Jamia Millia Islamia MCA Definite Integration Previous Year Question Paper and Solution with PDF

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Jamia Millia Islamia MCA Previous Year Questions (PYQs)

Jamia Millia Islamia MCA Definite Integration PYQ

Qus : 1 Jamia Millia Islamia MCA PYQ

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

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2021 PYQ

Solution

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

Qus : 2 Jamia Millia Islamia MCA PYQ

$\displaystyle \int_{0}^{1000} e^{\,x-[x]}\,dx$ is –

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA 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 $1000(e-1)$.

Qus : 3 Jamia Millia Islamia MCA PYQ

The value of $\displaystyle \int_{0}^{\pi/2}\sin^{4}x\,\cos^{4}x\,dx$ is –

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA 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 $J=\dfrac{1}{32}\cdot\dfrac{3\pi}{8}=\dfrac{3\pi}{256}$.

Qus : 4 Jamia Millia Islamia MCA PYQ

$\displaystyle \int_{0}^{1}\frac{x}{(1-x)^{3/4}}\,dx$ is equal to …

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2017 PYQ

Solution

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

Qus : 5 Jamia Millia Islamia MCA 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.

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2020 PYQ

Solution

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

Qus : 6 Jamia Millia Islamia MCA PYQ

$\displaystyle \int_{\frac{3\pi}{4}}^{\frac{7\pi}{4}} \dfrac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} \, dx$ is equal to

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA 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, $\int dx = x + C$. $\boxed{\text{Answer: (B) } x}$

Qus : 7 Jamia Millia Islamia MCA PYQ

$\displaystyle \int_{1}^{x}(1+\log t)^{2}\,dt$ is equal to …

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2016 PYQ

Solution

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

Qus : 8 Jamia Millia Islamia MCA PYQ

If $x>0$, then $\displaystyle \int |x|^{3} dx$ is equal to …

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2016 PYQ

Solution

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

Qus : 9 Jamia Millia Islamia MCA PYQ

$\displaystyle \int_{0}^{\frac{\pi}{4}}\sec^{2}x\sin x\,dx=a+\sqrt{2}$, find $a$

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA 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 $a=-1$.

Qus : 10 Jamia Millia Islamia MCA PYQ

$\displaystyle \int_{0}^{1}\frac{x}{(1-x)^{1/2}}dx$ is equal to …

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2016 PYQ

Solution

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

Qus : 11 Jamia Millia Islamia MCA PYQ

Evaluate the following integral: $ \int_{-2}^{2} \frac{3x^3 + 2|x| + 1}{x^2 + |x| + 1}\,dx $

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA 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 $= 3\log 7$.

Qus : 12 Jamia Millia Islamia MCA PYQ

Evaluate the following integral: $ \int_{-\pi/2}^{\pi/2} \log\!\left(\frac{2-\sin x}{2+\sin x}\right)\,dx $

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA 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 $0$.

Qus : 13 Jamia Millia Islamia MCA PYQ

$\displaystyle \int_{0}^{\pi} \sin^2 x,dx =$

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2024 PYQ

Solution

$\sin^2 x = \dfrac{1 - \cos 2x}{2} \Rightarrow \int_{0}^{\pi} \sin^2 x,dx = \dfrac{1}{2} \left[ x - \dfrac{\sin 2x}{2} \right]_{0}^{\pi} = \dfrac{1}{2}(\pi - 0) = \dfrac{\pi}{2}$

Qus : 14 Jamia Millia Islamia MCA PYQ

The area under a curve between two points is given by:

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2025 PYQ

Solution

Area $= \int_a^b f(x),dx$

Qus : 15 Jamia Millia Islamia MCA PYQ

The order of the differential equation $\frac{d^2 y}{dx^2} + 3\frac{dy}{dx} = 5$ is:

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2025 PYQ

Solution

Highest derivative $= \frac{d^2 y}{dx^2}$

Order $= 2$

Qus : 16 Jamia Millia Islamia MCA PYQ

Value of $\displaystyle \int_{0}^{\frac{\pi}{2}} (x^3 + x\cos x + \tan^3 x + 1) dx$ is:

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA 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 $I = \pi$.

Qus : 17 Jamia Millia Islamia MCA PYQ

The definite integral of $ x^2 $ from $ 0 $ to $ 2 $ is:

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2025 PYQ

Solution

$ \int_{0}^{2} x^2 dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{8}{3} $

Qus : 18 Jamia Millia Islamia MCA 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:

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA 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: $\sin\dfrac{y}{x} = \log x + \dfrac{1}{2}.$

Qus : 19 Jamia Millia Islamia MCA PYQ

The area under the curve $ y = x^2 $ from $ x = 0 $ to $ x = 1 $ is:

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2025 PYQ

Solution

$ \int_{0}^{1} x^2 dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{3} $

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