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Calculus1
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Theorem
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if f is integrable on [a,b], then \[\int\limits_{a}^{b} f(x)dx=\lim_{n \rightarrow \infty} \sum_{j=1}^{n}f(x_j)Deltax\] where \[\Delta(x)=\frac{ b-a }{ n } and x_j=a+j \Delta x\] evaluate the integral using the above theorem.
\[\int\limits_{a}^{b} f(x)dx=\lim_{n \rightarrow \infty} \sum_{j=1}^{n}f(x_j)\Delta x\] Where \(\Delta(x)=\frac{ b-a }{ n } \) and \(x_j=a+j \Delta x\) \[\qquad\qquad =\lim_{n \rightarrow \infty} \sum_{j=1}^{n}f\left(a+j \frac{ b-a }{ n }\right)\frac{ b-a }{ n } \]
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