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)dx∫302g(x)dx=2∫30g(x)dx =2[y]30 =2[5xx2+1]30 =2[5(3)32+1−0] =2(1510) =3
Given y=5xx2+1 and dydx=g(x), find the value of ∫302g(x)dx.
Solution:
Sincedydx=g(x), thus y=∫g(x)dx∫302g(x)dx=2∫30g(x)dx =2[y]30 =2[5xx2+1]30 =2[5(3)32+1−0] =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+2k2−352 =k2+2k−352
Find ∫k5(x+1)dx, in terms of k.
Solution:
∫k5(x+1)dx=[x22+x]k5 =(k22+k)−(522+5) =k2+2k2−352 =k2+2k−352
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+m∫52xdx=19 −2+m[x22]52=19 m2[x2]52=21 m2[25−4]=21 21m=42 m=2
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+m∫52xdx=19 −2+m[x22]52=19 m2[x2]52=21 m2[25−4]=21 21m=42 m=2