That's Pretty Long!

Calculus Level 5

{ x = 0 x . e 3 y = sin ⁡ ( π y ) \begin{cases} x=0 \\ \large x.e^{3y} = \sin(\pi y) \end{cases}

Let for j = 0 , 1 , 2 , … , n j = 0,1,2 ,\ldots, n . And let S j S_j be the area of the region bounded by the above curves and j ≤ y ≤ ( j + 1 ) j \leq y \leq (j+1) . Find ∣ ln ⁡ ( S 2016 S 2015 ) ∣ . \large \left | \ln \left (\frac{S_{2016}}{S_{2015}} \right) \right | .

Notation : ∣ ⋅ ∣ | \cdot | denotes the absolute value function .


The answer is 3.

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Vatsalya Tandon by Vatsalya Tandon , 21, India

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1 solution

Brian Moehring
Jun 26, 2018

We can directly find S j = ∫ j j + 1 e − 3 y ∣ sin ⁡ ( π y ) ∣ d y , S_j = \int_j^{j+1} e^{-3y}|\sin(\pi y)|\,dy, and therefore S j + 1 = ∫ j + 1 j + 2 e − 3 y ∣ sin ⁡ ( π y ) ∣ d y = ∫ j j + 1 e − 3 ( y + 1 ) ∣ sin ⁡ ( π ( y + 1 ) ) ∣ d y = e − 3 ∫ j j + 1 e − 3 y ∣ sin ⁡ ( π y ) ∣ d y = e − 3 S j S_{j+1} = \int_{j+1}^{j+2} e^{-3y}|\sin(\pi y)|\,dy = \int_j^{j+1} e^{-3(y+1)}|\sin(\pi(y+1))|\,dy = e^{-3}\int_j^{j+1} e^{-3y}|\sin(\pi y)|\,dy = e^{-3}S_j

It follows that ∣ ln ⁡ ( S 2016 S 2015 ) ∣ = ∣ ln ⁡ ( e − 3 ) ∣ = ∣ − 3 ∣ = 3 \left|\ln\left(\frac{S_{2016}}{S_{2015}}\right)\right| = \left|\ln\left(e^{-3}\right)\right| = |-3| = \boxed{3}

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