NIMCET 2027 Area Enclosed Between The Curves Definite Integration Previous Year Questions (PYQs) – Page 1 of 2
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Showing 11 questions
The area enclosed between the curves y2 = x and
y = |x| is
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Solution
Solving y2 = x and y = x, 
we get, y = 0, x = 0, y = 1, x = 1 Therefore,
The smaller area bounded by
$y = 2 - x$
and
$x^2 + y^2 = 4$
is:
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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 = $\pi - 2$.
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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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 $
The area enclosed within the curve $|X|+|Y| = 1$ (in square units) is
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Solution
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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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
$\boxed{\frac{19}{3}}$
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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Solution
The area enclosed between the curves $y^2=x$ and $y=|x|$ is
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Solution
Area between $y = 2 - x^2$ and $y = x^2$
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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 = $2(2 - \frac{2}{3}) = \frac{8}{3}$
The area enclosed between the curve y = sin x, y = cosx, $0\leq x\leq\frac{\pi}{2}$ is
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Solution
The area of region bounded by the lines $y = |x - 1|$ and $y = 3 - |x|$ is
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Solution
NIMCET 2027
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
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NIMCET 2027
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
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