Integration

Year 12s: what will be the first way you research your future options? >>

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lgs98jonee

ok...can anybody clarify whether these are correct...as i got none of these answers :-(((

i must admit it was late so numerical mistakes may have crept in. work through the methods for youself to see. alternatively differentiate the answers and see if they work.

Reply 21

m:)ckel

i just did no.2 quickly, and i got:

(e^2x / 21)[3sin3x + 2cos2x]

which looks completely different to mathz's, but oh well...

Reply 22

mathz

mockel

i just did no.2 quickly, and i got:

(e^2x / 21)[3sin3x + 2cos2x]

which looks completely different to mathz's, but oh well...

y=(e^2x / 21)[3sin3x + 2cos2x]
y'=(2e^(2x)[3sin 3x+2cos2x]/21+{e^(2x)(9cos3x-4sin3x)/21
=e^2x[6sin3x+4cos2x+9cos3x-4sin2x}/21
which is not e^(2x).cos3x . follow the method i used. i showed working as well as the answers.

Reply 23

m:)ckel

mathz

yeah, i tried differentiating mine and saw that it was wrong

i just realised where i went wrong....i got to the penultimate line (of your working) and for the last term, wrote (4/3)I instead of (4/9)I