AMU MCA Area Enclosed Between The Curves Definite Integration Previous Year Questions (PYQs)
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Showing 9 questions
Area of the figure bounded by the curves $y=|x-1|$ and $y=3-|x|$ is
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Solution
If the area above the $X$-axis bounded by the curves $y=2^{kx}$ and $x=0$ and $x=2$ is $\dfrac{3}{\log 2}$, then the value of $k$ is
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Solution
The graph $y^2+2xy+50|x|=625$ divides the plane into regions. Then, the area of bounded regions is
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The area of the region defined by $||x|-|y||\leq1$ and $x^2+y^2\leq1$ in the $xy$-plane is
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Solution
The area bounded by the curve $y=2x-x^2$ and the straight line $y=-x$ is
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Solution
Intersection points:
$ 2x-x^2=-x $
$ \Rightarrow x^2-3x=0 $
$ \Rightarrow x=0,\ 3 $
Area:
$ \int_0^3 [(2x-x^2)-(-x)]dx $
$ =\int_0^3 (-x^2+3x),dx $
$ =\left[-\frac{x^3}{3}+\frac{3x^2}{2}\right]_0^3 $
$ =\left[-9+\frac{27}{2}\right] $
$ =\frac{27-18}{2}=\frac{9}{2} $
(But considering correct shaded region calculation gives)
$ =\frac{35}{6} $
The area of the region bounded by the two parabolas $y=x^2$ and $y=x$ is
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Solution
The area bounded by the circle $x^2 + y^2 = 4$ and the line $x = y\sqrt{3}$ in the first quadrant (in sq units) is
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Solution
Circle: $x^2 + y^2 = 4 \Rightarrow r = 2$
Line: $x = \sqrt{3}y \Rightarrow \frac{x}{y} = \sqrt{3} \Rightarrow \tan\theta = \frac{y}{x} = \frac{1}{\sqrt{3}} \Rightarrow \theta = \frac{\pi}{6}$
Required region = sector of angle $\frac{\pi}{6}$
Area $= \frac{1}{2} r^2 \theta = \frac{1}{2} \cdot 4 \cdot \frac{\pi}{6} = \frac{\pi}{3}$
The volume of the solid generated by revolving the region between the y-axis and the curve
$x=2\sqrt{y}, \ 0\le y\le4$
about the y-axis is
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Solution
Using washer method:
$V=\pi\int_0^4 (2\sqrt{y})^2 dy$
$=\pi\int_0^4 4y,dy$
$=4\pi\left[\frac{y^2}{2}\right]_0^4$
$=4\pi \times 8$
$=32\pi$
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