Approximate the area under the function between a and b using a left hand sum with the given number of intervals.
f(x)=x^2
a=0 b=4 4 intervals

The length of each interval is (b-a)/i where i=number of intervals, a=starting value and b=ending value
so (4-0)/4=4/4=1
each interval is 1 unit in width

ok
the values of the heights are
f(0)
f(1)
f(2)
f(3)

f(0)=0
f(1)=1
f(2)=4
f(3)=9

1(0+1+4+9)
1(14)
14 is the aproximate area under the curve from a=0 to a=4

The answer to your question is b’(16,14)