First slide

Evaluation of definite integrals

Question

$Evaluate lim_{x \to \infty} \frac{{(\int_{0}^{x} e^{x^{2}} d x)}^{2}}{\int_{0}^{x} e^{2 x^{2}} d x}$

Moderate

A

0
B

2
C

3
D

4

Solution

$Since e^{x^{2}} > 0, e^{2 x^{2}} > 0 in [0, x], where x > 0,$ $\int_{0}^{x} e^{x^{2}} d x and \int_{0}^{x} e^{2 x^{2}} d x \to \infty as x \to \infty$
$\begin{matrix} L = lim_{x \to \infty} \frac{{(\int_{0}^{x} e^{x^{2}})}^{2}}{\int_{0}^{x} e^{2 x^{2}}} is of the form \frac{\infty}{\infty} . \\ Therefore, using L' Hopital's rule \\ \begin{aligned} L & = lim_{x \to \infty} \frac{2 \int_{0}^{x} e^{x^{2}} d x \frac{d}{d x} \int_{0}^{x} e^{x^{2}}}{e^{2 x^{2}}} \\ = lim_{x \to \infty} \frac{2 e^{x^{2}} \int_{0}^{x} e^{x^{2}} d x}{{(e^{x^{2}})}^{2}} \\ = 2 lim_{x \to \infty} \frac{\int_{0}^{x} e^{x^{2}} d x}{e^{x^{2}}} (\frac{\infty}{\infty} form) \\ \begin{aligned} (Using L'Hopital Rule) \end{aligned} \\ = 2 lim_{x \to \infty} \frac{e^{x^{2}}}{2 x e^{x^{2}}} \end{aligned} \\ = lim_{x \to \infty} \frac{1}{x} = 0 \end{matrix}$

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