The area of the region given by A = x , y : x 2 ≤ y ≤ min x + 2 , 4 - 3 x is

The graph of y = x 2 & y = min x + 2 , 4 - 3 x will be From the graph, required area A = ∫ - 1 1 2 x + 2 - x 2 d x + ∫ 1 2 1 4 - 3 x - x 2 d x = x 2 2 + 2 x - x 3 3 - 1 1 2 + 4 x - 3 x 2 2 - x 3 3 1 2 1 = 17 6

Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesMedium2 minPYQ_2022

MathematicsMediumsingle choice

The area of the region given by $A = \{(x, y) : x^{2} \leq y \leq \min \{x + 2, 4 - 3 x\}\}$ is

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Answer:

Solution:

The graph of $y = x^{2} & y = \min \{x + 2, 4 - 3 x\}$ will be
From the graph,
required area $A = \int_{- 1}^{\frac{1}{2}} (x + 2 - x^{2}) d x + \int_{\frac{1}{2}}^{1} (4 - 3 x - x^{2}) d x$
$= {(\frac{x^{2}}{2} + 2 x - \frac{x^{3}}{3})}_{- 1}^{\frac{1}{2}} + {(4 x - 3 \frac{x^{2}}{2} - \frac{x^{3}}{3})}_{\frac{1}{2}}^{1} = \frac{17}{6}$

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

⏱ 2mℹ️ Source: PYQ_2022

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