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Evaluate the following definite integrals as a limit of sums.

    Q1.    \int_a^b x dx

Answers (1)

best_answer

We know that,
\int_{a}^{b}f(x)dx = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+...+f(a+(n-1)h)]
\therefore \int_{a}^{b}xdx = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[a+(a+h)...(a+2h)..a+(n-1)h]
                      \\ = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[(a+a...a+a)_{n}+(h+2h+3h....(n-1)h)]\\ = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[na+h(1+2+3..+n-1)]\\ = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[na+h(\frac{n(n-1)}{2})]\\ = (b-a)\lim_{x\rightarrow \infty }[a+\frac{n-1}{2}h]\\ = (b-a)\lim_{x\rightarrow \infty }[a+\frac{(n-1)(b-a)}{2n}]\\ = (b-a)\lim_{x\rightarrow \infty }[a+\frac{(1-\frac{1}{n})(b-a)}{2}]\\ = (b-a)[a+\frac{(b-a)}{2}]\\ =(b-a)(b+a)/2\\ =\frac{(b^2-a^2)}{2}

This is how the integral is evaluated using limit of a sum

Posted by

manish

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