Chapter 4 Definite Integration Ex 4.1 Solutions

Let f(x) = 3x – 4, for 1 ≤ x ≤ 3 Divide the closed interval [1, 3] into n subintervals each of length h at the points 1, 1 + h, 1 + 2h, 1 + rh, ….., 1 + nh = 3 ∴ nh = 2 ∴ h = and as n → ∞, h → 0 Here, a = 1 ∴ f(a + rh) = f(1 + rh) = 3(1 + rh) – 4 = 3rh – 1

Let f(x) = x2, for 0 ≤ x ≤ 4 Divide the closed interval [0, 4] into n subintervals each of length h at the points 0, 0 + h, 0 + 2h, ….., 0 + rh, ….., 0 + nh = 4 i.e. 0, h, 2h, ….., rh, ….., nh = 4 ∴ h = as n → ∞, h → 0 Here, a = 0

Let f(x) = ex, for 0 ≤ x ≤ 2 Divide the closed interval [0, 2] into n equal subntervals each of length h at the points 0, 0 + h, 0 + 2h, ….., 0 + rh, ….., 0 + nh = 2 i.e. 0, h, 2h, ….., rh, ….., nh = 2 ∴ h = and as n → ∞, h → 0 Here, a = 0

Let f(x) = 3x2 – 1, for 0 ≤ x ≤ 2 Divide the closed interval [0, 2] into n subintervals each of length h at the points. 0, 0 + h, 0 + 2h, ….., 0 + rh, ……, 0 + nh = 2 i.e. 0, h, 2h, ….., rh, ….., nh = 2 ∴ h = and as n → ∞, h → 0 Here, a = 0 ∴ f(a + rh) = f(0 + rh) = f(rh) = 3(rh)2 – 1 = 3r2h2 – 1

Let f(x) = x3, for 1 ≤ x ≤ 3. Divide the closed interval [1, 3] into n equal su bintervals each of length h at the points 1, 1 + h, 1 + 2h, ……, 1 + rh, ……, 1 + nh = 3 ∴ nh = 2 ∴ h = and as n → ∞, h → 0 Here a = 1