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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Evaluate : \( \int_1^3 (2x^2+3)dx\) as the limit of sums.

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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 by sreemathi.v
 
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