The Student Room Group

Substitution for Definite Integrals

eq0034M.gif

A bit confused, where did the -198 and -70 numbers on the integral come from?
Reply 1
Original post by Airess3
eq0034M.gif

A bit confused, where did the -198 and -70 numbers on the integral come from?


substitution

u=2-8t2

t=3 u =...

t=5 u=...
(edited 9 years ago)
We have used the substitution u=2-8t^2 so you sub in t=3 to get the lower limit as 2-8(3)^2=-70 and similarly for the upper limit.
Thanks, i understand it now.
Would the answer have to be -0.26 or would it be fine by just leaving it as it is in the OP?
Reply 5
Original post by Airess3
Would the answer have to be -0.26 or would it be fine by just leaving it as it is in the OP?


What did the question say. If it doesn't specify anything, or specifies exact form, the OP answer is correct. If it asks for a certain about of dp or sf, you would need a decimal.
Original post by Phichi
What did the question say. If it doesn't specify anything, or specifies exact form, the OP answer is correct. If it asks for a certain about of dp or sf, you would need a decimal.


It just asked for me to compute the integral. So I guess the OP would be fine.
This is another question but still to do with the substitution of definite integrals. Is there a step(s) missing from the first to second line? I don't know how the equation changed so drastically?
Reply 8
Original post by Airess3
This is another question but still to do with the substitution of definite integrals. Is there a step(s) missing from the first to second line? I don't know how the equation changed so drastically?


Can you post images the right way up - I just dislocated my neck trying to read that :biggrin:

Where has that working come from? It's produced the right answer using a completely wrong piece of algebra!!

What it should say is that if x=tanθx = \tan \theta then

dx=sec2θdθdx = \sec^2 \theta d\theta

and 1+x2=sec2θ1 + x^2 = \sec^2 \theta

which simplifies things enormously :smile:
Original post by davros
Can you post images the right way up - I just dislocated my neck trying to read that :biggrin:

Where has that working come from? It's produced the right answer using a completely wrong piece of algebra!!

What it should say is that if x=tanθx = \tan \theta then

dx=sec2θdθdx = \sec^2 \theta d\theta

and 1+x2=sec2θ1 + x^2 = \sec^2 \theta

which simplifies things enormously :smile:


Haha, my lecturer did this example on the board.
Reply 10
Original post by Airess3
Haha, my lecturer did this example on the board.


You need to have a word!

Oh, and I've just noticed you've posted this question in another thread too - naughty! - and had the same explanation :smile:

Quick Reply