Calculus: Early Transcendentals, Chapter 5, 5.2, Section 5.2, Problem 9

You need to use the midpoint rule to approximate the interval. First, you need to find Delta x , such that:
Delta x = (b-a)/n
The problem provides b=8, a=0 and n = 4, such that:
Delta x = (8-0)/4 = 2
Hence, the following intervals of length 2 are: [0,2], [2,4], [4,6], [6,8].
Now, you may evaluate the integral such that:
int_0^8 sin (sqrt x) dx = Delta x(f((0+2)/2) + f((2+4)/2) + f((4+6)/2) + f((6+8)/2))
int_0^8 sin (sqrt x) dx = 2(f(1) + f(3) + f(5) + f(7))
int_0^8 sin (sqrt x) dx = 2(sin sqrt1 + sin sqrt3 + sin sqrt 5 + sin sqrt 7))
int_0^8 sin (sqrt x) dx = 2(0.017 + 0.030 + 0.039 + 0.046)
int_0^8 sin (sqrt x) dx = 2*0.132
int_0^8 sin (sqrt x) dx =0.264
Hence, approximating the definite integral, using the midpoint rule, yields int_0^8 sin (sqrt x) dx =0.264.