Why $\lim_{x \to \infty}\int_x^{x+1}\ e^{t^2} \, dt=\infty$?

Question

$$F(x)= \int_x^{x+1} e^{t^2}\,dt$$ why the $\lim_{x \to \infty}F(x)=$

$$\lim_{x \to \infty}\int_x^{x+1}\, e^{t^2}\,dt=\infty \text{?}$$

Answer 1

Note that $$ \int_{x}^{x+1}\ e^{t^2}{dt}\ge e^{x^2} $$ for $x\geq 0$ since the integrand is increasing on this interval.

Answer 2

Because\begin{align}\lim_{x\to\infty}\int_x^{x+1}e^{t^2}\,\mathrm dt&\geqslant\lim_{x\to\infty}\int_x^{x+1}e^t\,\mathrm dt\\&=\lim_{x\to\infty}e^{x+1}-e^x\\&=\lim_{x\to\infty}e^x(e-1)\\&=+\infty.\end{align}

Answer 3

Just for fun, here's another proof. We have $$ F(x) = \int_x^{x+1} f(t) dt, $$ with $f(t) = e^{t^2}$. By the Liebniz integral rule, we have $$ \frac{dF}{dx} = f(x+1) - f(x) = e^{x^2 + 2x + 1} - e^{x^2} = e^{x^2}(e^{2x+1} - 1). $$ Since $dF/dx$ is strictly positive and does not approach 0 as $x \to \infty$, $F(x) \to \infty$ as $x \to \infty$.

Answer 4

If $f'(x) > 0$ and $f(x) \to \infty$ as $x \to \infty$ then $\int_x^{x+1} f(t) dt \gt \int_x^{x+1} f(x) dt =f(x) \to \infty$.