Find $H'(2)$ given that $H(x) = \int_{2x}^{x^3-4} \frac{x}{1+\sqrt{t}}dt$


$H'(x) = \frac{d}{dx}\big(x\int_{2x}^{x^3-4} \frac{1}{1+\sqrt{t}}dt\big)$

= $\int_{2x}^{x^3-4} \frac{1}{1+\sqrt{t}} dt$ + $x\frac{d}{dx} \big(-\int_{1}^{2x} \frac{1}{1+\sqrt{t}}dt + \int_{1}^{x^3-4} \frac{1}{1+\sqrt{t}} dt\big)$

= $\int_{2x}^{x^3-4} \frac{1}{1+\sqrt{t}} dt + x\big(-\frac{2}{1+\sqrt{2x}} + \frac{3x^2}{1+\sqrt{x^3-4}} \big)$

$H'(2) = \int_{4}^{4} \frac{1}{1+\sqrt{t}}dt + (2)\big( -\frac{2}{1+\sqrt{4}} + \frac{12}{1+\sqrt{4}}\big)$

= $0 + \frac{20}{3} = \frac{20}{3}$

Is this right? Also another question: If I was given the same derivative of this but the first part of $H'(x)$ being : $\int_{2x}^{x^3-4} \frac{1}{1+\sqrt{t}} dt$ . How would I solve $H'$(any integer) if it didn't have the same a and b values, do I just use FTOC pt 1?

Also I have no idea If I could seperate the integrals using union interval property at 1. I just saw a textbook solution of another question do that and I just imitated it on this question.


Let's say I have 2 functions, $h(t)$ and $g(t)$ and $\int f(x)\text{d}x=F(x)$ so by the chain rule we can prove that $\frac{d}{dt}\int_{h(t)}^{g(t)}f(x)\text{d}x=\frac{d}{dt}\left (F(h(t))-F(g(t))\right )=f(h(t))*h'(t)-f(g(t))*g'(t)$ you do not need to split it up all though you can in the future, it can all be done in one step. For example $\int_{x^2}^{x^3-1}\frac{1}{t}\text{d}t=\frac{2x}{x^2}-\frac{3x^2}{x^3-1}$. And yes it appears as though everything is correct

  • $\begingroup$ Could you show me how this could all be done in one step? $\endgroup$ – user349557 Feb 4 '17 at 22:52
  • $\begingroup$ @user349557 see my edit $\endgroup$ – Teh Rod Feb 4 '17 at 22:56
  • $\begingroup$ @positrón0802 my bad yeah $\endgroup$ – Teh Rod Feb 4 '17 at 22:59

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