Jamia Millia Islamia MCA Definite Integration Previous Year Questions (PYQs) – Page 1 of 2

From $-1$ to $0$: $[x] = -1$ From $0$ to $1$: $[x] = 0$ So, $I = \int_{-1}^{0} x^2 e^{-1} dx + \int_{0}^{1} x^2 e^0 dx$ $= e^{-1}\int_{-1}^{0} x^2 dx + \int_{0}^{1} x^2 dx$ $= e^{-1}\left[\dfrac{x^3}{3}\right]{-1}^{0} + \left[\dfrac{x^3}{3}\right]{0}^{1}$ $= e^{-1}\left(\dfrac{1}{3}\right) + \dfrac{1}{3}$ $I = \dfrac{1}{3e} + \dfrac{1}{3}$

Solution: For $x \in [n,n+1)$, we have $[x]=n$. $\displaystyle \int_n^{n+1} e^{\,x-[x]}dx = e^{-n}\!\int_n^{n+1} e^{x}dx = e^{-n}(e^{n+1}-e^{n})=e-1.$ The interval $[0,1000)$ has $1000$ such unit pieces, so the total integral is $1000(e-1)$.

Solution: $\sin^{4}x\cos^{4}x=\big(\sin^{2}x\cos^{2}x\big)^2 =\left(\dfrac{\sin 2x}{2}\right)^{4} =\dfrac{1}{16}\sin^{4}2x.$ Thus $J=\displaystyle\int_{0}^{\pi/2}\sin^{4}x\cos^{4}x\,dx =\dfrac{1}{16}\!\int_{0}^{\pi/2}\!\sin^{4}2x\,dx =\dfrac{1}{32}\!\int_{0}^{\pi}\!\sin^{4}u\,du.$ Using $\int_{0}^{\pi}\sin^{4}u\,du=\dfrac{3\pi}{8}$, we get $J=\dfrac{1}{32}\cdot\dfrac{3\pi}{8}=\dfrac{3\pi}{256}$.

Let \(u=1-x\Rightarrow du=-dx\). Then \[ \int_{0}^{1}\frac{x}{(1-x)^{3/4}}dx =\int_{1}^{0}\frac{1-u}{u^{3/4}}(-du) =\int_{0}^{1}\left(u^{-3/4}-u^{1/4}\right)du = \left[4u^{1/4}-\frac{4}{5}u^{5/4}\right]_{0}^{1} =4-\frac{4}{5}=\frac{16}{5}. \] \(\boxed{\tfrac{16}{5}}\)

Required area $= \int_1^6 \dfrac{1}{x} \, dx = [\log_e x]_1^6 = \log_e 6 - \log_e 1 = \log_e 6.$

Given integral: $\int \dfrac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} \, dx$. We know $\sin 2x = 2\sin x \cos x$ and $1 + \sin 2x = (\sin x + \cos x)^2$. So, $\sqrt{1 + \sin 2x} = |\sin x + \cos x|$. Hence, integrand becomes $\dfrac{\sin x + \cos x}{|\sin x + \cos x|} = 1$. Therefore, $\int dx = x + C$. $\boxed{\text{Answer: (B) } x}$

Let $y=1+\log t \Rightarrow t=e^{y-1}$, $dt=e^{y-1}dy$. After simplification: $x((1+\log x)^{2}-2(1+\log x)+2)-1$.

For $x>0$, $|x|=x$, so $\int x^{3}dx=\dfrac{x^{4}}{4}+C$.

Let $u=\tan x$, then $\sin x=\dfrac{u}{\sqrt{1+u^{2}}}$ and $du=\sec^{2}x\,dx$. $\int_{0}^{1}\dfrac{u}{\sqrt{1+u^{2}}}du=\left[\sqrt{1+u^{2}}\right]_{0}^{1}=\sqrt2-1$. Hence $a=-1$.

Let $u=1-x \Rightarrow du=-dx$. Integral $=\int_{0}^{1}(u^{-1/2}-u^{1/2})du=\left[2u^{1/2}-\frac{2}{3}u^{3/2}\right]_{0}^{1}=2-\frac{2}{3}=\frac{4}{3}$.