Calculus: Early Transcendentals, Chapter 5, 5.2, Section 5.2, Problem 12

You need to evaluate the definite integral using the mid point rule, hence, first you need to evaluate Delta x:
Delta x = (b-a)/n => Delta x = (5-1)/4 = 1
You need to denote each of the 4 intervals, such that: [1,2],[2,3],[3,4],[4,5].
You need to evaluate the definite integral, such that:
int_1^5x^*e^(-x)dx = Delta x*(f((1+2)/2)+f((2+3)/2)+f((3+4)/2)+f((4+5)/2))
int_1^5x^*e^(-x)dx = 1*(f(3/2) + f(5/2) + f(7/2) + f(9/2))
int_1^5x^*e^(-x)dx = ((3/2)^2*e^(-3/2) + (5/2)^2*e^(-5/2) + (7/2)^2*e^(-7/2) + (9/2)^2*e^(-9/2))
int_1^5x^*e^(-x)dx = (9/4*e^(-3/2) + 25/4*e^(-5/2) + 49/4*e^(-7/2) + 81/4*e^(-9/2))
int_1^5x^*e^(-x)dx = (2.25*0.21 + 6.25*0.08+ 12.25*0.03 + 20.25*0.01)
int_1^5x^*e^(-x)dx = (0.4725 + 0.5+ 0.3675 + 2.025)
int_1^5x^*e^(-x)dx =3.3650
Hence, evaluating the definite integral, using the mid point rule, yields int_1^5x^*e^(-x)dx =3.3650.