Then lim n → ∞ [ e − 1/ A 1 + e − 1/ A 2 + e − 1/ A 3 + ⋯ n terms ] is

A r 1 = ( e π ) 3 r , so e − 1/ A r = q r with q = e − π 3 / e 3 , 0 < q < 1 . The sum is 1 − q q = 1/ q − 1 1 = ( e π 3 / e 3 − 1 ) − 1 .

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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 $lim_{n \to \infty} [e^{- 1/ A_{1}} + e^{- 1/ A_{2}} + e^{- 1/ A_{3}} + \dots n$ terms $]$ is

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Answer:

Solution:

$\frac{1}{A _{r}} = (\frac{π}{e})^{3} r$ , so $e^{- 1/ A_{r}} = q^{r}$ with $q = e^{- π^{3} / e^{3}}, 0 < q < 1$ . The sum is $\frac{q}{1 - q} = \frac{1}{1/ q - 1} =$ $(e^{π^{3} / e^{3}} - 1)^{- 1}$ .

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

⏱ 2mℹ️ Source: QB

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