Definite integral and substitution

Activity 1 . And definite integral by substitution.

Solve the following definite integral :



solution:
Apply the formula to the definidaabf integral ( x ) dx = F ( x ) | ab = F ( b ) - F ( a)

01 ( 3x2 +10 ) dx = [ 3x33 +10 x1 ] 01

01 ( 3x2 +10 ) dx = [ (1) 3 +10 ( 1) ] - [ (0) 3 +10 ( 0 ) ]

01 ( 3x2 +10 ) dx = 1 +10-0-0
01 ( 3x2 +10 ) dx = 11

conclusion :
Using the definite integral can clcular an exact value for an integral , as we did in the year in range (1, 0 ) to obtain a value of 11 .











Exercise 2 . Integration by substitution.
Solve the following integral by the substitution method :


 
solution:
For this case we use the integration of replacement, so we have :
u = x4 +1 du = 4x3dx
 Comprehensive 12x3dx instead we have to factor 4x3dx 12 ie multiply ( 4) (3 ) so that we have:

3 ( x4 +1) ( 4x3 ) dx

Replaced u = x4 +1 and du = 4x3dx with what you have :
3udu
We apply the formula for integration with what we have:
3udu +1 +11 = 3u1 + c
+ C = 3u22 3udu
Substituting the value of u with what we have :

( 12x3 ) ( x4 +1 ) dx = 3 ( x4 +1 ) 2 2 + c

conclusion :

Using integration by substitution facilitala obtaining results by making a change of variable which simplifies the procedure to be done, in the previous exercise we was easier to reach the right solution applying seen on the unit