Mathematics - Definite Integration Question with Solution | TestHub

(A) Let $g\left(x\right)=f\left(x\right)-3\cos 3x$

Now

${\int }_0^{\frac{\pi }{3}}g\left(x\right)dx={\int }_0^{\frac{\pi }{3}}f\left(x\right)dx-3{\int }_0^{\frac{\pi }{3}}\cos 3xdx$

$=0-{\left[\sin 3x\right]}_0^{\frac{\pi }{3}}=0$

Hence $g\left(x\right)=0$ has a root in $\left(0,\frac{\pi }{3}\right)$

(B) Let $h\left(x\right)=f\left(x\right)-3\sin 3x+\frac{6}{\pi }$

Now

${\int }_0^{\frac{\pi }{3}}h\left(x\right)dx={\int }_0^{\frac{\pi }{3}}f\left(x\right)dx-3{\int }_0^{\frac{\pi }{3}}\sin 3xdx+{\int }_0^{\frac{\pi }{3}}\frac{6}{\pi }dx$

$=0-{\left[-\cos 3x\right]}_0^{\frac{\pi }{3}}+{\left[\frac{6x}{\pi }\right]}_0^{\frac{\pi }{3}}$

$=0-2+2=0$

Hence $h\left(x\right)=0$has a root in $\left(0,\frac{\pi }{3}\right)$

(C) $\underset{x\to 0}{\lim }\frac{x{\int }_0^{x}f\left(t\right)dt}{1-e^{x^2}}$

$=\underset{x\to 0}{\lim }\left(\underset{-1}{\underbrace{\left(\frac{x^2}{1-e^{x^2}}\right)\times }}\underset{\mathrm{Apply} {\mathrm{L}}^{\text{'}}\mathrm{hospital} \mathrm{rule}}{\underbrace{\left(\frac{{\int }_0^{x}f\mathit{\left(}t\mathit{\right)}dt}{x}\right)}}\right)$

$=-1\underset{x\to 0}{\lim }\frac{f\left(x\right)}{1}=-1$

(D) $\underset{x\to 0}{\lim }\frac{\left(\sin x\right){\int }_0^{x}f\left(t\right)dt}{x^2}$

$=\underset{x\to 0}{\lim }\left(\underset{1}{\underbrace{\left(\frac{\sin x}{x}\right)}}\times \underset{\mathrm{Apply} {\mathrm{L}}^{\text{'}}\mathrm{hospital} \mathrm{rule}}{\underbrace{\left(\frac{{\int }_0^{x}f\mathit{\left(}t\mathit{\right)}dt}{x}\right)}}\right)$

$=1\underset{x\to 0}{\lim }\frac{f\left(x\right)}{1}=1$