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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Area Enclosed Between The Curves Definite Integration PYQ

MCA NIMCET PYQ
The area enclosed between the curves y2 = x and y = |x| is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Solving y2 = x and y = x, 
we get, y = 0, x = 0, y = 1, x = 1  Therefore, 



MCA NIMCET PYQ
The smaller area bounded by $y = 2 - x$ and $x^2 + y^2 = 4$ is:





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Line intersects circle at two symmetric points. 
Required smaller area = circular segment area. 
 Solve intersection: 
$y = 2 - x$ 
$x^2 + (2 - x)^2 = 4$ 
$x^2 + x^2 - 4x + 4 = 4$ 
$2x^2 - 4x = 0$ 
$2x(x - 2) = 0$ 
 So intersection points at $(0, 2)$ and $(2, 0)$. 
 Using known segment area formula → the smaller area = $\pi - 2$.
MCA NIMCET PYQ
If area between $ y=x^{2} $ and $ y=x $ is $ A $, then area between $ y=x^{2} $ and $ y=1 $ is:





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

$ A = \int_{0}^{1} (x - x^{2}), dx = \frac{1}{6} $ Area between $1$ and $x^{2}$: $ \int_{0}^{1} (1 - x^{2}) dx = \frac{2}{3} = 4A $
MCA NIMCET PYQ
The area enclosed within the curve $|X|+|Y| = 1$ (in square units) is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

MCA NIMCET PYQ
The area of the plane bounded by $y=\sqrt{x},\ x\in[0,1]$, $y=x^2,\ x\in[1,2]$, $y=-x^2+2x+4,\ x\in[0,2]$





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Area = $\displaystyle \int_0^1[( -x^2+2x+4)-\sqrt{x}]dx + \int_1^2[( -x^2+2x+4)-x^2]dx$ Evaluating gives $\boxed{\frac{19}{3}}$
MCA NIMCET PYQ
The area of the region bounded by x-axis and the curves defined by $y=tanx$, $-\frac{\pi}{3}\leq x\leq \frac{\pi}{3}$ and $y=cotx$, $-\frac{\pi}{6}\leq x\leq \frac{3\pi}{2}$ is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

MCA NIMCET PYQ
The area enclosed between the curves $y^2=x$ and $y=|x|$ is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

MCA NIMCET PYQ
Area between $y = 2 - x^2$ and $y = x^2$





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Solve intersection: $2 - x^2 = x^2$ $2x^2 = 2$ $x^2 = 1$ → $x = -1,\ 1$ Area = $\int_{-1}^{1} [(2 - x^2) - x^2] dx$ $= \int_{-1}^{1} (2 - 2x^2) dx$ $= 2\int_{-1}^{1} (1 - x^2) dx$ Compute: $\int_{-1}^{1} 1 dx = 2$ $\int_{-1}^{1} x^2 dx = \frac{2}{3}$ Area = $2(2 - \frac{2}{3}) = \frac{8}{3}$
MCA NIMCET PYQ
The area enclosed between the curve y = sin x, y = cosx, $0\leq x\leq\frac{\pi}{2}$ is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

MCA NIMCET PYQ
The area of region bounded by the lines $y = |x - 1|$ and $y = 3 - |x|$ is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

MCA NIMCET PYQ
Find the area bounded by the line y = 3 - x, the parabola y = x2 - 9 and 





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution


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