Aspire's Library

A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
  • Jamia Millia Islamia
    Previous Year Questions

Jamia Millia Islamia Previous Year Questions (PYQs)

Jamia Millia Islamia Definite Integration PYQ

Jamia Millia Islamia PYQ
Find the value of $I = \displaystyle\int_{-1}^{1} x^2 e^{[x]} dx$, where $[\,]$ denotes the greatest integer function.





Go to Discussion
Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2021 PYQ

Solution

From $-1$ to $0$: $[x] = -1$ From $0$ to $1$: $[x] = 0$ So, $I = \int_{-1}^{0} x^2 e^{-1} dx + \int_{0}^{1} x^2 e^0 dx$ $= e^{-1}\int_{-1}^{0} x^2 dx + \int_{0}^{1} x^2 dx$ $= e^{-1}\left[\dfrac{x^3}{3}\right]{-1}^{0} + \left[\dfrac{x^3}{3}\right]{0}^{1}$ $= e^{-1}\left(\dfrac{1}{3}\right) + \dfrac{1}{3}$ $I = \dfrac{1}{3e} + \dfrac{1}{3}$
Jamia Millia Islamia PYQ
$\displaystyle \int_{0}^{1000} e^{\,x-[x]}\,dx$ is –





Go to Discussion
Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2023 PYQ

Solution

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 $1000(e-1)$.
Jamia Millia Islamia PYQ
The value of $\displaystyle \int_{0}^{\pi/2}\sin^{4}x\,\cos^{4}x\,dx$ is –





Go to Discussion
Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2023 PYQ

Solution

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 $J=\dfrac{1}{32}\cdot\dfrac{3\pi}{8}=\dfrac{3\pi}{256}$.
Jamia Millia Islamia PYQ
\(\displaystyle \int_{0}^{1}\frac{x}{(1-x)^{3/4}}\,dx\) is equal to …





Go to Discussion
Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2017 PYQ

Solution

Let \(u=1-x\Rightarrow du=-dx\). Then \[ \int_{0}^{1}\frac{x}{(1-x)^{3/4}}dx =\int_{1}^{0}\frac{1-u}{u^{3/4}}(-du) =\int_{0}^{1}\left(u^{-3/4}-u^{1/4}\right)du = \left[4u^{1/4}-\frac{4}{5}u^{5/4}\right]_{0}^{1} =4-\frac{4}{5}=\frac{16}{5}. \] \(\boxed{\tfrac{16}{5}}\)
Jamia Millia Islamia PYQ
The area of the region bounded by the curve $y = \dfrac{1}{x}$, the x-axis, and between $x = 1$ to $x = 6$ is …… sq units.





Go to Discussion
Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2020 PYQ

Solution

Required area $= \int_1^6 \dfrac{1}{x} \, dx = [\log_e x]_1^6 = \log_e 6 - \log_e 1 = \log_e 6.$
Jamia Millia Islamia PYQ
$\displaystyle \int_{\frac{3\pi}{4}}^{\frac{7\pi}{4}} \dfrac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} \, dx$ is equal to





Go to Discussion
Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2020 PYQ

Solution

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, $\int dx = x + C$. $\boxed{\text{Answer: (B) } x}$
Jamia Millia Islamia PYQ
$\displaystyle \int_{1}^{x}(1+\log t)^{2}\,dt$ is equal to …





Go to Discussion
Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

Let $y=1+\log t \Rightarrow t=e^{y-1}$, $dt=e^{y-1}dy$. After simplification: $x((1+\log x)^{2}-2(1+\log x)+2)-1$.
Jamia Millia Islamia PYQ
If $x>0$, then $\displaystyle \int |x|^{3} dx$ is equal to …





Go to Discussion
Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

For $x>0$, $|x|=x$, so $\int x^{3}dx=\dfrac{x^{4}}{4}+C$.
Jamia Millia Islamia PYQ
$\displaystyle \int_{0}^{\frac{\pi}{4}}\sec^{2}x\sin x\,dx=a+\sqrt{2}$, find $a$





Go to Discussion
Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

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 $a=-1$.
Jamia Millia Islamia PYQ
$\displaystyle \int_{0}^{1}\frac{x}{(1-x)^{1/2}}dx$ is equal to …





Go to Discussion
Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

Let $u=1-x \Rightarrow du=-dx$. Integral $=\int_{0}^{1}(u^{-1/2}-u^{1/2})du=\left[2u^{1/2}-\frac{2}{3}u^{3/2}\right]_{0}^{1}=2-\frac{2}{3}=\frac{4}{3}$.
Jamia Millia Islamia PYQ
Evaluate the following integral: $ \int_{-2}^{2} \frac{3x^3 + 2|x| + 1}{x^2 + |x| + 1}\,dx $





Go to Discussion
Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2022 PYQ

Solution

Note $f(x) = \dfrac{3x^3 + 2|x| + 1}{x^2 + |x| + 1}$ is **odd + even combination**, so integrate separately. After simplifying using $|x|$ symmetry and limits $[-2,2]$, only even part contributes. Result $= 3\log 7$.
Jamia Millia Islamia PYQ
Evaluate the following integral: $ \int_{-\pi/2}^{\pi/2} \log\!\left(\frac{2-\sin x}{2+\sin x}\right)\,dx $





Go to Discussion
Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2022 PYQ

Solution

Let $f(x)=\log\!\left(\frac{2-\sin x}{2+\sin x}\right)$. Then $f(-x)=\log\!\left(\frac{2+\sin x}{2-\sin x}\right)=-f(x)$, so $f$ is odd. Integral over $[-\pi/2,\pi/2]$ of an odd function is $0$.
Jamia Millia Islamia PYQ
$\displaystyle \int_{0}^{\pi} \sin^2 x,dx =$





Go to Discussion
Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2024 PYQ

Solution

$\sin^2 x = \dfrac{1 - \cos 2x}{2} \Rightarrow \int_{0}^{\pi} \sin^2 x,dx = \dfrac{1}{2} \left[ x - \dfrac{\sin 2x}{2} \right]_{0}^{\pi} = \dfrac{1}{2}(\pi - 0) = \dfrac{\pi}{2}$
Jamia Millia Islamia PYQ
Value of $\displaystyle \int_{0}^{\frac{\pi}{2}} (x^3 + x\cos x + \tan^3 x + 1) dx$ is:





Go to Discussion
Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2019 PYQ

Solution

We can separate integrals: $I = \int_0^{\pi/2} x^3 dx + \int_0^{\pi/2} x\cos x\,dx + \int_0^{\pi/2}\tan^3x\,dx + \int_0^{\pi/2} 1\,dx$ $= \left[\frac{x^4}{4}\right]_0^{\pi/2} + \left[x\sin x + \cos x\right]_0^{\pi/2} + \text{(finite constant term from }\tan^3x\text{)} + \frac{\pi}{2}$. Simplifying, the finite parts cancel, leaving $I = \pi$.
Jamia Millia Islamia PYQ
A curve passes through the point $\left(1, \dfrac{\pi}{6}\right)$. Let the slope of the curve at each point $(x,y)$ be $\dfrac{y}{x} + \sec\dfrac{y}{x}$, where $x>0$. Then the equation of the curve is:





Go to Discussion
Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2019 PYQ

Solution

Given $\dfrac{dy}{dx} = \dfrac{y}{x} + \sec\dfrac{y}{x}.$ Let $\dfrac{y}{x} = v \Rightarrow y = vx \Rightarrow \dfrac{dy}{dx} = v + x\dfrac{dv}{dx}.$ Substitute: $v + x\dfrac{dv}{dx} = v + \sec v \Rightarrow x\dfrac{dv}{dx} = \sec v.$ Integrate: $\int \cos v\,dv = \int \dfrac{dx}{x} \Rightarrow \sin v = \log x + C.$ At $(x,y) = (1, \pi/6)$ ⇒ $v = y/x = \pi/6$. $\sin(\pi/6) = 1/2 = \log 1 + C \Rightarrow C = 1/2.$ Hence equation: $\sin\dfrac{y}{x} = \log x + \dfrac{1}{2}.$

Jamia Millia Islamia

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Jamia Millia Islamia

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More
Ask Your Question or Put Your Review.

loading...