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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

    Attachment 728202728208

    also for this e^(x^2) am i right to split it up into (e^3)*(e^(x^2)) ?
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    (Original post by ihatePE)

    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
    No because we have

    \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 \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, \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.
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    May I ask where you got these C3 questions from? I'd like to check them out and do them myself.
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    (Original post by RDKGames)
    No because we have

    \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 \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, \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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