Let (2^ 1-a + 2^ 1+a ) , f(a) , (3^a + 3^ -a ) be in A.P. and be the minimum value of f(a) . Then the value of the integral _ _e( -1) ^ _e( ) dx (e^ 2x - e^ -2x ) is :

Step-by-step solution

Since (2^ 1-a + 2^ 1+a ) , f(a) , and (3^a + 3^ -a ) are in A.P., we have: 2f(a) = 2^ 1-a + 2^ 1+a + 3^a + 3^ -a f(a) = (2^ -a + 2^a) + 1 2 (3^a + 3^ -a ) Using the AM-GM inequality, x + 1 x 2 for x > 0 . 2^a + 2^ -a 2 and 3^a + 3^ -a 2 . The minimum value of f(a) occurs when a = 0 . = f(0) = (2^0 + 2^0) + 1 2 (3^0 + 3^0) = 2 + 1 = 3 . Now, we evaluate the integral: I = _ _e 2 ^ _e 3 dx e^ 2x - e^ -2x = _ _e 2 ^ _e 3 e^ 2x e^ 4x - 1 dx Let e^ 2x = t 2e^ 2x dx = dt . When x = _e 2 , t = e^ 2 _e 2 = 4 . When x = _e 3