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Question 1: If 4 (x 2  − 2 )dx = 15. Then find the value (values) of x          that satisfy the Mean-Value

 

1

 

Theorem for Integrals for the given function on the closed interval.

 

Solution:

 

4

f ( x )dx = f ( x * )( b a)

 

1

 

4

( x 2  − 2)dx = ( x* ) 2  − 2  (41)

 

 

1

15    = ( x* ) 2  − 2  3

 

 

 

153 = ( x* ) 2  − 2

5  + 2 = ( x* )2

 

7   = ( x* )2

 

( x* ) 2  = 7

 

x*  =  7

 

 

Question 2: Find the area of the region bounded by the curve  y = x 2 2x and the line y = 3 .

 

Answer:

 

y = x 2  − 2 x        y = 3

 

L. H .S equal then R.H.S is equal

 

x 2  − 2 x = 3

 

x 2  − 2 x 3 = 0

 

x 2  − 3x + x 3 = 0

 

x ( x 3) + 1( x 3) = 0

 

( x 3)( x + 1)

= 0

x 3 = 0

,   x + 1 = 0

x = 3

,

x = −1


y = 3 is greater function from  y = x 2  − 2 x

 

3

A      =   3 ( x 2  − 2 xdx

 

 

1

 

3

A      =   3 x 2  + 2 x  dx

 

 

1


 

 

x 3

2 x2   3

 

 

 

 

 

A = 3 x

 

 

 

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

2   1

 

 

 

 

 

 

 

x3

 

3

 

 

 

 

 

A = 3 x

 

 

 

+ x2

 

 

 

 

 

3

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

(3)3

 

 

 

 

 

 

(1)3

A = 3(3)

 

 

 

+ (3)2

 

3(

1)

 

3

 

3

 

 

 

 

 

 

 

 

 

 

A  = 99+9 − −3+1+1 3

 

A  =9 12 3


A=913+2

A= 271+6

3

 

A=32

3



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