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I have two problems, they're not from a book so I can't check the answer for one of them and the other I'm not sure on what to do. $$ {d\over dx}{\int^{1}_{x^{2}}} {\sqrt{t^{2}+1}} {dt} $$ $$=-{d\over dx} {\int^{x^2}_{0^{}}} {\sqrt{t^{2}+1}} {dt} $$ $$u=x^2, u'=2x $$ $$=-{\int^{u}_{0^{}}}{\sqrt{t^{2}+1}} *2x $$ $$=-2x \sqrt{x^{4}+1} $$ The other one is $${d\over dx} {\int^{x}_{-x}} {\sqrt{1+t^2}}dt$$ Can I go about it the same way or does the $\int^{x}_{-x}$ require me to do an extra step?

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  • $\begingroup$ You will have another term, yes. Hint: $\int_{-x}^x = \int_0^x + \int_{-x}^0$. $\endgroup$
    – BaronVT
    Dec 19, 2014 at 18:53

3 Answers 3

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I don't understand the passages you did in the first exercise. However, if for the second your problem is having two $x$ as the extrema of integration, you can always use the linearity of the integral and the derivative to split it as

$$\dfrac{d}{dx}\int^{x}_{-x}f(t)dt=\dfrac{d}{dx}\left(\int^{x}_{0}f(t)dt+\int^{0}_{-x}f(t)dt\right)$$

Then you apply the fundamental theorem of calculus to both parts.

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    $\begingroup$ To explain the passages: the OP is taking $F(x^2) = \int_0^{x^2} f(t)dt$ in which case $\frac{d}{dx} -F(x^2) = -f(u) \cdot u'(x)$ with $u(x) = x^2$. $\endgroup$
    – A.S
    Dec 19, 2014 at 19:01
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Your answer is correct. For the second problem, it's probably best if you split up the integral as follows:

$$\int_{-x}^x f(t) dt = \int_{-x}^0 f(t) dt + \int_0^x f(t)dt$$

You can use the exact same method that you used for the first problem now.

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the integral $\int_{-x}^x \sqrt{1+t^2} \ dt= 0$ for all $x,$ so is the derivative. i made a mistake. i wanted to believe $\sqrt{1+x^2}$ is odd.

edit in response to the comment by user Urgye:

$ d \left( \int_{-x}^x \sqrt{1+t^2} \ dt \right)= 2 d \left(\int_0^x \sqrt{1+t^2}\ dt\right)=2\sqrt{1+x^2}\ dx.$

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  • $\begingroup$ It seems to me that the integrand is positive and at least 1 so the integral is at least $2x$. $\endgroup$
    – Urgje
    Dec 19, 2014 at 20:35
  • $\begingroup$ @Urgye, yes. i made a mistake. i will edit is. $\endgroup$
    – abel
    Dec 19, 2014 at 20:41
  • $\begingroup$ @Urgye, thanks for the gentle reminder. yes, i made a mistake. i will edit my answer to reflect your comment. $\endgroup$
    – abel
    Dec 19, 2014 at 20:47

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