Then A 1 , A 2 , A 3 , … are in

The curves x = 6 π 3 r y 2 and x = 6 e 2 y meet at y = 0 and y 0 = π 3 r e 2 . A r = ∫ 0 y 0 ( 6 e 2 y − 6 π 3 r y 2 ) d y = 3 e 2 y 0 2 − 2 π 3 r y 0 3 = π 6 r 2 e 6 . So A r = ( π e ) 3 r 1 , and the reciprocals A r 1 = ( e π ) 3 r are in A.P. Hence A 1 , A 2 , … are in H.P.

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Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesMedium2 minQB

MathematicsMediumsingle choice

Passage / Comprehension

Let $A_{r} (r \in N)$ be the area of the bounded region whose boundary is defined by $(6 π^{3} r y^{2} - x) (6 e^{2} y - x) = 0$ .

Then $A_{1}, A_{2}, A_{3}, \dots$ are in

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Answer:

Solution:

The curves $x = 6 π^{3} r y^{2}$ and $x = 6 e^{2} y$ meet at $y = 0$ and $y_{0} = \frac{e ^{2}}{π ^{3} r}$ . $A_{r} = \int_{0}^{y_{0}} (6 e^{2} y - 6 π^{3} r y^{2}) d y =$ $3 e^{2} y_{0}^{2} - 2 π^{3} r y_{0}^{3} = \frac{e ^{6}}{π ^{6} r ^{2}}$ . So $A_{r} = (\frac{e}{π})^{3} \frac{1}{r}$ , and the reciprocals $\frac{1}{A _{r}} = (\frac{π}{e})^{3} r$ are in A.P. Hence $A_{1}, A_{2}, \dots$ are in H.P.

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

⏱ 2mℹ️ Source: QB

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