Definite Integral

Let $\int$f (x) dx = F(x) +c. Then the definite integral of f(x) between the limits 'a' and 'b', denoted by $\underset{a}{\overset{b}{\int }}$ f(x) dx is given by $\underset{a}{\overset{b}{\int }}$ f(x) dx= F(b) - F(a)

Properties of definite integral

a) $\underset{a}{\overset{b}{\int }}$ f(x) dx= $\underset{a}{\overset{b}{\int }}$ f(y) dy

b) $\underset{a}{\overset{b}{\int }}$ f(x) dx= $\underset{a}{\overset{b}{-\int }}$ f(x) dx

c) $\underset{a}{\overset{b}{\int }}$ f(x) dx= $\underset{a}{\overset{c}{\int }}$ f(x) dx + $\underset{c}{\overset{b}{\int }}$ f(x) dx, a$<$c$<$b

d) $\underset{a}{\overset{b}{\int }}$ f(x) dx= $\underset{a}{\overset{b}{\int }}$ f(a+b-x) dx

e) $\underset{0}{\overset{a}{\int }}$ f(x) dx= $\underset{0}{\overset{a}{\int }}$ f(a-x) dx

f) $\underset{-a}{\overset{b}{\int }}$ f(x) dx= $\{2\underset{\underset{0}{0}}{\overset{a}{\int }}$f(x) dx

if f(x) is even

if f(x) is odd

g) $\underset{0}{\overset{2a}{\int }}$f(x) dx= $\{2\underset{\underset{0}{0}}{\overset{a}{\int }}$f(x) dx

if f(2a-x)= f(x)

if f(2a-x)= -f(x)

1. The area included between the curves y= f(x), the x-axis and the ordinates x = a and x = b, b $>$ a is $\underset{a}{\overset{b}{\int }}$ydx.

2. The area included between the curves x = f(y), the y-axis and the adscissae y = c and y = d (d $>$c) is equal to $\underset{c}{\overset{d}{\int }}$xdy

3. The area between the curves y = f(x) and y = g(x) such that y = f(x) lies above the curve y = g(x) and both are above x- axis and the ordinates x= a and x= b, (b $>$a) is $\underset{a}{\overset{b}{\int }}$[f(x) -g(x)]dx.

Note

1. The area of the region bounded by $y^2$=4ax and $x^2$= 4ay is $\frac{16\mathrm{ab}}{3}$ sq. units.

2. The area of the region bounded by $y^2$=4ax and y = mx is $\frac{8a^2}{{3m}^3}$ sq. units.

3. The area of the region bounded by $y^2$=4ax and its latus rectum is $\frac{{8a}^2}{3}$ sq. units.

4. The area of the region bounded by one arch of sin (ax) or cos (ax) and x- axis is $\frac{2}{a}$

q. units.

5. Area of the ellipse $\frac{x^2}{a^2}$+$\frac{y^2}{b^2}$= 1 is $\pi$ ab sq. units.