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
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=14∫k5g(x)dx+∫k52dx=14 6+[2x]k5=14 2(k−5)=8 k−5=4 k=9
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=14∫k5g(x)dx+∫k52dx=14 6+[2x]k5=14 2(k−5)=8 k−5=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)=2822−2k2−7k=282k2+7k+6=0(2k+3)(k+2)=0k=−32 or k=−2
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)=2822−2k2−7k=282k2+7k+6=0(2k+3)(k+2)=0k=−32 or k=−2