneous equations with the points given

That is exactly what i did.

k/6(0-1)³ +c = 0
k/6(1-1)³ +c = 0

I got -k/6 +c =0, and c=2

, and that doesn't give the right answer.
Reply 5
Avatar for Srathmore
Srathmore
OP
11
Undry1
integrating gives y = k/6 (2x-1)3 + C

thereafter use simultaneous equations with the points given

That is exactly what i did.

k/6(0-1)³ +c = 0
k/6(1-1)³ +c = 2

I got -k/6 +c =0, and c=2

, and that doesn't give the right answer.
Reply 6
Avatar for yusufu
yusufu
16
integrating gives y = k/6 (2x-1)3 + C

thereafter use simultaneous equations with the points given

That is exactly what i did.

k/6(0-1)³ +c = 0
k/6(1-1)³ +c = 2 herein lies your error. equation is 2x-1 not x-1


I got -k/6 +c =0, and c=2
, and that doesn't give the right answer.
Reply 7
Avatar for Srathmore
Srathmore
OP
11
Undry1
integrating gives y = k/6 (2x-1)3 + C

thereafter use simultaneous equations with the points given

That is exactly what i did.

k/6(0-1)³ +c = 0
k/6(1-1)³ +c = 2 herein lies your error. equation is 2x-1 not x-1


I got -k/6 +c =0, and c=2
, and that doesn't give the right answer.


Sorted. Any luck on the other one?
Reply 8
Avatar for Srathmore
Srathmore
OP
11
Bump.
Reply 9
Avatar for yusufu
yusufu
16
Srathmore
Sorted. Any luck on the other one?

∫(x-2)-2 -1 dx between 3 and 4

= [ -(x-2)-1 - x]
= [-2 -4] - [-1 - 3] = -2 = 2

∫(x-2)-2 -1 dx between -2 and 0

= [ -(x-2)-1 - x]
= [-(-4) +2] - [-(-2) + 0]
= 4

adding areas gives 6
Reply 10
Avatar for Srathmore
Srathmore
OP
11
Undry1
∫(x-2)-2 -1 dx between 3 and 4

= [ -(x-2)-1 - x]
= [-2 -4] - [-1 - 3] = -2 = 2

∫(x-2)-2 -1 dx between -2 and 0

= [ -(x-2)-1 - x]
= [-(-4) +2] - [-(-2) + 0]
= 4

adding areas gives 6


Snap.

The answer says 1.75? So are we agreed that the asnwers are wrong?
Reply 11
Avatar for yusufu
yusufu
16
Srathmore
Snap.

The answer says 1.75? So are we agreed that the asnwers are wrong?

if you look at the graph, when x is 0 y is -1.25
the area between 0 and -2 must be greater than 2*1.25
so i would say the answers are wrong. :smile:
Reply 12
Avatar for Srathmore
Srathmore
OP
11
Did it again, and got 2.25 as the answer?
Reply 13
Avatar for yusufu
yusufu
16
Srathmore
Did it again, and got 2.25 as the answer?

who did it again?
Reply 14
Avatar for Srathmore
Srathmore
OP
11
Undry1
who did it again?


Me. you subbed in the numbers wrong.
Reply 15
Avatar for yusufu
yusufu
16
which numbers?
Reply 16
Avatar for yusufu
yusufu
16
oh i see. yeah i did do it wrong. :redface: ill do it again.
Reply 17
Avatar for yusufu
yusufu
16
it seems i concur with their answer this time. 1.75

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