# Evaluate : $$\int_1^3 (2x^2+3)dx$$ as the limit of sums.

## 1 Answer

Toolbox:
• $\int\limits_a^b f(x)dx=\lim\limits_{h\to 0}h[f(a)+f(a+h)+...f(a+(n-1)h)]$
• where $h=\large\frac{b-a}{n}$
• $\int \limits_a^b f(x)dx=\large\frac{(b-a)}{n}$$\lim\limits_{n \to \infty} [f(a)+f(a+h)+f(a+(n-1)h)] Step 1: \int\limits_a^b f(x)dx=\lim\limits_{h\to 0}h[f(a)+f(a+h)+...f(a+(n-1)h)] where h=\large\frac{b-a}{n} Given I=\int\limits_1^3 (2x^2+3)dx a=1,b=3 \therefore h=\large\frac{2}{n} Step 2: \int\limits_{1^3}(2x^2+3)dx=\lim\limits_{h\to 0}h[f(1)+f(1+h)+f(1+2h)+...f(1+(n-1)h)] \qquad\qquad\;\;\;\;\;=\lim\limits_{h\to 0}h[2(1)^2+3]+[2(1+h)^2+3]........+[2(1+(n-1)h^2)+3] \qquad\qquad\;\;\;\;\;=\lim\limits_{h\to 0}h[2(1)^2+(1+h)^2+(1+2h)^2+....(1+(n-1)h)^2]+3n] \qquad\qquad\;\;\;\;\;=\lim\limits_{h\to 0}h[2\{n+2h(1+2+3+.....(n-1)+h^2(1^2+2^2+3^2+......(n-1)^2]+3n] \qquad\qquad\;\;\;\;\;=\lim\limits_{h\to 0}h[2[n+2h\large\frac{n(n-1)}{2}$$+h^2\large\frac{n(n-1)(2n-1)}{6}]$$+3n 1^2+2^2+3^2+.......(n-1)^2=\large\frac{n(n-1)(n-2)}{6} 1+2+3+........(n-1)=\large\frac{n(n-1)}{2} Step 3: \qquad\qquad\;\;\;\;\;=\lim\limits_{h\to 0}h[2n+2hn(n-1)+2h^2\large\frac{n(n-1)(2n-1)}{6}]$$+3n]$
Substituting for $h=\large\frac{2}{n}$
$\qquad\qquad\;\;\;\;\;=\lim\limits_{h\to \infty}[2n+2.\large\frac{2}{h}$$n(n-1)+2(\large\frac{4}{n^2})\frac{n(n-1)(2n-1)}{6}$$+3n]$
$\qquad\qquad\;\;\;\;\;=\lim\limits_{h\to \infty}\large\frac{2}{h}$$[5n+\large\frac{4(n-1)}{n}+\frac{8}{3}\frac{(n-1)(2n-1)}{n^2}] \qquad\qquad\;\;\;\;\;=\lim\limits_{h\to \infty}[10+8(1-\large\frac{1}{n})+\frac{8}{3}($$1-\large\frac{1}{h})$$(2-\large\frac{1}{h})]$
Step 4:
Applying the limits we get,
$10+8+\large\frac{16}{3}=\frac{70}{3}$sq.units
answered Oct 2, 2013

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