Use the left sums with n=5 to estinate the interval

Question

Interval [1,20]

((5/2)x^2+x^-2+4)dx

Answer 1

So the width of deltaX is (20-1)/5 = 19/5 = 3.8

So x=1, 4.8, 8.6, 12.4, 16.2; the partitions are [1, 4.8]

[4.8, 8.6]

[8.6, 12.4]

[12.4, 16.2]

[16.2, 20]

f(1) = 5/2 + 1+ 5 = 2.5 +1 +5 = 8.5

f(4.8) = (5/2)(4.8)^2 + 1/(4.8)^2 + 4 = 61.6434027

f(8.6)= (5/2)(8.6)^2 + 1/(8.6)^2 + 4 = 188.913520822065981611682-

f(12.4) = (5/2)(12.4)^2 + 1/(12.4)^2 + 4 =388.4065036420395....

f(16.2) = (5/2)(16.2)^2 + 1/(16.2)^2 + 4= 660.1009765625...

the total of these function heights is 1307.564403804383257

multiplying by the width of 3.8 gives an approximate area of 4968.7447344566563795

Integrating:

(5/6)x^3 +(-1) x^-1 + 4x

x=20: (5/6)20^3 - 1/20 + 80 = 5*20*20*20/6 - 1/20 + 80

= 6666 and 2/3 -1/20 + 80

= 6742.783

x=1 (5/6)*1^3 - 1 + 4 = 3 and 5/6 = 23/6

the actual area is approximately 6748.95

if you graph the function on desmos, you will see the reason for the discrepency.

The reason is that the function grows very fast.

If the right endpoint of the interval is Xn and the left endpoint is Xn-1 then

f(xN) is MUCH greater than f(Xn-1).

In other words, the left endpoint is not the best choice in this case