# CPT Chapter Integral Calculus With Anand.Duration:5 hrs 26mins

 Intergration Duration (min:sec) Introduction 08:36 Index 06:24 What is Integration 03:23 Basic Formulae 11:14 Integration of Constant 03:40 Example Basic Formulae 06:06 Problems on Basic Formulae 15:18 Special Variable 25:56 Method of Substitution & Examples 15:07 Problems on Method of Substitution 26:29 Integration by Parts & Example 11:18 Problems Integration by Parts 19:29 Special cases- Integration by Parts 06:17 Partial Fraction 08:52 Problem on Partial Fraction 21:00 Special Substitution 28:07 What is Definite Integration (DI) 03:54 Properties of DI 06:23 Examples of DI 06:14 Problems 1 DI 26:07 Problems 2 DI 21:06 Problems 3 DI 20:35 Slope Problem 06:06 Mis Problem 12:57 Summary 05:28 Total 5:26:06

Few things which we have covered

What is Integration?
Basic Formulae
Special Variable
Actual Substitution
Integration by parts
Parts Special case
Partial Fraction
Special Substitution
Definite Integration
Properties of Definite Integration

What is Integration?
Integration is reverse of Differentiation
ex-

Symbol
∫x2dx

Integration of a constant-

∫ a dx =∫a.x0 dx

∫ 7 dx =

Addition & Subtraction to a constant-

3+k

3-k

Example-

∫(√x +1/√x)dx
(a) 2x1/2(1/3 x-1)
(b) 2x1/2(1/3 x+1)
(c) 2(1/3x + x1/2)
(d) None of these.

Evaluate = ∫5x2 dx :
(a) 5/3x3+ k (b) 5x3/3+k (c) 5x3 (d) none of these

Integration of 3 – 2x – x4 will become
(a) – x2 – x5 / 5 (b) 3x - x2 - x5 /5+ k (c) 3x -x2 +x5/5 +k (d) none of these

Given f(x) = 4x3 + 3x2 – 2x + 5 and ∫ f(x) dx is
(a) x4 + x3 – x2 + 5x (b) x4 + x3 – x2 + 5x + k
(c) 12x2 + 6x – 2x2 (d) none of these

Special Variables-

This is applied when by taking a linear function of x i.e ax or ax+b as X , we can apply any of the standard formulae

Here we apply standard formula and divide the answer with derivative of ax+b i.e a

Use method of substitution to integrate the function f(x)=(4x+5)6 and the answer is
(a) 1/28(4x+5)7+k
(b) (4x+5)7/7+k
(c) (4x+5)7/7
(d) none of these

Integrate (x+a)n and the result will be
(a) (x+a)n+1/n+1 + k
(b) (x+a)n+1/n+1
(c) (x+a)n+1
(d) none of these

Method of Substitution or Change of Variable-

Substitution
When we have a function & it’s derivatives
∫ (5x2+4x+3)4 (10x+4)dx

Note - Trick to identify which one is function & which is derivative

Evaluate: ∫dx/√(x2+a2)
(a) 1/2 - log (x+√x2+a2) + C
(b) log (x+√x2+a2) +C
(C) log (x √x2+a2)+ C
(d) 1/2log (x √x2+a2)+ C

Square Partial Integration-

Using method of partial fraction to evaluate
∫ (x+5) dx/(x +1) (x + 2)2 we get
(a) 4 log (x + 1) - 4log(x + 2)+3/x+2+k
(b) 4 log (x + 2) – 3/x +2)+k
(c) 4 log (x + 1) – 4 log(x+2)
(d) none of these

Special Substitution-

∫d( x2+1) / √x2+2 is equal to
(a) x/2 (√x2 +2)+k
(b) √x2 +2 +k
(c) 1/(x2+2)3/2 +k
(d) none of these

Definite Integration-

The value of ∫(1+logx)/x dx is:[Given Loge =1]
(a) 1/2
(b) 3/2
(c) 1
(d) 5/2

Summary-

What is Integration?
Basic Formulae
Special Variable
Actual Substitution
Integration by parts
Parts Special case
Partial Fraction
Special Substitution
Definite Integration
Properties of Definite Integration

>

# CPT Exam Exposure

Take Quiz Show you appreciate by sharing!

Fields marked with * are required