# Question about the Fundamental Theorem of Calculus with variable integration limit

If $$F(x)=\int_2^{x^3}\sqrt{t^2+t^4}dt$$

a.) The integral of $$F(x)$$ is $$3x^2\sqrt{x^2+x^4}$$.

b.) The derivative of $$F(x)$$ does not exist.

c.) $$F'(x)=3x^2\sqrt{x^6+x^{12}}$$.

I can't seem to find the answer. I found that $$F(x)=\int_2^{x^3}\sqrt{t^2+t^4}dt= \frac{(x^6+1)^{3/2}}{3}-\frac{5 \sqrt{5}}{3}$$

and $$\frac{d}{dx}F(x)=\int_2^{x^3}\sqrt{t^2+t^4}dt=3x^5\sqrt{x^6+1}$$

so, to me the answer is not provided. Any help would be much appreciated.

• Please think about why you think you found a closed form expression for $F$. That integral has no nice answer - the problem is really about the FTC. What you've written suggests that there's some algebra you got wrong, independent of calculus. Feb 21, 2019 at 23:37

Let's say $$G(x)$$ is an antiderivative of $$\sqrt{x^2+x^4}$$. (we know about its existence from the fundamental theorem of calculus). So now from Newton-Leibniz formula we have $$F(x)=G(x^3)-G(2)$$. Now if we differentiate by $$x$$ using the chain rule we get:

$$F'(x)=3x^2G'(x^3)=3x^2\sqrt{x^6+x^{12}}$$

• Thank you Mark! Feb 21, 2019 at 18:17

Hint: If $$F(x)= \int_a^{g(x)} f(t) dt$$ , then $$F’(x) = f(g(x))g’(x)$$

• Thank you Jose! Feb 21, 2019 at 18:17
• Your welcome, notice that with this method is not necessary to calculate an undefined integral, which is useful because is not always possible (see $\int e^{-x^2} dx$) @RyanPennell Feb 21, 2019 at 18:57

By the way, aside from the good answers provided by Mark and JoseSquare, note that you actually did compute $$F(x)$$ and $$F'(x)$$ correctly! The result you obtained for $$F'(x)$$ is equivalent to the expression given in choice (c).

• I believe the OP's answer for $F'(x)$ is only the same as choice (c) when $x\ge0$... Feb 22, 2019 at 2:08