A. Definition of Integration

In Class XI, you have learned the concept of derivative. Comprehensionon the derivative concept you can use to understandintegration concept. For that, try to determine the following derivative functions:

f 1( x )=3 x3+3 f 2( x )=3 x3+7 f 3( x )=3 x3−1 f 4( x )=3 x3−10 f 5( x )=3 x3−99

Note that these functions have the general form , f ( x )=3 x3+c with c is constant. Each function has a derivative f ' ( x )=9 x2

. Thus, the derivative function f ( x )=3 x3+c is f ' ( x )=9 x2.

Now,what if you have to define the function f (x) of f ' (x) is known?Determine the function f (x) from f ' (x), means determining antiderivative of f ' (x). Thus, the integration is the antiderivative (Antidiferensial) or the inverse operation of the differential.

If F (x) is general function y that is common F '(x )=f (x ) , then F ( x ) is antiderivative or integral of f (x).

Integration function f (x) with respect to x is denoted as follows:

∫ f ( x )dx=F ( x)+c

With :

∫ = integration

f ( x ) = function integration

F ( x ) = integration common function

c = constanta

Now, consider the derivative of the following functions ;

g1 ( x )=x, be obtained g1 ' ( x )=1

So, if g1 ' ( x )=1,then g1 ( x)= ∫ g1' (x ) dx=x+c

g2 ( x )=12

x2 , be obtained g2 ' ( x )=x

So,if g2 ' ( x )=x, then g2 ( x)= ∫ g2' ( x ) dx=1

2x2+c

g3 ( x )=13

x3 , be obtained g3 ' ( x )=x

So,if g3 ' ( x )=x, then g3 ( x )= ∫ g3 ' ( x )dx=13

x3+c

g4 ( x )=16

x6 ,be obtained g4 ' ( x )=x5

So,if g4 ' ( x )=x5, then g4 ( x )= ∫ g4 ' ( x )dx=16

x6+c

Of this description, it appears that if g' (x )=xn, then g ( x )= 1n+1

xn+1+cor

can be written ∫ xn dx= 1n+1

xn+1+c , n ≠−1.

For example, the derivative function f ( x )=3 x3+c is f ' ( x )=9 x2.This means, antiderivative of f ' ( x )=9 x2 is f ( x )=3 x3+c or written ∫ f ' ( x ) dx=3 x2+c.This description illustrates the following relationship.

If f ' ( x )=xn,then f ( x )= 1n+1

xn+1+c ,n ≠−1, with c is a constant.

Example:

1. Find the derivative of each of the following functions :

Answers:

2. Find the antiderivative x if known:

Answers:

B. Indefinite Integrals

In the previous part, you have known that the integral is an antiderivative. So, if there

is a function F(x) that can differential at intervals [ a , b ] so that d ( F ( x ))

dx=f ( x ),the

antiderivative of f (x) is F (x) + c.Mathematically, written

∫ f ( x ) dx=F (x )+cwhere, ∫ dx= symbol of stated integral antiderivative operation

f(x) = integrand functions, namely functions which sought antiderivativec = constant

For example, you can write

Because,

So you can look at indefinite integral as representatives of the whole family of functions (one antiderivative for each value constant c. The definition can be used to provethe following theorems which will help in the execution of arithmeticintegrals.Theorem 1

If n is a rational number and n ≠−¿1,then ∫ xn dx= 1n+1

xn+1+c where

c is a constant.

Theorem 2

If f the integral function and k is a constant, then ∫k f(x) dx = k ∫ f (x) dx

Theorem 3

If f and g is integral functions, then ∫ ( f ( x )+g ( x ) ) dx= ∫ f ( x )dx+ ∫ g ( x ) dx

Theorem 4

If f and g is integral functions, then ∫ ( f ( x )−g ( x ) ) dx= ∫ f ( x ) dx− ∫ g ( x )dx

Theorem 5

Substitution Integrals Rule

If u is a function which can differential and r is a numbers which no zero, then

∫ (u ( x ) )r u' ( x ) dx= 1n+1 ( u (x ) )r+1+c, where c is a constant and r≠−1

Theorem 6

Partial Integrals Rule

If u and v is a functions which can differential, then ∫ udv=uv− ∫ vdu

Theorem 7

Trigonometri Integrals Rule

∫ sin dx=−cos x+c ∫ cos dx=sin x+c

∫ 1cos2 x

dx=tan x+c

Where c is a constant

Prove Theorem 1

For prove theorem 1,we can differential xn+1+cwhich be found at right space the following ;

ddx

( xn+1+c )=(n+1 ) xn …. multiply twospace with 1n+1

1n+1

. ddx

( xn+1+c )= (n+1 ) xn . 1n+1

ddx [ xn+1

n+1+c ]=xn

So, ∫ xn dx= 1n+1

xn+1+c

Prove Theorem 3 and 4

For prove theorem 4,we can differential ∫ f ( x ) dx ± ∫ g ( x ) dxwhich be found at right space the following ;

ddx

∫ f ( x ) dx± ∫ g ( x ) dx= ddx [∫ f ( x ) dx ]± [∫ g ( x ) dx ]= f (x)± g (x)

ddx

∫ f ( x ) dx ± ∫ g ( x ) dx=f (x)± g(x )

So,

∫ (f ( x ) ± g ( x ))dx= ∫ f (x )dx ± ∫ g (x ) dx

1. Find integral from ∫ (3 x2−3 x+7 ) dx !Answers:∫ (3 x2−3 x+7 ) dx=3 ∫ x2 dx−3 ∫ x dx+∫7 dxtheorema 2,3 and 4

¿ 32+1

x2+1− 31+1

x1+1+7 x+c theorem 1

¿ x3−32

x2+7 x+c

So, ∫ (3 x2−3 x+7 ) dx=x3−32

x2+7 x+c

Prove Theorem 6

In the class XI, you have know derivative of two times product of functions

f ( x )=u ( x ) . v (x) is ddx [u ( x ) . v (x )]=u ( x ) . v' ( x )+v (x ) . v ' (x)

It will prove that partial integral rule with formula them. Method them is with differential two equation it the following :

∫ ddx [u ( x ) . v ( x ) ]= ∫ u ( x ) . v ' ( x )+ ∫ v ( x ) . v '(x )dx

u ( x ) . v( x)= ∫ u ( x ) . v ' ( x )+ ∫ v ( x ) . v '( x)dx

∫ u ( x ) . v ' ( x )=u ( x ) . v ( x )− ∫ v (x ) . v ' (x)dx

Because v’(x) dx= dv and u’(x)dx=du

So,the equation can be written ∫ u dv=uv−∫ vdu

B.1 Substitution Integral Rule

Substitution Integral Rule is like which be written at Theorem 5. This rule was used for to solve the problem in integration which not can to solve with base formulas what already learn. For remainder it, example the following it

Example ;

1. Find the integral from

Answers;

a. Supposing that: u=9-x2 then du =-2x dx

So,

b. Supposing that u= √ x =x12

with the result that

c. Supposing that u= 1- 2x2 and du = -4x dx

dx =du

−4 x

so the integral can be written the following

Substitution u= 1- 2x2 to equation 12u-3+ c

So,

Prove theorem 7

In the class XI, you have learn derivative trigonometric function, is ddx

(sin x )=cos x

ddx

(cos x )=−sin x ,and ddx

(tan x )=sec2 x

The following this we can prove trigonometric integral rule to use formulas. This method is with integration two space the following;

From ddx

(sin x )=cos xbe found ∫ cos dx=sin x+c

From ddx

(cos x )=−sin x be found ∫ sin dx=−cos x+c

From ddx

( tan x )=sec2 x be found ∫ sec2 dx=tan x+c

B.2 space integral with √a2−x2,√a2+x2 and √ x2+a2

Integration spaces √a2−x2,√a2+x2 and √ x2+a2 can be work with substitution with x = a sin t, x= a tan t, x = a sec t. So can be found spaces the following it ;

Right angle for integral trigonometric substitution;

( i ) √a2−x2=acos x,(ii ) √a2+x2=a sect , (iii)√ x2−a2 =a tan x

1. Find each integral the following it:

Answers:For to work this integral, you must change sin(3x+1)cos(3x+1) in the double angle trigonometric formulas