Short Question 8 – 10


Question 8:
Given y=5xx2+1 and dydx=g(x), find the value of 302g(x)dx.

Solution:
Sincedydx=g(x), thus y=g(x)dx302g(x)dx=230g(x)dx  =2[y]30  =2[5xx2+1]30  =2[5(3)32+10]  =2(1510)  =3



Question 9:
Find k5(x+1)dx, in terms of k.

Solution:
k5(x+1)dx=[x22+x]k5  =(k22+k)(522+5)  =k2+2k2352  =k2+2k352



Question 10:
Given that=52g(x)dx=2. Find(a) the value of 25g(x)dx,(b) the value of m if 52[g(x)+m(x)]dx=19

Solution:
(a) 25g(x)dx=52g(x)dx                     =(2)                     =2

(b) 52[g(x)+m(x)]dx=19     52g(x)dx+m52xdx=19              2+m[x22]52=19                      m2[x2]52=21                   m2[254]=21                            21m=42                                m=2