# Definite Integral

• If we construct the graph of the integrand y = f(x), then in the case f(x) ≥ 0 the integral will be numerically equal to the area bounded by the curve y = f(x), the axis of x and the straight lines x = a and x = b.
• Also \int_{a}^{b} f(x)dx=\mid F(x)\mid^{^b}_{_a}=F(b)-f(a), where F(x) is an anti derivative of f(x). If f(x) ≥ g(x) on [a, b], then the area bounded by y = f(x) and y = g(x) and the lines x = a and x = b is \int_{a}^{b} [f(x)-g(x)]dx.

### View the Topic in this video From 10:54 To 43:18

Disclaimer: Compete.etutor.co may from time to time provide links to third party Internet sites under their respective fair use policy and it may from time to time provide materials from such third parties on this website. These third party sites and any third party materials are provided for viewers convenience and for non-commercial educational purpose only. Compete does not operate or control in any respect any information, products or services available on these third party sites. Compete.etutor.co makes no representations whatsoever concerning the content of these sites and the fact that compete.etutor.co has provided a link to such sites is NOT an endorsement, authorization, sponsorship, or affiliation by compete.etutor.co with respect to such sites, its services, the products displayed, its owners, or its providers.

• \int_{a}^{b} f(x)dx=\lim_{h \rightarrow 0}h[f(a)+f(a+h)+...+f(a+(n-1)h]\

⇒  \ \int_{a}^{b} f(x) dx=(b-a)\lim_{n \rightarrow \infty}\frac{1}{n}[f(a)+f(a+h)+...+f(a+(n-1)h)]\

where\ h=\frac{b-a}{n}\rightarrow 0\ as\ n\rightarrow \infty

• \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} f \left(\frac{r}{n}\right) = \int_{o}^{1} f(x) dx
• \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f \left(\frac{r}{n}\right) = \int_{o}^{1} f(x) dx
• \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{a(n-1)} f \left(\frac{r}{n}\right) = \int_{o}^{a} f(x) dx
• \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{an} f \left(\frac{r}{n}\right) = \int_{o}^{a} f(x) dx