# Fundamental Theorem of Calculus 1 - definite integral

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?

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

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.

• 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 $u(x) = x^2$.
– A.S
Dec 19, 2014 at 19:01

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.

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:

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

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