Chapter 6 Definite Integration Ex 6.1 Solutions

Let I = Let ∴ x = A(x + 2) + B(x + 3) Put x + 3 = 0, i.e. x = -3, we get -3 = A(-1) + B(0) ∴ A = 3 Put x + 2 = 0, i.e. x = -2, we get -2 = A(0) + B(1) ∴ B = -2

Let I = = = a2 + a – 0 = a2 + a ∴ I = 2 gives a2 + a = 2 ∴ a2 + a – 2 = 0 ∴ (a + 2)(a – 1) = 0 1 ∴ a + 2 = 0 or a – 1 = 0 ∴ a = -2 or a = 1.

Let I = = = = (a3 + a2 + a) – (1 + 1 + 1) = a3 + a2 + a – 3 ∴ I = 11 gives a3 + a2 + a – 3 = 11 ∴ a3 + a2 + a – 14 = 0 ∴ (a3 – 8) + (a2 + a – 6) = 0 ∴ (a – 2)(a2 + 2a + 4) + (a + 3)(a – 2) = 0 ∴ (a – 2)(a2 + 2a + 4 + a + 3) = 0 ∴ (a – 2)(a2 + 3a + 7) = 0 ∴ a – 2 = 0 or a2 + 3a + 7 = 0 ∴ a = 2 or a = The latter two roots are not real. ∴ they are rejected. ∴ a = 2.

Let I = = Put 3x = t ∴ 3 dx = dt ∴ dx = When x = 1, t = 3 × 1 = 3 When x = 2, t = 3 × 2 = 6