Mathematics -Class XII
Unit III -Calculus
Chapter VII -Integrals
Topic -Definite Integrals
Sub Topic -Definite Integral as
the Limit of a Sum –
Definition and proof
Outline:
DEFINITE INTEGRAL
AS THE LIMIT OF SUM
(PART-I)
DEFINITION
Let f be a continuous function defined on the closed interval [a, b],
As n ∞
is called definite integral of f from a to b.
The definite integral
is the area bounded by the curve
y= f(x)., the ordinates x=a , x=b and the x-axis
To evaluate this area , consider the region PRSQP between the curve, x-axis and the ordinates at x=a and x=b.
Y
0
X
Q
P
R
S
x=a
x=b
Divide the interval [a,b] into n equal subintervals denoted by
Y
0
X
Xn=b
X0=a
X1
X2
X3
xr-1
xr
Q
P
R
S
A
B
L
D
M
C
[x0,x1],[x1,x2],….[xn-1,xn] , where x0=a, x1=a+h, x2=a+2h…, xr=a+rh , r=1,2,3,…n
and xn=b=a+nh, or h= (b-a)/n. Also as
The region PRSQP under consideration is the sum of n sub regions where each sub region is defined on the subintervals [xr-1,xr], r=1,2,3,…n. hence from fig.
Area of the rectangle (ABLC)<area of the region (ABDCA)< area of rectangle(ABDM )
Y
0
X
A
B
L
C
M
D
xr-1
xr
As xr-xr-1 0,i.e ,h 0 the three areas are nearly equal .
Here sn & Sn are sum of the areas of all lower rectangles and higher rectangles over subintervals [xr-1,xr], r=1,2,3,…n.
Therefore sn<area of the region PRSPQ<Sn
or
As , all the three areas are equal
Applications
Ans.
SOLUTION BY SECOND FORMULA
Here a=0, b=2 , h=2/n
Solution contd…