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Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesMedium2 minQB

MathematicsMediumsingle choice

The area bounded by the curves $y = 2 - ∣ x - 1∣, y = sin x; x = 0$ and $x = 2$ is

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Answer:

Solution:

On [0, 1] the line is $y = 1 + x$ and on [1, 2] it is $y = 3 - x$ , both above $sin x$ . Area $= \int_{0}^{1} (1 + x - sin x) d x +$ $\underline{\int_{1}^{2} (3 - x - sin x) d x = (\frac{3}{2} + cos 1 - 1) + (\frac{3}{2} + cos 2 - cos 1) = 2 + cos 2 = 1 + 2 cos^{2} 1.}$

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

⏱ 2mℹ️ Source: QB

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