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, m=2

(b)

(6x2+1)dx=13 when x=1.2(1)3+1+c=13           3+c=13                c=10



Question 6:
It is given that k5g(x)dx=6, and k5[g(x)+2]dx=14,find the value of k.

Solution:
k5[g(x)+2]dx=14k5g(x)dx+k52dx=14               6+[2x]k5=14                2(k5)=8                     k5=4                          k=9



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

Solution:
2k(4x+7)dx=28[2x2+7x]2k=288+14(2k2+7k)=28222k27k=282k2+7k+6=0(2k+3)(k+2)=0k=32 or k=2