Sine and few friends

Calculus Level 3

2017 π 2018 π sin ( x ) + sin ( x ) + sin ( x ) d x = ? \large \int_{-2017\pi}^{2018 \pi} \sin(x)+|\sin(x)|+\sin(|x|) \ dx = ?

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Graph credit: Math is Fun


The answer is 8070.

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1 solution

Chew-Seong Cheong
Feb 19, 2018

Let I = 2017 π 2018 π sin x + sin x + sin x d x = I 1 + I 2 + I 3 I = \displaystyle \int_{-2017 \pi}^{2018 \pi} \sin x + |\sin x| + \sin |x| \ dx = I_1 + I_2 + I_3 respectively.

We note that s i n x sin x has a period or cycle of 2 π 2\pi . As an odd function, its area under the curve over one cycle is 0 and half a cycle has positive and the other half has equal area but negative. The absolute value of area of half a cycle is 0 π sin x d x = cos x 0 π = 2 \displaystyle \int_0^\pi \sin x \ dx = - \cos x \bigg|_0^\pi = 2 . Then we have I s ( n ) = 0 n π sin x d x = { 2 if n is odd 0 if n is even n N \displaystyle I_s(n) = \int_0^{n\pi} \sin x \ dx = \begin{cases} 2 & \text{if }n \text{ is odd} \\ 0 & \text{if }n \text{ is even} \end{cases} \quad \forall \ n \in \mathbb N

I 1 = 2017 π 2018 π sin x d x = 2017 π 0 sin x d x + 0 2018 π sin x d x Since sin x is odd = 0 2017 π sin x d x + I s ( 2018 ) = I s ( 2017 ) + 0 = 2 \begin{aligned} I_1 & = \int_{-2017\pi}^{2018 \pi} \sin x \ dx \\ & = {\color{#3D99F6}\int_{-2017\pi}^0 \sin x \ dx} + \int_0^{2018 \pi} \sin x \ dx & \small \color{#3D99F6} \text{Since }\sin x \text{ is odd} \\ & = {\color{#3D99F6}- \int_0^{2017\pi} \sin x \ dx} + I_s(2018) \\ & = - I_s(2017) + 0 \\ & = - 2 \end{aligned}

I 2 = 2017 π 2018 π sin x d x All half-cycles are positive. = ( 2018 ( 2017 ) ) I s ( 1 ) = 8070 \begin{aligned} I_2 & = \int_{-2017\pi}^{2018 \pi} |\sin x| \ dx & \small \color{#3D99F6} \text{All half-cycles are positive.} \\ & = \left(2018-(-2017)\right) I_s(1) \\ & = 8070 \end{aligned}

I 3 = 2017 π 2018 π sin x d x = 2017 π 0 sin x d x + 0 2018 π sin x d x Since sin x is even = 0 2017 π sin x d x + 0 2018 π sin x d x = I s ( 2017 ) + I s ( 2018 ) = 2 \begin{aligned} I_3 & = \int_{-2017\pi}^{2018 \pi} \sin |x| \ dx \\ & = {\color{#3D99F6}\int_{-2017\pi}^0 \sin |x| \ dx} + \int_0^{2018 \pi} \sin |x| \ dx & \small \color{#3D99F6} \text{Since }\sin |x| \text{ is even} \\ & = {\color{#3D99F6}\int_0^{2017\pi} \sin x \ dx} + \int_0^{2018 \pi} \sin x \ dx \\ & = I_s(2017) + I_s(2018) \\ & = 2 \end{aligned}

Therefore, I = I 1 + I 2 + I 3 = 2 + 8070 + 2 = 8070 I = I_1+I_2+I_3 = -2+8070+2 = \boxed{8070} .

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