Definite and Indefinite Integration - Formulas, Properties, Problems

In calculus, integration and differentiation are the two most important concepts. Integration originated during the course of finding the area of a plane figure, whereas differentiation is a process of finding a function that outputs the rate of change of one variable with respect to another variable. Integration is the reverse of differentiation. It is also called the antiderivative. In this section, students will learn about the list of definite and indefinite integration important formulas, how to use integral properties to solve integration problems, integration methods and much more.

Indefinite Integration

The indefinite integral is defined as a function that describes an area under the function’s curve from an undefined point to another arbitrary point. Read more.

Standard Formulas for Indefinite Integration

$\begin{array}{l} \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c \end{array}$
$\begin{array}{l} \int a^{x} d x = \frac{a^{x}}{ℓ n a} + c \end{array}$
$\begin{array}{l} \int e^{x} d x = e^{x} + c \end{array}$
$\begin{array}{l} \int \frac{1}{x} d x = ℓ n x + c \end{array}$
$\begin{array}{l} \int \sin x d x = - \cos x + c \end{array}$
$\begin{array}{l} \int \cos x d x = \sin x + c \end{array}$
$\begin{array}{l} \int \sec^{2} x d x = \tan x + c \end{array}$
$\begin{array}{l} \int \cos e c^{2} x d x = - \cot x + c \end{array}$
$\begin{array}{l} \int \sec x \tan x d x = \sec x + c \end{array}$
$\begin{array}{l} \int \cos e c x \cot x d x = - \cos e c x + c \end{array}$
$\begin{array}{l} \int \frac{1}{\sqrt{1 - x^{2}}} d x \begin{array}{c} = \\ = \end{array} {\begin{array}{c} \sin^{- 1} x + c \\ - \cos^{- 1} x + c \end{array} \end{array}$
$\begin{array}{l} \int \frac{1}{1 + x^{2}} d x \begin{array}{c} = \\ = \end{array} {\begin{array}{c} \tan^{- 1} x + c \\ - \cot^{- 1} x + c \end{array} \end{array}$
.
$\begin{array}{l} \frac{1}{x \sqrt{x^{2} - 1}} = {\begin{array}{c} \sec^{- 1} x + c \\ - \cos e c^{- 1} x + c \end{array} \end{array}$

⇒ Also Read – Introduction to Integration

Methods of Integration

Substitution method
Integration by partial fractions
By parts
Euler substitution
Reduction method

Properties of Indefinite Integration

$\begin{array}{l} k \int f (x) d x = k \int f (x) \end{array}$
$\begin{array}{l} \int (f (x) + g (x)) d x = \int f (x) d x + \int g (x) d x \end{array}$

Below are some Illustrations based on integration properties and the substitution method:

Illustration 1: Solve

\begin{array}{l} \int \frac{{(1 + x)}^{3}}{\sqrt[3]{x^{2}}} d x \end{array}

.

Solution:

\begin{array}{l} \int \frac{{(1 + x)}^{3}}{\sqrt[3]{x^{2}}} d x = \int \frac{1 + 3 x + 3 x^{2} + x^{3}}{x^{2 / 3}} d x \end{array}

\begin{array}{l} = \int (x^{- 2 / 3} + 3 x^{1 / 3} + 3 x^{4 / 3} + x^{7 / 3}) d x \end{array}

\begin{array}{l} = 3 x^{1 / 3} + \frac{9}{4} x^{4 / 3} + \frac{9}{7} x^{7 / 3} + \frac{3}{10} x^{10 / 3} + c \end{array}

Illustration 2: Solve

\begin{array}{l} \int \sec^{2} x \cos e c^{2} x d x \end{array}

.

Solution:

\begin{array}{l} \int \sec^{2} x \cos e c^{2} x d x = \int \frac{\sin^{2} x + \cos^{2} x}{\sin^{2} x \cos^{2} x} d x \end{array}

\begin{array}{l} = \int \sec^{2} x + \cos e c^{2} x d x \end{array}

\begin{array}{l} = \tan x - \cot x + c \end{array}

Illustration 3. Solve

\begin{array}{l} \int 2 x \sin (x^{2}) d x \end{array}

.

Solution: Put x² = t.

Then, 2x dx = dt

\begin{array}{l} \int \sin t d t = - \cos t + c \end{array}

\begin{array}{l} = - \cos x^{2} + c \end{array}

Illustration 4.

\begin{array}{l} \int \sin^{3} x \cos^{5} x d x \end{array}

Solution:

\begin{array}{l} \int \sin^{2} x \cos^{5} x \sin x d x \end{array}

\begin{array}{l} = \int (1 - \cos^{2} x) \cos^{5} x \sin x d x \end{array}

put cos x = t

Therefore, -sin x dx = dt

So, the new equation is:

\begin{array}{l} \int (1 - t^{2}) t^{5} (- d t) \end{array}

\begin{array}{l} = \int t^{5} - t^{7} d t \end{array}

\begin{array}{l} = - (\frac{t^{6}}{6} - \frac{t^{8}}{8}) \end{array}

Resubstituting the value, we get

\begin{array}{l} - \frac{\cos^{6} x}{6} + \frac{\cos^{8} x}{8} + c \end{array}

.

Important Formulae Set for Indefinite Integration

$\begin{array}{l} \int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} (\frac{x}{a}) + c \end{array}$
$\begin{array}{l} \int \frac{d x}{\sqrt{x^{2} - a^{2}}} = ℓ n | x + \sqrt{x^{2} - a^{2}} | \end{array}$
$\begin{array}{l} \int \frac{d x}{\sqrt{x^{2} + a^{2}}} = ℓ n | x + \sqrt{x^{2} + a^{2}} | \end{array}$
$\begin{array}{l} \int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) . \end{array}$
$\begin{array}{l} \int \frac{1}{a^{2} - x^{2}} d x = \frac{1}{2 a} ℓ n | \frac{a + x}{a - x} | . \end{array}$
$\begin{array}{l} \int \frac{1}{x^{2} - a^{2}} d x = \frac{1}{2 a} ℓ n | \frac{x - a}{x + a} | . \end{array}$
$\begin{array}{l} \int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} (\frac{x}{a}) . \end{array}$
$\begin{array}{l} \int \sqrt{a^{2} + x^{2}} d x = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} ℓ n | x + \sqrt{x^{2} + a^{2}} | \end{array}$
$\begin{array}{l} \int \sqrt{x^{2} - a^{2}} d x = \frac{x}{2} \sqrt{x^{2} - a^{2}} + \frac{a^{2}}{2} ℓ n | x + \sqrt{x^{2} - a^{2}} | \end{array}$

Types of Substitutions

Type I

\begin{array}{l} \int \sin^{m} x \cos^{n} x d x \end{array}

Rule:

If from either m and n, one of them is odd, then substitute for even power.
If both are odd, then substitute either of them.
If both even use trigonometric identities.

Type II: Substitution of Trigonometric Functions.

$\begin{array}{l} \sqrt{a^{2} - x^{2}} \to x = a \sin θ \end{array}$
$\begin{array}{l} \sqrt{x^{2} + a^{2}} \to x = a \tan θ \end{array}$
$\begin{array}{l} \sqrt{x^{2} - a^{2}} \to x = a \sec θ \end{array}$
$\begin{array}{l} \sqrt{\frac{x - a}{b - x}} \to x = a \cos^{2} θ + b \sin^{2} θ \end{array}$
$\begin{array}{l} \sqrt{\frac{x - a}{x - b}} \to x = a \sec^{2} θ - b \end{array}$

Type III:

\begin{array}{l} \int \frac{p x + q}{a x^{2} + b x + c} d x & \int \frac{p x + q}{\sqrt{a x^{2} + b x + c}} d x & \int (p x + q) \sqrt{a x^{2} + b x + c} d x \end{array}

We express px + q as:

\begin{array}{l} m {(a x^{2} + b x + c)}^{1} + n \end{array}

Then, this gets changed to standard integral.

Biquadratic Substitutions

$\begin{array}{l} I = \int f (x + \frac{1}{x}) (1 - \frac{1}{x^{2}}) d x \end{array}$

$\begin{array}{l} Put x + \frac{1}{x} = t \end{array}$
$\begin{array}{l} I = \int f (x - \frac{1}{x}) (1 + \frac{1}{x^{2}}) d x \end{array}$

$\begin{array}{l} Put (x - \frac{1}{x}) = t \end{array}$

Solved Problems

Problem 1: Solve

\begin{array}{l} \int \frac{d x}{\sqrt{x^{2} + 2 x + 2}} \end{array}

Solution:

\begin{array}{l} \int \frac{d x}{\sqrt{x^{2} + 2 x + 2}} = \int \frac{d x}{\sqrt{{(x + 1)}^{2} + 1}} \end{array}

\begin{array}{l} = \log | (x + 1) + \sqrt{{(x + 1)}^{2} + 1} | . \end{array}

Problem 2: Solve

\begin{array}{l} \int \sqrt{5 x^{2} + 2 x + 3} d x \end{array}

Solution:

\begin{array}{l} \int \sqrt{5 x^{2} + 2 x + 3} d x \end{array}

here

\begin{array}{l} 5 x^{2} + 2 x + 3 = 5 [{(x + \frac{1}{5})}^{2} + {(\sqrt{\frac{14}{25}})}^{2}] \end{array}

\begin{array}{l} \int \sqrt{5 x^{2} + 2 x + 3} d x = \sqrt{5} \int \sqrt{{(x + \frac{1}{5})}^{2} + {(\sqrt{\frac{14}{25}})}^{2}} \end{array}

\begin{array}{l} = \sqrt{5} \frac{(x + \frac{1}{5})}{2} \sqrt{{(x + \frac{1}{5})}^{2} + {(\sqrt{\frac{14}{15}})}^{2}} + \frac{14}{25.2} \log | (x + \frac{1}{5}) + \sqrt{{(x + \frac{1}{5})}^{2} + \frac{14}{15}} | \end{array}

Problem 3: Evaluate

\begin{array}{l} I = \int \frac{x + 1}{x^{2} + 3 x + 4} d x \end{array}

Solution:

\begin{array}{l} I = \int \frac{x + 1}{x^{2} + 3 x + 4} d x \end{array}

\begin{array}{l} x + 1 = a {(x^{2} + 3 x + 4)}^{1} + b \end{array}

\begin{array}{l} = 2 a x + (3 a + b) \end{array}

[Using Type III (mentioned above)]

\begin{array}{l} ∴ a = \frac{1}{2} b = \frac{- 1}{2} \end{array}

\begin{array}{l} \Rightarrow I = \int \frac{1}{2} \frac{{(x^{2} + 3 x + 4)}^{1}}{x^{2} + 3 x + 4} - \frac{1}{2} (\frac{1}{x^{2} + 3 x + 4}) \end{array}

\begin{array}{l} = \frac{1}{2} \log (x^{2} + 3 x + 4) - \frac{1}{2} \int \frac{1}{{(x + \frac{3}{2})}^{2} + \frac{7}{4}} \end{array}

\begin{array}{l} = \frac{1}{2} \log (x^{2} + 3 x + 4) - \frac{1}{2} \frac{1}{\sqrt{\frac{7}{4}}} \tan^{- 1} \frac{(x + \frac{3}{2})}{\sqrt{\frac{7}{4}}} \end{array}

\begin{array}{l} = \frac{1}{2} \log | x^{2} + 3 x + 4 | - \frac{1}{\sqrt{7}} \tan^{- 1} (\frac{2 x + 3}{\sqrt{7}}) \end{array}

Problem 4: Solve

\begin{array}{l} \int \frac{1 + x^{2}}{1 + x^{4}} d x \end{array}

Solution:

\begin{array}{l} \int \frac{1 + x^{2}}{1 + x^{4}} d x = \int \frac{(1 + \frac{1}{x^{2}})}{(\frac{1}{x^{2}} + x^{2})} d x \end{array}

\begin{array}{l} = \int \frac{(1 + \frac{1}{x^{2}}) d x}{{(x - \frac{1}{x})}^{2} + 2} \end{array}

\begin{array}{l} = \int \frac{d (x - \frac{1}{x})}{{(x - \frac{1}{x})}^{2} + 2} \end{array}

\begin{array}{l} = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{(x - \frac{1}{x})}{\sqrt{2}}) \end{array}

Integration by Partial Fraction

Integrals of rational functions can be evaluated by splitting them into partial fractions.

\begin{array}{l} \frac{f (x)}{g (x)} \end{array}

Where, f and g are polynomials is called a rational function.

If the degree of f < the degree of g, it is called a proper fraction.

Otherwise, it is an improper fraction.

Then

\begin{array}{l} \frac{f (x)}{g (x)} = h (x) + \frac{d (x)}{g (x)}, \end{array}

degree of d < degree of g.

Cases:

1. When g(x) is expressed as the product of non-repeating linear factors,

\begin{array}{l} g (x) = (x - a_{1}) (x - a_{2}) \dots . (x - a_{n}) \end{array}

then

\begin{array}{l} \frac{f}{g} = \frac{A_{1}}{x - a_{1}} + \frac{A_{2}}{x - a_{2}} + \dots . \frac{A_{n}}{x - a_{n}} . \end{array}

2. Some factors of g are repeating, then

\begin{array}{l} g (x) = {(x - a)}^{k} (x - a_{1}) (x - a_{2}) \dots \end{array}

\begin{array}{l} \frac{f}{g} = \frac{A_{1}}{(x - a)} + \frac{A_{2}}{{(x - a)}^{2}} + \dots . . \frac{A_{k}}{{(x - a)}^{n}} + \dots \end{array}

3. If g(x) has a quadratic term, then

\begin{array}{l} \frac{f}{g} = \frac{A x + B}{a x^{2} + b x + c} \end{array}

Where, A & B are constants determined by comparing coefficients.

Let us understand with the help of examples:

Example 1: Solve

\begin{array}{l} \int \frac{d x}{(x + 1) (x - 2)} \end{array}

Solution:

\begin{array}{l} \int \frac{d x}{(x + 1) (x - 2)} = \int \frac{a}{x + 1} + \frac{b}{x - 2} d x \end{array}

\begin{array}{l} = \int \frac{- \frac{1}{3}}{x + 1} + \int \frac{\frac{1}{2}}{x - 2} d x \end{array}

\begin{array}{l} = - \frac{1}{3} ℓ n | x + 1 | + \frac{1}{2} ℓ n | x - 2 | \end{array}

Example 2: Solve

\begin{array}{l} \int \frac{2 x^{2} - 1}{(x - 1) {(x + 1)}^{2}} d x \end{array}

Solution:

\begin{array}{l} \int \frac{2 x^{2} - 1}{(x - 1) {(x + 1)}^{2}} d x = \int \frac{A}{x - 1} + \frac{B}{(x + 1)} + \frac{C}{{(x + 1)}^{2}} \end{array}

\begin{array}{l} 2 x^{2} - 1 = A {(x + 1)}^{2} + B (x + 1) (x + 1) + C (x - 1) \end{array}

Put x = 1 ⇒ A = 1/2

x = -1 ⇒ C – -1/2

B = 3/2.

\begin{array}{l} I = \int \frac{\frac{1}{2}}{(x - 1)} d x + \frac{3}{2} \int \frac{d x}{(x + 1)} + (\frac{- 1}{2}) \int \frac{d x}{{(x - 2)}^{2}} \end{array}

,

\begin{array}{l} = \frac{1}{2} ℓ n | x - 1 | + \frac{3}{2} (l n | x + 1 |) + \frac{1}{2 (x + 2)} + c \end{array}

,

Example 3: Solve

\begin{array}{l} \int \frac{d x}{(x + 2) (x^{2} + 1)} \end{array}

Solution:

\begin{array}{l} \int \frac{d x}{(x + 2) (x^{2} + 1)} = \frac{A}{x + 2} + \frac{B x + C}{x^{2} + 1} \end{array}

Comparing by x = -2 ⇒ A = 1/5

And by x² coefficient B = -1/2 and C = 2/5,

On substituting the values and integrating, we have

\begin{array}{l} I = \frac{1}{5} ℓ n | x + 2 | - \frac{1}{10} ℓ n (x^{2} + 1) + \frac{2}{5} \tan^{- 1} (x) \end{array}

Integration of Trigonometric Functions

Type 1:

\begin{array}{l} I = \int \sin^{m} x \cos^{n} x d x \end{array}

1. If m –odd, put cos x = t

2. If n odd, put sin x = t

3. If m and n rationales, then put tan x = t

4. If both are even, then use the reduction method.

Let us understand with the help of an example:

\begin{array}{l} \int \frac{\cos^{3} x}{\sin^{6} x} d x = \int \frac{1 - t^{2}}{t^{6}} d t \end{array}

Where, t = sin x

\begin{array}{l} = \int t^{- 6} - t^{- 4} d t \end{array}

\begin{array}{l} = - \frac{1}{5 s i n^{5} x} + \frac{1}{3 \sin^{3} x} + c \end{array}

Type 2:

\begin{array}{l} \int \frac{d x}{a \cos x + b \sin x + c} \end{array}

Put t = tan (x/2)

Illustration:

\begin{array}{l} \int \frac{d x}{2 + \sin x} \end{array}

\begin{array}{l} \Rightarrow t = \tan (\frac{x}{2}) \end{array}

\begin{array}{l} d x = \frac{2 d t}{1 + t^{2}} \end{array}

\begin{array}{l} = \int \frac{\frac{2 d t}{1 + t^{2}}}{2 + \frac{2 t}{1 + t^{2}}} \end{array}

\begin{array}{l} \Rightarrow \int \frac{d t}{t^{2} + t + 1} \end{array}

\begin{array}{l} = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 t + 1}{\sqrt{3}}) \end{array}

\begin{array}{l} = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 \tan \frac{x}{2} + 1}{\sqrt{3}}) + c \end{array}

Type 3

\begin{array}{l} \int \frac{d x}{a \cos^{2} x + b \sin^{2} x}, \int \frac{d x}{a + b \sin^{2} x} \end{array}

\begin{array}{l} \int \frac{1}{a + b \cos^{2} x} d x, \int \frac{1}{{(a \sin x + b \cos x)}^{2}} d x \end{array}

or

\begin{array}{l} \int \frac{1}{a + b \sin^{2} x + \cos^{2} x} d x \end{array}

Rule:

Divide both the numerator & denominator by cos²x.

Illustration:

\begin{array}{l} \int \frac{1}{3 - 4 \sin^{2} x} d x = \int \frac{\frac{1}{\cos^{2} x}}{\frac{3}{\cos^{2} x} - \frac{4 \sin^{2} x}{\cos^{2} x}} \end{array}

\begin{array}{l} = \int \frac{\sec^{2} x}{3 \sec^{2} x - 4 \tan^{2} x} d x \end{array}

\begin{array}{l} = \int \frac{d t}{3 (1 + t^{2}) - 4 t^{2}} \end{array}

(Since, by tan x = t, sec²x dx = dt)

\begin{array}{l} = \int \frac{d t}{3 - t^{2}} \end{array}

\begin{array}{l} = \int \frac{1}{{(\sqrt{3})}^{2} - t^{2}} \end{array}

\begin{array}{l} = \frac{1}{2 \sqrt{3}} ℓ n | \frac{\sqrt{3} + t}{\sqrt{3} - t} | \end{array}

\begin{array}{l} = \frac{1}{2 \sqrt{3}} \log | \frac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x} | \end{array}

Indefinite Integration – Video Lesson

Indefinite Integration - Video Lesson

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Indefinite-Integration Problems

Indefinite-Integration Problems

77,284

Definite Integrals

Definite Integrals

Choose n divide [a, b] in n parts of width h = (b – a)/n partition of the interval, putting (n – 1) parts in between

\begin{array}{l} \int_{a}^{b} f (x) d x = lim_{h \to 0} h \sum_{n = 0}^{n - 1} f (a + r h) \end{array}

80,755

Definition of Definite Integral as the Limit of a Solution

\begin{array}{l} = lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + \dots . . + f (a + (n - 1) h)] \end{array}

\begin{array}{l} n h = b - a \end{array}

Integrals using limits

Evaluate as the Limit of a Sum

\begin{array}{l} I = \int_{a}^{b} x d x \Rightarrow f (x) = x s o h = \frac{b - a}{n} \end{array}

f(a) = a

f(a + h) = a + h

\begin{array}{l} I = h [a + (a + h) + \dots . (a + (n - 1) h)] \end{array}

look for nh

\begin{array}{l} = h [(n a) + h + 2 h + 3 h \dots . (n - 1) h] \end{array}

\begin{array}{l} = h n a + \frac{h^{2} n (n - 1)}{2} \end{array}

\begin{array}{l} = (b - a) a + \frac{h^{2} n^{2}}{2} - \frac{{(b - a)}^{2}}{2} = \frac{b^{2} - a^{2}}{2} \end{array}

\begin{array}{l} \sum_{r = 1}^{n} n = \frac{n (n + 1)}{2} \end{array}

\begin{array}{l} \sum_{r = 1}^{n} n^{2} = \frac{n (n + 1) (2 n + 1)}{6} \end{array}

\begin{array}{l} \sum_{r = 1}^{n} n^{3} = {[\frac{n (n + 1)}{2}]}^{2} \end{array}

GP …

\begin{array}{l} a + a r + a r^{2} \dots . a r^{n - 1} = \frac{a (1 - r^{n})}{1 - r} \end{array}

\begin{array}{l} \sin α + \sin (α + β) + \sin (α + 2 β) \dots . \sin (α + (n - 1) β) \end{array}

\begin{array}{l} = \frac{\sin (n \frac{β}{2})}{\sin (\frac{β}{2})} \sin [\frac{α + α + (n - 1) β}{2}] \end{array}

\begin{array}{l} = \frac{\sin (n \frac{d i f f}{2})}{\sin (\frac{d i f f}{2})} \sin (\frac{1 s t + l a s t}{2}) \end{array}

\begin{array}{l} = \frac{\sin (\frac{n d i f f}{2})}{\sin (\frac{d i f f}{2})} \cos (\frac{1 s t + l a s t}{2}) \end{array}

\begin{array}{l} 1 - \frac{1}{2^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + \frac{1}{5^{2}} - \frac{1}{6^{2}} = \frac{π^{2}}{12} \end{array}

\begin{array}{l} 1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \frac{1}{5^{2}} \dots . = \frac{π^{2}}{12} \end{array}

\begin{array}{l} \int_{a}^{b} e^{x} d x = lim_{n \to \infty} h [e^{a} + e^{a + h} + e^{a + 2 h} \dots e^{a + (n - 1) h}] \end{array}

\begin{array}{l} = h e^{a} [1 + e^{n} + e^{2 h} + \dots .] \end{array}

\begin{array}{l} = e^{a} [e^{n h} - 1] \frac{h}{(e^{n} - 1)} \end{array}

\begin{array}{l} = e^{b} - e^{a} \end{array}

\begin{array}{l} \int_{a}^{b} \sin x d x = lim_{n \to \infty} h [\sin a + \sin (a + n) + \dots \sin (a + (n - 1) h)] \end{array}

\begin{array}{l} = lim_{n \to \infty} h \frac{\sin (\frac{n h}{2})}{\sin (\frac{h}{2})} [\sin (\frac{2 a + (n - 1) h}{2})] \end{array}

\begin{array}{l} = \frac{\sin (\frac{b - a}{2})}{(\frac{1}{h}) \sin (\frac{h}{2})} [\sin (a + \frac{n h}{2} - \frac{h}{2})] = 2 \sin (\frac{b - a}{2}) \sin (\frac{a + b}{2}) \end{array}

\begin{array}{l} = \cos (\frac{b - a - (a + b)}{2}) - \cos (\frac{(b - a) + (a + b)}{2}) \end{array}

= cos a – cos b

\begin{array}{l} \int_{0}^{2} (x^{2} + 1) d x = \frac{14}{3} \end{array}

\begin{array}{l} \int_{0}^{2} (3 x^{2} + 2 x + 1) d x = 3 \end{array}

\begin{array}{l} \int_{0}^{\frac{π}{2}} \cos x d x = 1 \end{array}

or

\begin{array}{l} \int_{0}^{\frac{π}{2}} \sin^{2} x d x = h [\frac{1 - \cos 2 x}{2} + \frac{1 - \cos^{(2 n + 2 n)}}{2} \dots \dots] \end{array}

\begin{array}{l} = \frac{1}{2} (b - a) - \frac{1}{4} (\sin 2 b - \sin 2 n) \end{array}

Limits Using Definite Integration

\begin{array}{l} \int_{a}^{b} f (x) d x = lim_{h \to 0} \sum_{r = 1}^{n} f (a + r h) \end{array}

Special case a = 0 and b = 1

\begin{array}{l} \int_{0}^{1} f (x) d x = lim_{h \to 0} \frac{1}{n} \sum_{r = 1}^{n} f (\frac{r}{n}) \end{array}

\begin{array}{l} \int_{0}^{k} f (x) d x = lim_{h \to 0} \frac{1}{n} \sum_{r = 1}^{k n} f (\frac{r}{n}) \end{array}

\begin{array}{l} \sum \to M u l t i p l e o f n \int \end{array}

\begin{array}{l} \frac{1}{n} \to d x \frac{r}{n} \Rightarrow x \end{array}

How to solve integration with the help of limits?

\begin{array}{l} lim_{n \to \infty} [\frac{1}{n a} + \frac{1}{n a + 1} + \frac{1}{n a + 2} + \dots \frac{1}{n b}] \end{array}

\begin{array}{l} = lim_{n \to \infty} \sum_{r = 0}^{(b - a)} \frac{1}{n a + r} \end{array}

\begin{array}{l} = \sum_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{(b - a) n} \frac{1}{a + \frac{r}{n}} = \int_{0}^{b - a} \frac{d x}{a + x} = {[\log (a + x)]}_{a}^{b - a} = \log (\frac{b}{a}) \end{array}

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Frequently Asked Questions

Q1

What do you mean by integration?

Integration is the method of finding the antiderivative of a function.

Q2

What are the rules of integration?

There are many rules of integration, namely the power rule, the sum and difference rules, the exponential rule, the reciprocal rule, the constant rule, the substitution rule, and the rule of integration by parts, etc.

Q3

Why is integration important?

We utilise integration in Mathematics to find areas, volumes, displacement, etc. Integral calculus originated from the concept of integration in calculus.

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