The least value of the function F(x)=∫0x   (3 sin u+4 cos u) duon the interval (5π/4, 4π/3] is

We have F'(x)=3 sin x+4 cos x. Since sin x and cos x assume negative values in the third quadrant, we have F′(x)<0 for all x∈(5π/4, 4π/3) so F(x) assumes the least value at the point x = 4π/3.Thus the least value is F(4π/3)=∫04π/3 (3sin⁡ u+4cos⁡ u)du=(−3cos⁡ u+4sin⁡ u)04π/3=−3cos⁡ 4π3+4sin⁡ 4π3−(−3)=92−432=9−432.