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  • Evaluate the following definite integrals as limit of sums. Integral 2 to 3 x^2 d x

Evaluate the following definite integrals as limit of sums.

    Q3.    \int_2^3 x^2dx

Answers (1)

best_answer

We know that

\\\int_{a}^{b}f(x)dx=(b-a)\lim_{n\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]\\ h = \frac{b-a}{n}
here  a = 2 and b = 3 , so h = 1/n


\int_{2}^{3}x^2dx=(3-2)\lim_{x\rightarrow \infty }\frac{1}{n}[f(2)+f(2+\frac{1}{n})+f(2+\frac{2}{n})+....+f(2+\frac{(n-1)}{n})]                     \\=(1)\lim_{x\rightarrow \infty }\frac{1}{n}[2^2+(2+\frac{1}{n})^2+......+(2+\frac{(n-1)}{n})^2]\\ =\lim_{x\rightarrow \infty }\frac{1}{n}[(2^2+2^2+...2^2)_{n}+(\frac{1}{n})^2+(\frac{2}{n})^2+....(\frac{n-1}{n})^2+4(\frac{1}{n}+\frac{2}{n}+.....+\frac{n-1}{n})\\ =\lim_{x\rightarrow \infty }\frac{1}{n}[4n+\frac{n(n-1)(2n-1)}{6n^2}+\frac{4}{n}.\frac{n(n-1)}{2}]\\ =\lim_{x\rightarrow \infty }\frac{1}{n}[4n+(1-\frac{(1-\frac{1}{n})(2n-1)}{6})+\frac{4(n-1)}{2}]
\\=\lim_{x\rightarrow \infty }\frac{1}{n}[4n+(1-\frac{n(1-\frac{1}{n})(2-\frac{1}{n})}{6})+\frac{4(n-1)}{2}]\\ =\lim_{x\rightarrow \infty }\frac{1}{n}.n[4+(1-\frac{(1-\frac{1}{n})(2-\frac{1}{n})}{6})+2-\frac{2}{n}]\\ =4+\frac{2}{6}+2 =\frac{19}{3}

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manish

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