Find natural number n so that area bounded by y = n x 2 , y = 2 1 n x 2 and y 2 − 4 y + 3 = 0 is greatest.

Area = ∫ 1 3 ( n 2 y − n y ) d y is greatest when n is least

Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesVariable Area ( greatest and Least ValuesHard2 minQB

MathematicsHardinteger

Find natural number $n$ so that area bounded by $y = n x^{2}, y = \frac{1}{2} n x^{2}$ and $y^{2} - 4 y + 3 = 0$ is greatest.

Answer:

Solution:

Area $= \int_{1}^{3} (\frac{2 y}{n} - \frac{y}{n}) d y$

is greatest when $n$ is least

Stream:JEESubject:MathematicsTopic:Area Under CurvesSubtopic:Variable Area ( greatest and Least Values

⏱ 2mℹ️ Source: QB

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