Syllabus:
\[\text{Integration formulae of standard functions as inverse operation of}\ \hspace{10cm}\\ \text{differentiation – Bernoulli’s formula – Definite integrals}\ \hspace{10cm}\\ \text{Area and volume using integration – Simple problems.}\ \hspace{5cm}\]
Introduction:
In differentiation, if a function is given, then we know how to find its derivative. Let us take up the reverse problem. i.e, if the derivative of a function is given, we can find the function. This is to find a function whose derivative is given as a function f(x). A function F(x) thus found is called “the integral of f(x)” and the process of finding it is called integration. F(x) is also called anti derivative (or) primitive of f(x) w.r.t.x. Thus integration can be regarded as the reverse (or) inverse process of differentiation.
In symbol we write ∫ f(x) dx = F(x). This is read as integral of f(x) with respect to x’ (or) integral f(x)dx. f(x) is called the integrand and the differential dx indicates that x is the variable of the integration.
List of Formulae:
\[1.\int x^n \ dx= \frac{x^n+1}{n+1} +c\]
\[2.\int 1\ dx = x + c\]
\[3.\int \frac{1}{x}\ dx = log x + c\]
\[4.\int e^x \ dx= e^x +c\]
\[5.\int sin x \ dx= – cos x +c\]
\[6.\int cosx \ dx= sin x +c\]
\[7.\int sec^2x \ dx= tan x +c\]
\[8.\int cosec^2x \ dx= – cot x +c\]
\[9.\int sec x\ tan x\ dx= sec x +c\]
\[10.\int cosec x\ cot x\ dx= – cosec x +c\]
\[11.\int sin ax \ dx= -\frac{1}{a} cos ax +c\]
\[12.\int cos ax \ dx= \frac{1}{a} sin ax +c\]
\[\color {royalblue} {Properties}:\ \hspace{20cm}\]
\[1.\ \text{The process of differentiation and integration neutralizes each other.}\ \hspace{7cm}\\ i.e,\ \frac{d}{dx}\ (\int f(x)\ dx)\ =\ f(x)\]
\[2.\ \text{The integral of the product of a constant and a function is equal to the}\ \hspace{7cm}\\ \text{product of the constant and the integral}\ i.e,\ \int k f(x) dx\ =\ k\ \int f(x)\ dx\]
\[3.\ \text{If u,v,w… (finite number) are functions of x then}\ \hspace{12cm}\\ \int (u + v – w + …) dx\ =\ \int u\ dx\ +\ \int v\ dx\ -\ \int w\ dx\ +\ …..\]
INTEGRATION – DECOMPOSITION METHOD:
\[\color {purple} {Example\ 1\ .}\ \color {red}{Evaluate\ :} \int(x^2 -x + 5)\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(x^2 -x + 5)\ dx = \frac{x^3}{3} – \frac{x^2}{2} + 5x + c\]
\[\color {purple} {Example\ 2\ .}\ \color {red}{Evaluate\ :} \int(x^3\ -\ x\ -\ 2)\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(x^3\ -\ x\ -\ 2)\ dx = \frac{x^4}{4}\ -\ \frac{x^2}{2}\ -\ 2x\ + c\]
\[\color {purple} {Example\ 3\ .}\ \color {red}{Evaluate\ :} \int(\frac{100}{x}+100)\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(\frac{100}{x}+100)\ dx = 100\int \frac{1}{x}\ dx + 100\int 1\ dx\]
= 100 log x +100 x + c
\[\therefore\ \boxed{\int(\frac{100}{x}+100)\ dx = 100\ log\ x\ +\ 100\ x\ +\ c}\]
\[\color {purple} {Example\ 4\ .}\ \color {red}{Evaluate\ :} \int(x^2 + \frac{3}{x})\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(x^2 + \frac{3}{x})\ dx = \frac{x^3}{3} + 3log x + c\]
\[\color {purple} {Example\ 5\ .}\ \color {red}{Evaluate\ :} \int(2 + x)^2\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(2 + x)^2\ dx\ =\int(4+ 4x + x^2)\ dx\]
\[=4\int(1)dx + 4\int(x)dx + \int(x^2) dx\]
\[=4x + 4 \frac{x^2}{2} + \frac{x^3}{3} + c\]
\[=4x + 2 x^2 + \frac{x^3}{3} + c\]
\[\therefore\ \boxed{\int(2 + x)^2\ dx\ =\ 4x + 2 x^2 + \frac{x^3}{3} + c}\]
\[\color {purple} {Example\ 6\ .}\ \color {red}{Evaluate\ :} \int(x + 3 ) ( x + 2)\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(x + 3 ) ( x + 2)\ dx\]
\[= \int( x^2 +\ 3x\ +\ 2x\ +\ 6)\ dx\]
\[= \int(x^2\ +\ 5x\ +\ 6)\ dx\]
\[ = \frac{x^3}{3}\ +\ 5\frac{x^2}{2}\ +\ 6x\ +\ c\]
\[\therefore\ \boxed{\int(x + 3 ) ( x + 2)\ dx\ =\ \frac{x^3}{3}\ +\ 5\frac{x^2}{2}\ +\ 6x\ +\ c}\]
\[\color {purple} {Example\ 7\ .}\ \color {red}{Evaluate\ :} \int(x^2 + x + 1) ( x^2 – x + 1)\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(x^2 + x + 1) ( x^2 – x + 1)\ dx\]
\[= \int(x^4 – x^3 + x^2 + x^3 – x^2 + x + x^2 – x + 1)\ dx\]
\[= \int(x^4 + x^2 + 1)\ dx\]
\[ = \frac{x^5}{5} + \frac{x^3}{3} + x + c\]
\[\therefore\ \boxed{\int(x^2 + x + 1) ( x^2 – x + 1)\ dx\ =\ \frac{x^5}{5} + \frac{x^3}{3} + x + c}\]
\[\color {purple} {Example\ 8\ .}\ \color {red}{Evaluate\ :}\ \int(2 sin x + 7)\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(2 sin x + 7)\ dx =2\int sin x\ dx + 7\int 1\ dx\]
\[=- 2 cos x + 7x + c\]
\[\therefore\ \boxed{\int(2 sin x + 7)\ dx\ =\ – 2 cos x + 7x + c}\]
Trigonometry related formulae:
\[1.\ sin^2 x + cos^2 x = 1\]
\[2.\ tan^2 x = sec^2 x – 1\]
\[3.\ cot^2 x = cosec^2 x – 1\]
\[\color {purple} {Example\ 9:}\ \color {red}{Evaluate:} \int \frac{sin^2 x}{ 1- cos x} \ dx\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int \frac{sin^2 x}{ 1- cos x} \ dx = \int\frac{1 – cos^2 x}{ 1- cos x} \ dx\]
\[= \int\frac{(1+ cos x)(1 – cos x)}{1- cos x}\ dx\]
\[=\int ( 1 + cos x)\ dx\]
\[= x + sin x +c\]
\[\therefore\ \boxed{\int \frac{sin^2 x}{ 1- cos x} \ dx\ =\ x + sin x +c}\]
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\[\color {purple} {Example\ 10:}\ \color {red}{Evaluate:}\ \int \sqrt{1\ +\ sin\ 2x}\ dx\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int \sqrt{1\ +\ sin\ 2x} \ dx = \int \sqrt{sin^2\ x\ +\ cos^2\ x\ +\ 2\ sin\ x\ cos\ x} \ dx\]
\[= \int \sqrt{(sin\ x\ +\ cos\ x)^2}\ dx\]
\[= \int (sin\ x\ +\ cos\ x)\ dx\]
\[= \int\ sin\ x\ dx\ +\ \int cos\ x\ dx\]
\[=\ -\ cos\ x\ +\ sin\ x\ +\ c\]
\[\therefore\ \boxed{\int \sqrt{1\ +\ sin\ 2x}\ dx\ =\ -\ cos\ x\ +\ sin\ x\ +\ c}\]
\[\color {purple} {Example\ 11:}\ \color {red}{Evaluate:} \int tan^2 x \ dx\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int tan^2 x \ dx = \int(sec^2 x – 1)\ dx\]
\[=\int sec^2 x\ dx – \int 1\ dx\]
= tan x – x+ c
\[\therefore\ \boxed{\int tan^2 x \ dx\ =\ tan\ x\ -\ x\ + c}\]
Definition of Definite Integrals:
\[\int_{a}^{b} f(x)\ dx = F(b) – F(a)\]
Example 1:
\[\color{red}{Evaluate: \int_1^2 \frac{dx}{x}}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int_1^2 \frac{1}{x}\ dx = log\ x \Biggr]_{1}^{2}\]
\[=[log\ 2 – log\ 1]\]
\[= log\ 2 – 0\]
\[= log\ 2\]
\[\boxed{\int_1^2 \frac{1}{x}\ dx = log\ 2}\]
Example 2:
\[\color{red}{Evaluate: \int_0^1 (x^2\ +\ x\ +\ 1\ 2)\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int_0^1 (x^2\ -\ 2)\ dx = \frac{x^3}{3}\ -\ 2x \Biggr]_{0}^{1}\]
\[= [(\frac{1^3}{3}\ -\ 2(1))\ -\ 0]\]
\[= \frac{1}{3}\ -\ 2\]
\[= \frac{1 – 6}{3}\]
\[= \frac{-5}{3}\]
\[\boxed{\int_0^1 (x^2\ -\ 2)dx\ = \frac{-5}{3}}\]
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