Question

The value of $\underset{h\to 0}{\mathrm{Lim}}\frac{\underset{a}{\overset{x+h}{\int }}\frac{(\sin t{)}^{2014}}{t^{2013}+1}dt-\underset{a}{\overset{x}{\int }}\frac{(\sin t{)}^{2014}}{t^{2013}+1}dt}{h}$ is equal to

• A. $\frac{2013(\sin x{)}^{2013}}{2014(x{)}^{2013}+1}$
• B. $\frac{(\sin x{)}^{2015}}{(x{)}^{2014}+1}$
• C. $\frac{(\sin x{)}^{2014}}{x^{2013}+1}$
• D. $\frac{2014(\sin x{)}^{2013}}{2013\left(x^{2012}+1\right)}$

Answer 1

Answer: (C)

$\underset{h\to 0}{\mathrm{Lim}}\frac{\underset{a}{\overset{x+h}{\int }}\frac{(\sin t{)}^{2014}}{t^{2013}+1}dt-\underset{a}{\overset{x}{\int }}\frac{(\sin t{)}^{2014}}{t^{2013}+1}dt}{h}$
differentiate w.r.t. h
$\underset{h\to 0}{\mathrm{Lim}}\frac{\frac{(\sin (x+h){)}^{2014}}{(x+h{)}^{2013}+1}-0}{1}=\frac{(\sin x{)}^{2014}}{x^{2013}+1}$