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

Anika Dogra , 3 Years ago
Grade 12
anser 1 Answers
Abhishek Singh

Last Activity: 3 Years ago

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*}
 

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