Finding the minimum of an integral

Question

If 0≤b≤2, for what value of b is ∫₀^b cos(e^x) dx a minimum?

Answer 1

The fundamental theorem of calculus says that the derivative of F(b) = ∫₀^b cos(e^x) dx is equal to the integrand f(b) = cos(e^b).

Setting F'(b) = f(b) = 0 will give you the extreme points:

cos(e^b) =0 --> e^{b =} j*pi/2 --> with j = 1, 3, 5, ... (odd positive integers)

--> b = log(j*pi/2) (the domain range of b being [0,2] and the domain of the log function will convince you that negative values of j are not acceptable)

The only acceptable values are j = 1 and j = 3 since b > 2 for j >= 5.

To find the minimum, compute the second derivative and determine its sign at the points b_1 = log(1*p/2) and b_2 = log(3*pi/2):

F''(b) = -sin(e^b)e^b , which is negative at b_1 (local maximum) and positive at b_2 (local minimum).

In conclusion, the value of b for which the integral is minimum is b = log(3*pi/2).