The result foun by dividing the interval into 10 subintervals and then adding up

Question

The result foun by dividing the interval into 10 subintervals and then adding up the areas of the inscribed rectangles is 2.28

The result found by dividing the interval into 10 subintervals and then adding up the areas of the circumscribed rectangles is 3.08

The exact result found by evaluating the antiderivative at the bounds is approximately 2.67

Why is the mean of the sums of the inscribed rectangles and circumscribed retangles greater than the exact value

Three results for finding the area under the curve Y = x^2 between x=0 and x=2, are shown below The result foun by dividing the interval into 10 subintervals and then adding up the areas of the inscribed rectangles is 2.28 The result found by dividing the interval into 10 subintervals and then adding up the areas of the circumscribed rectangles is 3.08 The exact result found by evaluating the antiderivative at the bounds is approximately 2.67 Why is the mean of the sums of the inscribed rectangles and circumscribed retangles greater than the exact value

Explanation / Answer

y = x^2

a = 0 , b = 2, n = 10

delta(x) = (b - a)/n = 2/10 = 0.2

Inscribed :

We use the left end sum

0^2 + 0.2^2 + 0.4^2 + 0.6^2 + 0.8^2 + 1^2 + 1.2^2 + 1.4^2 + 1.6^2 + 1.8^2
= 11.4

11.4 * 0.2 = 2.28

Cirsumscribed :

Right end sum

0.2^2 + 0.4^2 + 0.6^2 + 0.8^2 + 1^2 + 1.2^2 + 1.4^2 + 1.6^2 + 1.8^2 + 2^2
= 15.4

15.4 * 0.2 = 3.08

Exact result :

Integral 0 to 2 x^2

x^3/3 from 0 to 2

(2)^3/3 - (0)^3/3

8/3 = 2.67

Mean :

(2.28 + 3.08) / 2 = 2.68

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