If _ -a ^a 1 (1+x^2)(1+e^x) dx = 6 for some a > 0 , then the value of a is

Step-by-step solution

Let I = _ -a ^a 1 (1+x^2)(1+e^x) dx Using the property of definite integrals _ -a ^a f(x) dx = ₀^a (f(x) + f(-x)) dx , we have: I = ₀^a ( 1 (1+x^2)(1+e^x) + 1 (1+(-x)^2)(1+e^ -x ) ) dx I = ₀^a 1 1+x^2 ( 1 1+e^x + e^x e^x+1 ) dx I = ₀^a 1 1+x^2 dx I = [ ⁻¹ x]₀^a = ⁻¹ a Given I = 6 , we get: ⁻¹ a = 6 a = ( 6 ) = 1 3 Answer: 1 3