]]>
LearnNext
##### Get a free home demo of LearnNext

Available for CBSE, ICSE and State Board syllabus.
Call our LearnNext Expert on 1800 419 1234 (tollfree)
OR submit details below for a call back

clear

# Definite Integration

2,980 Views
Have a doubt? Clear it now.
live_help Have a doubt, Ask our Expert
format_list_bulleted Take this Lesson Test

#### Definite Integration - Lesson Summary

Let y = f(x) be a continuous function defined on [a, b] .

The shaded area can be represented as ∫ab f(x) dx

Consider region ABDCA.The area of this whole region be denoted by TA.

Divide the [a,b] into some small sub-intervals, each of the same width. Now, choose the least value of f(x) in each sub-interval, and construct rectangles.

The sum of the areas of the rectangles, whose area is less than the total area bounded by the curve, is called the lower sum.

Now, choose the greatest value of f(x) in each sub-interval, and construct rectangles with that as its height.

The sum of the areas of the rectangles, whose area is greater than the total area bounded by the curve, is called the upper sum.

The larger the number of rectangles we draw, the smaller will be the difference between the upper and the lower sums. So, finally, the better approximation of the area under the given curve can be calculated. This approximation will be equal to the limit of either the upper sum or the lower sum.

1) The difference between the upper and lower sums approaches zero as the number of rectangles in the area subdivided approaches infinity.

2) The area under the curve is equal to the limit of either the upper sum or the lower sum as the number of sub divisions tends to infinity.

3) The area under the curve is represented as the limit of a sum.

Divide the interval [a, b] into n equal subintervals, each of width h, denoted by

[x0,x1], [x1,x2],........, [xr-1,xr],......, [xn-1,xn]

Here, x0 = a,

x1 = x0 + h = a + h,

x2 = x1 + h = a + h + h = a + 2h,

x3 = x2 + h = a + 2h + h = a + 3h,

................... xr = a + rh, xn = b = a + nh or

n = (b - a)/h       As n → ∞,

⇒ (b-a)/h

⇒ h → 0

Area of the region containing lower rectangles < Area of the region ABDCA < Area of the region containing bigger rectangles

Let the lower sum be denoted by Ln.

⇒ Ln = f(x0) x h + f(x1) x h + f(x2) x h + ..... + f(xr-1) x h + f(xr) x h + ..... + f(xn-1) x h

⇒ Ln = r= 0 n-1 f(xr) x h .......(1)

Let Un denote the upper sum.

⇒ Un = f(x1) x h + f(x2) x h + ..... + f(xr-1) x h + f(xr) x h + ..... + f(xn) x h

⇒ Un = ∑r= 1 n f(xr) x h -----(2)

We have denoted the total area bounded under the curve by TA.

Hence, we have Ln ≤ TA ≤ Un

r= 0 n-1 f(xr) x h ≤ TA ≤ ∑r= 1 n f(xr) x h

T= limn→∞r= 0 n-1 f(xr) x h

Or

T= limn→∞r= 1 n f(xr) x h

The limit of the sum, as the width of the each sub-interval approaches zero and the number of sub intervals approaches infinity, is called the "definite integral."

T= limn→∞r= 0 n-1 f(xr) x h = limn→∞r= 1 n f(xr) x h

A = ∫ ab f(x) dx

A = ∫ ab f(x) dx = limh→0 h[f(a) + f(a+h)+ .... + f(a + (n-1)h)]

A = ∫ ab f(x) dx = (b - a) limn→∞ x 1/n[f(a) + f(a+h)+ .... + f(a + (n-1)h)]

Where h = (b-a)/n ; as n → ∞

The value of the definite integral of a function in a particular interval depends on the function and the interval, but not on the variable of integration.

ab f(x) dx = ∫ ab f(t) dt = ab f(u) du

This variable of integration is called a "dummy variable.