Soalan 11:
Diberi=∫52g(x)dx=−2. Cari(a) nilai bagi ∫25g(x)dx,(b) nilai bagi m jika ∫52[g(x)+m(x)]dx=19
Penyelesaian:
(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]=2121m=42m=2
Diberi=∫52g(x)dx=−2. Cari(a) nilai bagi ∫25g(x)dx,(b) nilai bagi m jika ∫52[g(x)+m(x)]dx=19
Penyelesaian:
(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]=2121m=42m=2
Soalan 12:
a) Cari nilai bagi ∫1−1(3x+1)3dx.(b) Nilaikan ∫431√2x−4 dx.
Penyelesaian:
a) ∫1−1(3x+1)3dx=[(3x+1)44(3)]1−1 =[(3x+1)412]1−1 =112[44−(−2)4] =112(256−16) =20
(b) ∫431√2x−4 dx=∫431(2x−4)12 dx=∫43(2x−4)−12 dx=[(2x−4)−12+112(2)]43=[√2x−4]43=[√2(4)−4−√2(3)−4]=2−√2
a) Cari nilai bagi ∫1−1(3x+1)3dx.(b) Nilaikan ∫431√2x−4 dx.
Penyelesaian:
a) ∫1−1(3x+1)3dx=[(3x+1)44(3)]1−1 =[(3x+1)412]1−1 =112[44−(−2)4] =112(256−16) =20
(b) ∫431√2x−4 dx=∫431(2x−4)12 dx=∫43(2x−4)−12 dx=[(2x−4)−12+112(2)]43=[√2x−4]43=[√2(4)−4−√2(3)−4]=2−√2
Soalan 13:
Diberi y=x22x−1, tunjukkandydx=2x(x−1)(2x−1)2. Seterusnya, nilaikan ∫2−2x(x−1)4(2x−1)2 dx.
Penyelesaian:
y=x22x−1dydx=(2x−1)(2x)−x(2)(2x−1)2 =4x2−2x−2x2(2x−1)2 =2x2−2x(2x−1)2 =2x(x−1)(2x−1)2 (tertunjuk)∫2−22x(x−1)(2x−1)2 dx=[x22x−1]2−218∫2−22x(x−1)(2x−1)2 dx=18[x22x−1]2−214∫2−2x(x−1)(2x−1)2 dx=18[(222(2)−1)−((−2)22(−2)−1)] =18[(43)−(4−5)] =18(3215) =415
Diberi y=x22x−1, tunjukkandydx=2x(x−1)(2x−1)2. Seterusnya, nilaikan ∫2−2x(x−1)4(2x−1)2 dx.
Penyelesaian:
y=x22x−1dydx=(2x−1)(2x)−x(2)(2x−1)2 =4x2−2x−2x2(2x−1)2 =2x2−2x(2x−1)2 =2x(x−1)(2x−1)2 (tertunjuk)∫2−22x(x−1)(2x−1)2 dx=[x22x−1]2−218∫2−22x(x−1)(2x−1)2 dx=18[x22x−1]2−214∫2−2x(x−1)(2x−1)2 dx=18[(222(2)−1)−((−2)22(−2)−1)] =18[(43)−(4−5)] =18(3215) =415