Integral of The Year

Calculus Level 3

∫ − 1 1 6 x 2 + 6 202 0 x + 1 d x = ? \int_{-1}^{1}\frac{6x^2+6}{2020^x+1}dx = \ ?


The answer is 8.

Rofiud Darojad by Rofiud Darojad , 23, Indonesia

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1 solution

Chew-Seong Cheong
Jan 26, 2020

I = ∫ − 1 1 6 x 2 + 6 ( 202 0 x + 1 ) d x By reflection ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x = 1 2 ∫ − 1 1 ( 6 x 2 + 6 ( 202 0 x + 1 ) + 6 x 2 + 6 ( 202 0 − x + 1 ) ) d x = 1 2 ∫ − 1 1 ( 6 x 2 + 6 ( 202 0 x + 1 ) + ( 6 x 2 + 6 ) ( 202 0 x ) ( 1 + 202 0 − x ) ) d x = ∫ − 1 1 ( 3 x 2 + 3 ) d x = x 3 + 3 x ∣ − 1 1 = 8 \begin{aligned} I & = \int_{-1}^1 \frac {6x^2+6}{(2020^x+1)} dx & \small \blue{\text{By reflection }\int_a^b f(x) \ dx = \int_a^b f(a+b-x)\ dx} \\ & = \frac 12 \int_{-1}^1 \left(\frac {6x^2+6}{(2020^x+1)} + \frac {6x^2+6}{(2020^{-x}+1)} \right) dx \\ & = \frac 12 \int_{-1}^1 \left(\frac {6x^2+6}{(2020^x+1)} + \frac {(6x^2+6)(2020^x)}{(1+2020^-x)} \right) dx \\ & = \int_{-1}^1 (3x^2 + 3) \ dx \\ & = x^3 + 3x \ \bigg|_{-1}^1 \\ & = \boxed 8 \end{aligned}

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Can you please provide me an intuitive proof of the reflection formula?

Atomsky Jahid - 1 year, 4 months ago

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I = ∫ a b f ( a + b − x ) d x Let u = a + b − x ⟹ d u = − d x = − ∫ b a f ( u ) d u when x = a ⟹ u = b and x = b ⟹ u = a = ∫ a b f ( u ) d u Replace u with x = ∫ a b f ( x ) d x \begin{aligned} I & = \int_a^b f(a+b-x) \ dx & \small \blue{\text{Let }u=a+b-x \implies du = - dx} \\ & = - \int_b^a f(u) \ du & \small \blue{\text{when }x = a \implies u = b \text{ and }x=b \implies u = a} \\ & = \int_a^b f(u)\ du & \small \blue{\text{Replace }u \text{ with }x} \\ & = \int_a^b f(x)\ dx \end{aligned}

Chew-Seong Cheong - 1 year, 4 months ago

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