To be honest that doesn't look like an integral that could be given to a 2U student without first prompting them to differentiate . I might be missing something but I don't see how you could do that using only 2U methods.
how does one integrate
∫x(x+2)e^x dx
answer ends up x^2e^x + c
thanks in advance
To be honest that doesn't look like an integral that could be given to a 2U student without first prompting them to differentiate . I might be missing something but I don't see how you could do that using only 2U methods.
Last edited by fan96; 15 Mar 2018 at 8:08 PM.
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
_{1}∫^{(3√3)} t^{2} dt cos(3π/9) = log(^{3}√e) | Integral t^{2} dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.
it's actually the second part of a question,
it originally asked us to differentiate x^2e^x
and then hence find ∫x(x+2)e^x dx , i knew the answer should be x^2e^x because that was basically given to us in the first part, but getting to it was what i was stuck with.
but thank you
Oh. Just expand the integrand then.
You'll get , and then the integration should be easy.
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
_{1}∫^{(3√3)} t^{2} dt cos(3π/9) = log(^{3}√e) | Integral t^{2} dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.
Well anyway
My wrking is not 2u but if u r interested in the working out here it is :
hey, what textbook did you get this question from
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