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3.7.4 Integration, SPM Practice (Question 9 – 12)


Question 9:
Given y= 5x x 2 +1  and  dy dx =g( x ), find the value of  ∫ 0 3 2g( x )dx.

Solution:
Since dy dx =g( x ), thus y= ∫ g( x ) dx ∫ 0 3 2g( x )dx=2 ∫ 0 3 g( x )dx   =2 [ y ] 0 3   =2 [ 5x x 2 +1 ] 0 3   =2[ 5( 3 ) 3 2 +1 −0 ]   =2( 15 10 )   =3


Question 10:
Find  ∫ 5 k ( x+1 )dx, in terms of k.

Solution:
∫ 5 k ( x+1 )dx=[ x 2 2 +x ] 5 k   =( k 2 2 +k )−( 5 2 2 +5 )   = k 2 +2k 2 − 35 2   = k 2 +2k−35 2


Question 11:
Given that= ∫ 2 5 g(x)dx=−2 . Find (a) the value of  ∫ 5 2 g(x)dx, (b) the value of m if  ∫ 2 5 [ g(x)+m( x ) ]dx=19

Solution:
(a)  ∫ 5 2 g(x)dx= − ∫ 2 5 g(x)dx                      =−( −2 )                      =2

(b)  ∫ 2 5 [ g(x)+m( x ) ]dx=19       ∫ 2 5 g(x)dx+m ∫ 2 5 xdx=19               −2+m [ x 2 2 ] 2 5 =19                        m 2 [ x 2 ] 2 5 =21                     m 2 [ 25−4 ]=21                             21m=42                                 m=2


Question 12:
(a) Find the value of  ∫ −1 1 ( 3x+1 ) 3 dx. (b) Evaluate  ∫ 3 4 1 2x−4  dx.

Solution:
a)  ∫ −1 1 ( 3x+1 ) 3 dx=[ ( 3x+1 ) 4 4( 3 ) ] −1 1                            = [ ( 3x+1 ) 4 12 ] −1 1                            = 1 12 [ 4 4 − ( −2 ) 4 ]                            = 1 12 ( 256−16 )                            =20

(b)  ∫ 3 4 1 2x−4  dx= ∫ 3 4 1 ( 2x−4 ) 1 2  dx                             = ∫ 3 4 ( 2x−4 ) − 1 2  dx                             = [ ( 2x−4 ) − 1 2 +1 1 2 ( 2 ) ] 3 4                             = [ 2x−4 ] 3 4                             =[ 2( 4 )−4 − 2( 3 )−4 ]                             =2− 2


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