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Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesMedium2 minPYQ_2020

MathematicsMediumsingle choice

Given: $f (x) = \{\begin{matrix} \begin{matrix} x, 0 \leq x < \frac{1}{2} \\ \frac{1}{2}, x = \frac{1}{2} \end{matrix} \\ 1 - x, \frac{1}{2} < x \leq 1 \end{matrix}$
and $g (x) = {(x - \frac{1}{2})}^{2}, x \in R .$ Then, the area (in sq. units) of the region bounded by the curves, $y = f (x)$ and $y = g (x)$ between the lines $2 x = 1$ and $2 x = \sqrt{3},$ is:

Question diagram: Given: f x = x , 0 ≤ x < 1 2 1 2 , x = 1 2 1 - x , 1 2 < x ≤

Options:

Answer:

Solution:

Required area = Area of trapezium ABCD $- \int_{1 / 2}^{\sqrt{3} / 2} {(x - \frac{1}{2})}^{2} d x$
$= \frac{1}{2} (\frac{\sqrt{3} - 1}{2}) (\frac{1}{2} + 1 - \frac{\sqrt{3}}{2}) - \frac{1}{3} {({(x - \frac{1}{2})}^{3})}_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}$
$= \frac{\sqrt{3}}{4} - \frac{1}{3}$

Stream:JEESubject:MathematicsTopic:Area Under CurvesSubtopic:Area bounded by two curves

⏱ 2mℹ️ Source: PYQ_2020

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