Integral Calculus> I am unable to solve the following questi...

I am unable to solve the following questions in integration. Please help.

Anika Dogra , 4 Years ago

Grade 12

1 Answers

Abhishek Singh

This can be done using L’Hospital rule [here both numerator and denominator tend to 0 ]

Now numerator can be differentiated using Leibnitz rule

$\frac{\mathrm{d} }{\mathrm{d} x}\int_{0}^{x^2}sin(\sqrt{t})dt = \frac{\mathrm{d} }{\mathrm{d} x}(x^2).sin(\sqrt{x^2}) - 0$ $\frac{\mathrm{d} }{\mathrm{d} x}\int_{0}^{x^2}sin(\sqrt{t})dt = \frac{\mathrm{d} }{\mathrm{d} x}(x^2).sin(\sqrt{x^2}) - 0 = 2x.sin(x)$

Hence $\begin{align*} \lim_{x\rightarrow 0} \frac{\int_{0}^{x^2} sin(\sqrt{t})dt}{x^3} &=\lim_{x\rightarrow 0} \frac{2x.sin(x)}{3x^2} \\ &= \lim_{x\rightarrow 0} \frac{2}{3}*\frac{sin(x)}{x}\\ &= 2/3 \end{align*}$