If ₀^2 1 (a+4x-2x^2)(1+3^ 2-2x ) dx = 1 2 5 ( 5 +1 2 ) , where a > 0 , then the value of a is equal to

Step-by-step solution

Let I = ₀^2 1 (a+4x-2x^2)(1+3^ 2-2x ) dx Using the property _a^b f(x) dx = _a^b f(a+b-x) dx , we replace x with 2-x . The quadratic term becomes: a+4(2-x)-2(2-x)^2 = a+8-4x-2(4-4x+x^2) = a+4x-2x^2 . The exponent becomes: 2-2(2-x) = 2-4+2x = -(2-2x) . I = ₀^2 1 (a+4x-2x^2)(1+3^ -(2-2x) ) dx = ₀^2 3^ 2-2x (a+4x-2x^2)(1+3^ 2-2x ) dx Adding the two integrals gives: 2I = ₀^2 1 a+4x-2x^2 dx 2I = 1 2 ₀^2 1 a 2 +2x-x^2 dx = 1 2 ₀^2 1 a+2 2 -(x-1)^2 dx Let k^2 = a+2 2 . Then: 2I = 1 2 ₀^2 1 k^2-(x-1)^2 dx 2I = 1 2 [ 1 2k |