_ - 4 ^ 4 ( x+ 4 2- 2 x ) d x is equal to

Step-by-step solution

Given, I = _ - 4 ^ 4 x+ 4 2- ^2 x d x I = _ - 4 ^ 4 x+ 4 2- ^2 x d x+ 4 _ - 4 ^ 4 1 2- ^2 x d x Let, I₁= _ - 4 ^ 4 x 2- ^2 x d xf(x)= x 2- ^2 x f(-x)= -x 2- 2(-x) = -x 2- 2 x =-f(x)f(x) is an odd function. Thus, I₁=0 Let I₁= 4 _ - 4 ^ 4 1 2- 2 x d xg(x)= 4(2- 2 x) g(-x)= 4(2- 2(-x)) = 4(2- 2 x) =g(x) g(x) is an even function Thus, I ₂=2 4 ₀^ 4 d x 2- 2 x Now, I = I ₂= 4 ₀^ 4 d x 2- 1- ^2 x 1+ ^2 x aligned & I = 4 ₀^ 4 1+ ^2 x 2+ ^2 x-1+ ^2 x d x & I = 4 ₀^ 4 ^2 x 1+3 ^2 x d x aligned Put x=t ^2 x d x=d t Limits : x