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Someone please help me with Calculus.
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I am literally destroying things in my room in my rage over this problem.
it is to find the integral from 1 to 0 of:
(x)*(e^2x).
I know you integrate by parts, but it just keeps going on forever when I try to figure it out.
PLEASE HELP ME.
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Its ((e^2)*(x^3))/3
I don't have a calc to work out the answer from here though, sorry
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^ dude your introuble if your struggling with this!! it gets a lot harder!! just wait till you start BC topics and shit. oh boy fond memories of calc...
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i think that is wrong. the indefinat intagral is (1/2)Xe^(2x)-(1/4)e^2x
evaluated from 0 to 1 =(1/4)e^2+(1/4) = 2.097
u = x dv = e^(2x)dx
du = dx v = (1/2)e^(2x)
uv- int (v du)= (1/2)Xe^(2x)+int (.5 e^2x) = (1/2)Xe^(2x)+ 1/4 e^2x
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my eyes just started to bleed.
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are you intergrating from 1 to zero or zero to 1? I think calc only gets easier. After you learn the basics calc 2 I guess is alittle harder but 3 and diffy q are cake.
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diff eq is the hardest i think, multyvarible is the easyest
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I am almost 100% positive that the intergral works out to be (xe^2x)-e^2x. I'll let you plug in the numbers yourself, this should be a good enough start. If you need to know how to do it, its /fg'=fg-/f'g
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you forgot the coeficents 1/2 and 1/4
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close enough for how late it is. I'm not a big fan of calculus anywas, but thanks for pointing out my mistake. Never hurts to learn more.
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good memories of first year engineering, haha
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i would help if i were in that class
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calc ab is hard enough for me i can't do it and my teacher explains nothing emo is my only choice now i hate everything
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find the anti-derivative, then do g(1)-g(0) (fundamental theorem of Calculous)
All times are Eastern (-5)
