Let g(x) = int_0^{|x|^{3/4}} t^{2/3} sin frac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We are given $g(x) = \int_0^{|x|^{3/4}} t^{2/3} \sin \frac{1}{t} \, dt$. Since the integrand $f(t) = t^{2/3} \sin \frac{1}{t}$ is bounded near $t=

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) We are given $g(x) = \int_0^{|x|^{3/4}} t^{2/3} \sin \frac{1}{t} \, dt$. Since the integrand $f(t) = t^{2/3} \sin \frac{1}{t}$ is bounded near $t=0$ (because $|\sin(1/t)| \leq 1$,so $|f(t)| \leq t^{2/3}$),the integral exists. We want to evaluate $\lim_{x \rightarrow 0} \frac{g(x)}{x}$. Since $g(0) = \int_0^0 \dots = 0$,this is a $0/0$ form. Using the squeeze theorem or direct estimation: $|g(x)| = \left| \int_0^{|x|^{3/4}} t^{2/3} \sin \frac{1}{t} \, dt \right| \leq \int_0^{|x|^{3/4}} |t^{2/3} \sin \frac{1}{t}| \, dt \leq \int_0^{|x|^{3/4}} t^{2/3} \, dt$. Evaluating the integral: $\int_0^{|x|^{3/4}} t^{2/3} \, dt = \left[ \frac{t^{5/3}}{5/3} \right]_0^{|x|^{3/4}} = \frac{3}{5} (|x|^{3/4})^{5/3} = \frac{3}{5} |x|^{5/4}$. Thus,$\left| \frac{g(x)}{x} \right| \leq \frac{\frac{3}{5} |x|^{5/4}}{|x|} = \frac{3}{5} |x|^{1/4}$. As $x \rightarrow 0$,$\frac{3}{5} |x|^{1/4} \rightarrow 0$. Therefore,by the squeeze theorem,$\lim_{x \rightarrow 0} \frac{g(x)}{x} = 0$.

Let $g(x) = \int_0^{|x|^{3/4}} t^{2/3} \sin \frac{1}{t} \, dt$,for all real $x$. Then,$\lim_{x \rightarrow 0} \frac{g(x)}{x}$ is equal to

Similar Questions

If $f(x) = \begin{cases} \frac{6 x^2 + 1}{4 x^3 + 2 x + 3}, & 0 < x < 1 \\ x^2 + 1, & 1 \le x \le 2 \end{cases}$ then $\int_0^2 f(x) dx =$

The integral $\int_{1}^{e} \left( \left( \frac{x}{e} \right)^{2x} - \left( \frac{e}{x} \right)^{x} \right) \log_{e} x \, dx$ is equal to

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin (x-[x]) \, dx=$

If $f(x) = \int_{-1}^{x} |t| dt$,then for any $x \geq 0$,$f(x)$ is equal to

$\int_0^{\pi /2} e^x \sin x \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module