0

really stuck on this problem, my textbook doesn't have ANYTHING like it.

The only instruction is to find the dy/dx of the interval: $$y=x\int_2^{x^2}\sin(t^3)\,\mathrm dt $$

Thanks for any help!

Hagen von Eitzen
  • 374,180
Joe Caraccio
  • 163
  • 6
  • Please use MathJax to format problems into your question so you don't have to link to outside images. Putting the problem inside your question makes it just a bit easier for everyone else. Thanks! – Noble Mushtak Feb 02 '16 at 00:05
  • 1
    If we define $f(x)=\int_a^x g(t),\mathrm dt$ for some continuous $g$, do you know a theorem that tells you what $f'(x)$ is? Once you have that, what is the derivative of $x\mapsto xf(x^2)$? Saying your textbook has NOTHING like this is saying that you didn't cover the fundamental theorem of integration/differentiation, not the chain rule, and not the product rule for differentiation. – Hagen von Eitzen Feb 02 '16 at 00:06

1 Answers1

2

This may seem tricky at first but by the fundamental theorem of calculus, $$F(b)-F(a)=\int^b_a f(x)~\mathrm dx$$ and by the Second Fundamental Theorem of calculus $$F(x)=\int_a^xf(t)~dt \Rightarrow F'(x)=f(x).$$ So, in this case, all you are really doing is taking the derivative of the right-hand side. $$y(x)=x\int_2^{x^2}\sin(t^3)~\mathrm dt$$ $$y'(x)=\int_2^{x^2}\sin(t^3)~\mathrm dt+x\cdot(2 x \sin(x^6))$$ and I'll let you do the rest.

Graham Kemp
  • 129,094
Rob
  • 181
  • 4
  • thanks! this was really helpful. by the rest do you mean just evaluate the integral part of it too? Just not entirely sure what the 'rest' is – Joe Caraccio Feb 02 '16 at 01:03
  • Yes, he/she means for you to evaluate the integral. After that, you should have your answer. – Matt Feb 02 '16 at 01:11