TheLivingMartyr Posted August 21, 2011 Posted August 21, 2011 I'm not fantastic at integration, but I can integrate any number of polynomials pretty easily. I was given the equation "y=x(3-x)" to integrate. so I naturally treated it as a factorised quadratic, expanded it, and did: Int.(-x2 + 3x)dx = -1/3x3 + 3/2x2 I therefore assumed that: Int.(-x2 + 3x)dx = Int.(x(3 - x))dx It turns out i was wrong, and that I can't just expand these brackets, as the actual answer is (according to an integral calculator): -((2x3 - 9x2)/6) I need to know why I can't just expand these brackets and integrate, and how to integrate a factorised quadratic like this one. Please help me I also don't think a "u" substitution would work any differently, although i'm not particularly knowledgeable when it comes to those, soooo...
DrRocket Posted August 21, 2011 Posted August 21, 2011 I'm not fantastic at integration, but I can integrate any number of polynomials pretty easily. I was given the equation "y=x(3-x)" to integrate. so I naturally treated it as a factorised quadratic, expanded it, and did: Int.(-x2 + 3x)dx = -1/3x3 + 3/2x2 I therefore assumed that: Int.(-x2 + 3x)dx = Int.(x(3 - x))dx It turns out i was wrong, and that I can't just expand these brackets, as the actual answer is (according to an integral calculator): -((2x3 - 9x2)/6) I need to know why I can't just expand these brackets and integrate, and how to integrate a factorised quadratic like this one. Please help me I also don't think a "u" substitution would work any differently, although i'm not particularly knowledgeable when it comes to those, soooo... -1/3x3 + 3/2x2 = -((2x3 - 9x2)/6)
ajb Posted August 21, 2011 Posted August 21, 2011 [math]-\frac{(2x^{3}- 9 x^{2})}{6} = \frac{3}{2}x^{2} - \frac{1}{3}x^{3}[/math]
TheLivingMartyr Posted August 21, 2011 Author Posted August 21, 2011 so it does.... hmmm, now i just have to work out why I was getting the wrong answer when working out the area under it. Thankyou the thing is if I evaluate the integral: "-1/3x3 + 3/2x2" with the values x=0 and x=3 then i get a different area to if I evaluate the integral: "-((2x3 - 9x2)/6)" with the same values. I realise that the fractions are equal, but why then do i get different areas? I'm at a loss
Fuzzwood Posted August 21, 2011 Posted August 21, 2011 (edited) I don't get different areas. what are the answers you are getting? Answer should be 9/2 Edited August 21, 2011 by Fuzzwood
TheLivingMartyr Posted August 21, 2011 Author Posted August 21, 2011 oh..... fiddlesticks -.-" Simple syntax mistake on my calculator caused this whole mishap. I put "3*32" as equalling "92" instead of "3*9" I haven't made many posts here, but I'm pretty sure I've made myself look like an utter fool :')
mississippichem Posted August 21, 2011 Posted August 21, 2011 oh..... fiddlesticks -.-" Simple syntax mistake on my calculator caused this whole mishap. I put "3*32" as equalling "92" instead of "3*9" I haven't made many posts here, but I'm pretty sure I've made myself look like an utter fool :') No worries. Show me a scientist or mathematician who never makes mistakes, I'll show you a liar . If I had a dollar for every calculator syntax mistake I've made...well let's just say I wouldn't need any more grant money thats for sure.
TheLivingMartyr Posted August 22, 2011 Author Posted August 22, 2011 Hah, thanks! Anyway, it was nice to end the night having resolved my problem
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