Short Question 5 – 7


Question 5:
Given (6x2+1)dx=mx3+x+c, where m and c are constants, find(a) the value of m.(b) the value of c if (6x2+1)dx=13 when x=1.

Solution:
(a)
(6x2+1)dx=mx3+x+c6x33+x+c=mx3+x+c2x3+x+c=mx3+x+cCompare the both sides,

(b)

( 6 x 2 +1 )dx=13 when x=1. 2 ( 1 ) 3 +1+c=13            3+c=13                 c=10



Question 6:
It is given that  5 k g(x)dx=6 , and  5 k [ g( x )+2 ]dx =14, find the value of k.

Solution:
5 k [ g( x )+2 ]dx =14 5 k g( x )dx + 5 k 2dx =14                6+ [ 2x ] 5 k =14                 2( k5 )=8                      k5=4                           k=9



Question 7:
Given  k 2 (4x+7)dx=28 , calculate the possible value of k.

Solution:
k 2 (4x+7)dx=28 [ 2 x 2 +7x ] k 2 =28 8+14( 2 k 2 +7k )=28 222 k 2 7k=28 2 k 2 +7k+6=0 ( 2k+3 )( k+2 )=0 k= 3 2  or k=2