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CPT Chapter Integral Calculus

CA CPT Exam  

With Anand.Duration:5 hrs 26mins

IntergrationDuration (min:sec)
Introduction08:36
Index06:24
What is Integration03:23
Basic Formulae11:14
Integration of Constant03:40
Example Basic Formulae06:06
Problems on Basic Formulae15:18
Special Variable25:56
Method of Substitution & Examples15:07
Problems on Method of Substitution26:29
Integration by Parts & Example11:18
Problems Integration by Parts19:29
Special cases- Integration by Parts06:17
Partial Fraction08:52
Problem on Partial Fraction21:00
Special Substitution28:07
What is Definite Integration (DI)03:54
Properties of DI06:23
Examples of DI06:14
Problems 1 DI26:07
Problems 2 DI21:06
Problems 3 DI20:35
Slope Problem06:06
Mis Problem12:57
Summary05:28
Total5:26:06

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 


Notes

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