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the area of the region bo...
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The area of the region bounded by the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is
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The area of an ellipse is calculated as \( \pi \times a \times b \), where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.
COMEDK UGET - 2025
COMEDK UGET
Updated On:
May 5, 2026
\( \pi ab \) sq units
\( \pi^2 ab \) sq units
\( \pi ab^2 \) sq units
\( \pi a b^2 \) sq units
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Top Questions on applications of integrals
$A_1$ is the area bounded by $y=x^2+2$, $x+y=8$, and the $y$-axis in the first quadrant, and $A_2$ is the area bounded by $y=x^2+2$, $y^2=x$, $x=0$ and $x=2$ in the first quadrant. Find $(A_1-A_2)$.
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Let
\[ f(x)=\int \frac{1-\sin(\ell n t)}{1-\cos(\ell n t)} \, dt \]
and
\[ f\left(e^{\pi/2}\right)=-e^{\pi/2} \]
then find $f\left(e^{\pi/4}\right)$.
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The area (in sq. units) of the region, given by the set $\{(x, y) \in R \times R \mid x \ge 0, 2x^2 \le y \le 4 - 2x\}$ is :
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The area of the region bounded by $y-x=2$ and $x^2=y$ is equal to :
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