Revise notes Class 12 integrals

Revise notes
Class 12 Maths
Integrals

Topics

1. Introduction
2. Integration as an Inverse Process of Differentiation
3. Methods of Integration
4. Integrals of Some Particular Functions
5. Integration by Partial Fractions
6. Integration by Parts
7. Definite Integral
8. Fundamental Theorem of Calculus
9. Evaluation of Definite Integrals by Substitution
10. Some Properties of Definite Integrals

Definite Integral

A definite integral is denoted by _aʃ^b f(x) dx, where a is called the lower limit of the integral and b is called the upper limit of the integral. The definite integral is introduced either as the limit of a sum or if it has an anti derivative F in the interval [a, b], then its value is the difference between the values of F at the end points, i.e., F(b) - F(a).

Class_12_Maths_Integrals_Definite_Integration

Definite integral as the limit of a sum:

The definite integral _aʃ^b f(x) dx is the area bounded by the curve y = f (x), the ordinates x = a, x = b and the x-axis. To find this area Let us consider the region PQRSP between this curve and the ordinates x = a and x = b as shown in the given figure.

Now, divide the interval [a, b] into n equal sub intervals denoted by [x₀, x₁], [x₁, x₂], ……..,[x_r-1, x_r], ………..

[x_n-1, x_n] where x₀ = a, x₁ = a + h, x₂ = a + 2h, ……….,x_r = a + rh, x_n = b = a + nh or h = (b - a)/n

Again as n -> ∞, h -> 0

The region PRSQP under consideration is the sum of n sub regions, where each sub region is defined on sub intervals [x_r_{– 1}, x_r], r = 1, 2, 3, …, n.

Now, to find the area of the region PQRSQ is calculated as

Lim_x->∞ S_n = area of the region PQRSQ = _aʃ^b f(x) dx

= lim_h->0 h[f(a) + f(a + h) + ………….+ f{a + (n - 1)h}]

= (b - a) * lim_h->∞ (1/n)[f(a) + f(a + h) + ………….+ f{a + (n - 1)h}]

So, _aʃ^b f(x) dx = (b - a) * lim_h->∞ (1/n)[f(a) + f(a + h) + ………….+ f{a + (n - 1)h}]

Where h = (b - a)/n as n -> ∞

This is known as the definition of definite integral as the limit of sum.

Problem: Evaluate the definite integral as limit of sums

ʃ₀⁵ (x + 1) dx

Solution:

We know that

ʃ_a^b f(x) dx = (b - a)lim_n->∞ (1/n)[f(a) + f(a + h) + …………..+ f{a + (n - 1)h}], where h = (b - a)/n

Here a = 0, b = 5 and f(x) = x + 1

So, h = (5 - 0)/n = 5/n

Now, ʃ₀⁵ (x + 1) dx = (5 - 0)* lim_n->∞ (1/n)[f(0) + f(5/h) + …………..+ f{5(n - 1)/n}]

= 5 * lim_n->∞ (1/n)[1 + (5/n + 1) + …………..+ {1 + 5(n - 1)/n}]

= 5 * lim_n->∞ (1/n)[(1 + 1 + 1 + …….n times) + {5/n + 2 * 5/n + …………..+ 5(n - 1)/n}]

= 5 * lim_n->∞(1/n)[(1 + 1 + …….n times) + (5/n){1 + 2 + 3 +…………..+ (n - 1)}]

= 5 * lim_n->∞ (1/n)[n + (5/n) * n(n - 1)/2]

= 5 * lim_n->∞ (1/n)[n + 5(n - 1)/2]

= 5 * lim_n->∞ (n/n)[1 + 5(1 – 1/n)/2]

= 5 * [1 + 5(1 – 1/∞)/2]

= 5 * [1 + 5(1 – 0)/2]

= 5 * [1 + 5/2]

= 5 * (7/2)

= 35/2

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