Mathematics - Definite Integration Question with Solution | TestHub

(i) $I={\int }_0^{\pi /2}(\sin x{)}^{\cos x}\left(\cos x\cot x-\log (\sin x{)}^{\sin x}\right)dx={\int }_0^{\pi /2}\frac{d}{dx}(\sin x{)}^{\cos x}dx=1$ (ii) The point of intersection of $-4y^2=x$ and $x-1=-5y^2$ is $(-4,-1)$ and $(-4,1)$. Hence, required area $=2\left\{{\int }_0^1\left(1-5y^2\right)dy-{\int }_0^1-4y^{2}dy\right\}=\frac{4}{3}$. (iii) The point intersection of $y=3^{x-1}\log x$ and $y=x^x-1$ is $(1,0)$. Hence, $\frac{dy}{dx}=\frac{3^{x-1}}{x}+3^{x-1}\cdot \log 3\cdot \log x$ $\therefore {\left(\frac{dy}{dx}\right)}_{(1,0)}=1$ For $$\therefore {\left(\frac{dy}{dx}\right)}_{(1,0)}=1$$ If $\theta$ is angle between the curves, then $\tan \theta =0$. $$\because \theta =0^{\circ }$$ (iv) $$\begin{array}{rl}& \frac{dy}{dx}=\frac{2}{x+y}\Rightarrow \frac{dx}{dy}-\frac{x}{2}=\frac{y}{2}\\ & \Rightarrow x{e}^{-y/2}=\frac{1}{2}\cdot \int y\cdot e^{-y/2}dy\\ & \Rightarrow x+y+2=k{e}^{y/2}\\ & \Rightarrow k=3\\ & \therefore x+y+2=3e^{y/2}\\ & \end{array}$$ [passing through $(1,0)$ ]