Short Question 5 – 7 Posted on April 22, 2020 by user Question 5: Given ∫ ( 6 x 2 +1 )dx=m x 3 +x +c, where m and c are constants, find (a) the value of m. (b) the value of c if ∫ ( 6 x 2 +1 )dx=13 when x=1. Solution: (a) ∫ ( 6 x 2 +1 )dx=m x 3 +x +c 6 x 3 3 +x+c=m x 3 +x+c 2 x 3 +x+c=m x 3 +x+c Compare the both sides, ∴ m=2 (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( k−5 )=8 k−5=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 22−2 k 2 −7k=28 2 k 2 +7k+6=0 ( 2k+3 )( k+2 )=0 k=− 3 2 or k=−2