How we can solve that $\lim _{_{x\rightarrow \infty }}\int _0^x\:e^{t^2}dt$?

Question

How we can solve that $\lim _{_{x\rightarrow \infty }}\int _0^x\:e^{t^2}dt$ ?

P.S: This is my method as I thought: $\int _0^x\:\:e^{t^2}dt>\int _1^x\:e^tdt=e^x-e$ which is divergent, so all your answers, helped me to think otherwise, maybe my method help something else :D

Answer 1

Since $$ e^{x^2}=1+x^2+\frac{x^4}{2!}+\dotsb $$ we have that, for $x\ge0$, $e^{x^2}\ge1+x^2$. So $$ \int_{0}^{x}e^{t^2}\,dt\ge\int_{0}^x(1+t^2)\,dt=x+\frac{x^3}{3} $$ Can you finish?

Answer 2

This function diverges extremely fast. Notably, $e^{t^2}$ is monotone increasing with limit $\infty$ as $t \to \infty$. Thus your integral diverges (and it get very, very large very, very quickly).

Answer 3

$$e^{t^2}> e^t\text{ from 1 to $\infty$, and the part from 0 to 1 is finite.}$$ and the integral$ \int_1^\infty e^t $ diverges. Therefore, by the comparison test, it diverges too.

Answer 4

Or just use $e^{x^2} \ge 1$ on $[0,\infty)$ to see $\int_0^x e^{t^2}dt \ge x \to \infty.$

Answer 5

$$ \begin{align} \int_0^xe^{t^2}\,\mathrm{d}t &\ge\int_0^x\frac tx\,e^{t^2}\,\mathrm{d}t\\ &=\frac1{2x}\left(e^{x^2}-1\right) \end{align} $$ As $x\to\infty$, the function on the right goes to $\infty$ extremely fast.