The value of the integral ₀^ 2 ⁻¹( 2x) x , dx + 2 ₀^ 4 ⁻¹( 2x) x , dx is equal to

Step-by-step solution

Let I₁ = ₀^ 2 ⁻¹( 2x) x , dx Splitting the integral at 4 , we get: I₁ = ₀^ 4 ⁻¹( 2x) x , dx + _ 4 ^ 2 ⁻¹( 2x) x , dx In the second integral, substitute x = 2 - t dx = -dt . The limits change from 4 2 to 4 0 . _ 4 ^ 2 ⁻¹( 2x) x , dx = _ 4 ^0 ⁻¹( 2( 2 - t)) ( 2 - t) (-dt) = ₀^ 4 ⁻¹( 2t) t , dt Adding the two parts of I₁ gives: I₁ = ₀^ 4 ⁻¹( 2x) ( x + x) , dx Now, apply the property ₀^a f(x) , dx = ₀^a f(a-x) , dx on the interval [0, 4 ] : I₁ = ₀^ 4 ⁻¹( 2( 4 - x)) ( ( 4 - x) + ( 4 - x) ) , dx Since 2( 4 - x) = ( 2 - 2