I have the following limit:
$$\displaystyle \lim_{x \to \infty} \dfrac{\int_{0}^{2x}{xe^{x^2}dx}}{e^{4x^2}}$$
How do I evaluate the numerator? This is what I've tried doing:
$$\int_{0}^{2x}{xe^{x^2}dx}$$
$$=\dfrac{1}{2} \int_{0}^{2x}{e^{x^2}(2x.dx)}$$
$$=\dfrac{1}{2} \int_{0}^{2x}{e^{x^2}.d(x^2)}$$
$$=\dfrac{1}{2} \left[e^{x^2}\right]_{0}^{2x}$$
$$=\dfrac{1}{2} \left(e^{4x^2}-1\right)$$
Is this correct?
Also, is there a general way of evaluating the following (where $f(t)$ is just any real valued function and $n \in \mathbb Z$)?
$$\dfrac{d}{dx}\int_{0}^{x^n} {f(t).dt}$$
Do we have to do somehow evaluate the integral, apply the limits and only then take the derivative, or is there a different way of doing it?
(These are not 2 different questions, I'm asking about the derivative because I'm trying to use L'Hôpital's rule to find the limit.)
Best Answer
Yes your integral is correct. Then$$\lim_{x\to\infty}\frac{\int_0^{2x}te^{t^2}dt}{e^{4x^2}}=\frac{1}{2}\lim_{x\to\infty}\frac{e^{4x^2}-1}{e^{4x^2}}=\frac{1}{2}\lim_{x\to\infty}\left(1-\frac{1}{e^{4x^2}}\right)=\frac{1}{2}$$
For your last question, in general $$\frac{d}{dx}\int_0^{x^n}f(t)dt=nx^{n-1} f(x^n)$$
by the Fundamental Theorem of Calculus.