Question

Given f(x)=1+x+x2,

(a) Evaluate the definite integral A=∫10(1+x+x2)dx.

(b) Find the approximate An of the integral A by a sum, using n strips of equal width, and evaluate the sum.

(c) Find limn→∞An.

Answer 1

(a) $A=\int\limits_0^1 (1+x+x^2)\,dx = \left(x+\dfrac{x^2}{2} + \dfrac{x^3}{3} \right) \Big|_{0}^1 = \left(1+\dfrac12+\dfrac13\right) - 0 = 1\dfrac56 = \dfrac{11}{6}.$

(b) $A_n = \sum\limits_{i=1}^n \dfrac{f(\frac{i-1}{n})+f(\frac{i}{n})}{2}\cdot \dfrac1n = \dfrac1n\left( \dfrac{f(0)}{2} + f\left(\frac1n\right)+\ldots + f(\frac{n-1}{n}) + \dfrac{f(1)}{2}\right) = \dfrac1n\left(\dfrac12 + \sum\limits_{i=1}^{n-1} f(\frac{i}{n}) + \dfrac32 \right) = \dfrac1n\left(\sum\limits_{i=1}^{n-1} \dfrac{n^2+in+i^2}{n^2} + 2 \right) = \dfrac1n\left( \dfrac{n^2(n-1) + n(n(n-1)/2) + (n-1)n(2n-1)/6}{n^2} + \dfrac32\right) = \dfrac1n\left( \dfrac{11n^2-12n+1}{6n} + \dfrac32 \right)= \dfrac1n\left( \dfrac{11n^2+1}{6n}\right) = \dfrac{11}{6} + \dfrac{1}{6n^2} .$

(c) When $n\to \infty,$ $\dfrac{1}{6n^2}\to 0.$ Therefore, $A_n \to \dfrac{11}{6}.$