Show that the function defined bellow is bounded $$f(x) = \int_{x^2}^{x^4}e^{-t^2}dt\quad x\in\mathbb{R}$$
Since $e^{-t^2}$ is positive for any $t$ in $\Bbb R$ then I know that $f(x) > 0$ but I don't know how to find an upper bound of $f(x)$.
How can I show that $f(x) = \int_{x^2}^{x^4}e^{-t^2}dt$ is bounded on $\mathbb{R}$?
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$\begingroup$
calculus
real-analysis
integration
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0HINT: It is enough to show that $\lim_{x \to \infty} f(x) = 0$, and that $f$ is continuous. – 2017-01-08
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0Be careful that $f(0)=0$ and $f(.5)<0$ be the convention of Riemann integral. – 2017-01-08
2 Answers
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Note that $$ 0\leq |f(x)|\leq\int_{\mathbb{R}}e^{-x^2}\ dx=\sqrt{\pi}<\infty. $$
where $\int_{\mathbb{R}}e^{-x^2}\ dx$ is the well known Gaussian integral.
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\begin{align*} \int_{x^{2}}^{x^{4}}e^{-t^{2}}dt & \leq\int_{-\infty}^{\infty}e^{-t^{2}}dt & \text{(}e^{x}\text{ is nonnegative)}\\ & =\int_{-\infty}^{\infty}\frac{1}{\sqrt{2}}e^{-\frac{u^{2}}{2}}dt & \text{(substitute }t=u/\sqrt{2}\text{)}\\ & =\sqrt{\pi}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{u^{2}}{2}}dt & \text{}\\ & =\sqrt{\pi}&\text{(integrand is PDF of normal distribution)} \end{align*}