**Definite integrals as a limit of a sum, Fundamental Theorem of Calculus (without proof)**

If *f(x) *is a function defined for *a ≤ x ≤ b*, we divide the interval [*a,b*] into *n* subintervals of equal

width Δ*x = (b-a)/n*. We assume *x _{0} =a, x_{1}, x_{2},…, x_{n} (=b) *which denote the endpoints of these subintervals and we let

*x*be any sample points in these subintervals. Then,

_{1}^{ *}, x_{2}^{ *,}…, x_{n}^{* }**the definite integral of**is written in calculus as:

*f*from*a*to*b*provided this limit exists and it should provide the same value for all possible choices of sample points. If the limit exists, we say *f *is **integrable **on [*a,b*]. In this case, a is the lower limit of integration, and the number bis the upper limit of integration.

The precise meaning of the limit that defines the integral can be explained as follows:

For every number *ε*> 0 N is an integer which follows and for every integer *n*>*N* and for every choice of *x _{i}** in [

*x*

_{i}_{-1},

*x*] as shown below

_{i}If f is continuous and nonnegative on the closed interval [a, b], then the area of the region bounded by the graph of f, the x-axis, and the vertical lines x = a and x = b can be interpreted as:

If the graph is nonpositive from a to b then

Areas of common geometric figures:

If *f* is continuous on [*a*, *b*], or if *f* has only a finite number of jump discontinuities, then

*f* is integrable on [*a*, *b*].That is, the definite integral $\int_{a}^{b}{f(x)dx}$ exists.

Let f be a continuous function defined on the closed interval [a, b] and F bean antiderivative of f. Then,

$\int_{a}^{b}{f(x)dx}=[F(x)]_{a}^{b}=F(b)-F(a)$

Definition of Two Special Definite Integrals:

- If f is defined at a fixed point x = a, then
- If f is integrable on [a, b], then

**Example: **

For calculating $\int_{a}^{b}{f(x)dx}$, use the following steps:

Find the indefinite integral∫ f x dx ( ) . Let this be F(x). There is no need to keep integration constant C because if we consider F(x) + C instead of F(x), we get $\int_{a}^{b}{f(x)dx}=[F(x)]_{a}^{b}=F(b)-F(a)$

Thus, the arbitrary constant disappears in evaluating the value of the definite integral and then we can evaluate the value of integral by plugging in b and a.

**Example: Evaluate $\int_{0}^{1}{\frac{\tan ^{-1}x}{1+x^{2}}}$**

Use substitution method, let t = $\tan ^{-1}x$ then dt = $dt= \frac{1}{1+x^{2}}dx$. In this case, the new limits for t are $x=0\to t=0;x=1\to t=\frac{\pi }{4}$

$\int_{0}^{1}{\frac{\tan ^{-1}x}{1+x^{2}}}=\int_{0}^{\frac{\pi }{4}}{t dt[\frac{t^{2}}{2}]_{0}^{\frac{\pi }{4}}}$

$\frac{1}{2}\lbrack \frac{\pi ^{2}}{16}-0\rbrack =\frac{\pi ^{2}}{32}$