( _ _c 2 ^x d t e^t-1 = 6 x= )

Step-by-step solution

Given integral ( aligned & I= _ _e 2 ^x d t 2 e^t-1 = 6 (given) & _ _e 2 ^x e^ - t 2 1- (e^ -t / 2 )^2 d t= 6 & put e^ -t / 2 =u , at t= _e 2 u= 1 2 and at t=x , & u=e^ -x / 2 and e^ -t / 2 d t=-2 d u & so, I= _ 1 2 ^ e^ -x / 2 -2 d u 1-u^2 = 6 & [-2 ⁻¹ u ] _ 1 2 ^ e^ - x 2 = 6 ⁻¹ (e^ -x / 2 )- 4 =- 12 & ⁻¹ (e^ -x / 2 )= 6 e^ -x / 2 = 6 = 1 2 & - x 2 = _e ( 1 2 )=- _e(2) x=2 _e(2) . aligned ) Hence, option (1) is correct.