Question

I have an integration problem that I would like solved. I'll specify it using the equation editor in a minute. The equation editor doesn't seem to appear until after I do a post.
Okay, updating this - I have the equation editor now.
\[F(x) = \int\limits_{2}^{x^2}(t^{1/3})\sin(t^{2}-3t)\]

Answer 1

Refresh the page if the equation editor isn't showing up.

Answer 2

Okay, here is the equation. I don't know why ... but I can't see the equation editor in the original problem post box. I can only see it as an option in the reply box. So here is the integral:
\[F(x) = \int\limits_{{2}}^{x^2}(t ^{\frac{ 1 }{ 3 }})\sin(t ^{2}-3t)dt\]
I've tried integration by parts (but it becomes more and more complex instead of simplifying) and I've tried by substitution but I can't solve it. Perhaps there was a mistake in the original exam paper or perhaps there is a solution method/technique that I just haven't spotted yet.

Answer 3

Mine is ugly one but there is an outlet.
\[u = t^{\frac{4}{3}} \rightarrow t ^2 = u^{\frac{3}{2}} \\and ~~ du= \frac{4}{3}t^{\frac{1}{3}}dt. \]

Answer 4

then after applying , you have to use formula sin(a-b)= sin a cos b - sinb cos a, to solve the problem. Good luck

Answer 5

don't forget convert t = u^ ( 3/4) to replace inside sin (t^2 -3t)

Answer 6

@Aragusha Are you sure you are not supposed to find the derivative?

Answer 7

just wondering about an integration ...... i agree that htis is prolly a derivative of F(x) problem tho
\[sin(u)=\sum_0\frac{(-1)^{n}}{(2n+1)!}u^{2n+1}\]
\[sin(t^2-3t)=\sum_0\frac{(-1)^{n}}{(2n+1)!}(t^2-3t)^{2n+1}\]
\[t^{(1/3)~}sin(t^2-3t)=\sum_0\frac{(-1)^{n}}{(2n+1)!}(t^2-3t)^{2n+4/3}\]
\[\sum_0\frac{(-1)^{n}}{(2n+1)!}(t^2-3t)^{2(n+2/3)}\]

Answer 8

even the wolf is having issues with it :)

Answer 9

Hi guys, thank you all for your input. The question states ..... state the Fundamental Theorem of calculus and use it to DIFFERENTIATE the following function w.r.t. x ....(the equation being the one listed above). But does this not still mean I have to integrate the right hand side?

Answer 10

it does not

Answer 11

\[\int_{a(x)}^{b(x)}f(t)~dt=F(b(x))-F(a(x))\]
\[\frac d{dx} \int_{a(x)}^{b(x)}f(t)~dt=\frac d{dx}F(b(x))-F(a(x))\]

Answer 12

using the chain rule we get back to
\[=\frac d{dx}F(b(x))-F(a(x))=f(b(x))b'(x)-f(a(x))a'(x)\]

Answer 13

notice that regardless of the integration, we still ahve all the parts we need to define the end results

Answer 14

\[\frac{d}{dx}F(x) = \frac{d}{dx}\int\limits_{2}^{x^2}(t^{1/3})\sin(t^{2}-3t)\to\\ ~~~~~~~~~([x^2]^{1/3})\sin([x^2]^{2}-3(x^2))*2x-([2]^{1/3})\sin([2]^{2}-3(2))*2'\]
\[2x(x^{2/3})\sin(x^4-3x^2)\]

Answer 15

Hi Amistre, thanks for this. I didn't understand the question properly but your solution makes sense to me. It's easier than I thought ... once you know how ... which I didn't. I really appreciate your input and that of others and apologies to all who spent time on this for posing the question in a misleading way originally.

Answer 16

youre welcome, and good luck ;)

Answer 17

Thanks.