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integration c3

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part a) i worked out to be 0.7401
part b) i have turned the (sinx)^2 into 1-(cosx)^2 so am i right to do 1-0.7401 = 0.2599

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also for this e^(x^2) am i right to split it up into (e^3)*(e^(x^2)) ?
No because we have

0π3sin2x.dx0π31cos2x.dx=0π31.dx0π3cos2.dx\displaystyle \int_0^{\frac{\pi}{3}} \sin^2 x .dx \equiv \int_0^{\frac{\pi}{3}} 1 - \cos^2 x .dx = \int_0^{\frac{\pi}{3}}1 .dx - \int_0^{\frac{\pi}{3}} \cos^2 .dx

Hence we have 0π3cos2x.dx0π31.dx0π3sin2x.dx\displaystyle \int_0^{\frac{\pi}{3}} \cos^2 x .dx \equiv \int_0^{\frac{\pi}{3}} 1 .dx - \int_0^{\frac{\pi}{3}} \sin^2 x .dx

(though tbh, π31\frac{\pi}{3} \approx 1 so maybe this is the expected approach and hence you would then be correct)


As for the other question, indeed that's the correct split.
(edited 6 years ago)
May I ask where you got these C3 questions from? I'd like to check them out and do them myself.
Thank you
Reply 3
Original post by RDKGames
No because we have

0π3sin2x.dx0π31cos2x.dx=0π31.dx0π3cos2.dx\displaystyle \int_0^{\frac{\pi}{3}} \sin^2 x .dx \equiv \int_0^{\frac{\pi}{3}} 1 - \cos^2 x .dx = \int_0^{\frac{\pi}{3}}1 .dx - \int_0^{\frac{\pi}{3}} \cos^2 .dx

Hence we have 0π3cos2x.dx0π31.dx0π3sin2x.dx\displaystyle \int_0^{\frac{\pi}{3}} \cos^2 x .dx \equiv \int_0^{\frac{\pi}{3}} 1 .dx - \int_0^{\frac{\pi}{3}} \sin^2 x .dx

(though tbh, π31\frac{\pi}{3} \approx 1 so maybe this is the expected approach and hence you would then be correct)


As for the other question, indeed that's the correct split.


thanks, i always forget to integrate the constants!

i do the wjec board and they have every winter/summer papers from 2012 here

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