Question

In the following Stieltjes integrals, it should be useful to sketch the mass generating function g.

(a) Evaluate the integral ∫₀³ t²dg(t),
where g(t) =
⎧ 0 0 < t < 1,
{ 2t - 1, 1 ≤ t < 2,
⎧1, 2 ≤ t.

Answer 1

To evaluate ∫₀³ t²dg(t), we break the integral into three parts based on the different ranges of t. After evaluating each part separately, we can add up the results to find the final value of the integral, which is 25/3.

Step-by-step explanation:

To evaluate the integral ∫₀³ t²dg(t), where g(t) = ⎧ 0 0 < t < 1, { 2t - 1, 1 ≤ t < 2,⎧1, 2 ≤ t, we need to break the integral into three parts based on the different ranges of t: from 0 to 1, from 1 to 2, and from 2 to 3.

Part 1: From 0 to 1, g(t) = 0. Therefore, the integral ∫₀¹ t²dg(t) = ∫₀¹ t²(0)dt = 0.

Part 2: From 1 to 2, g(t) = 2t - 1. Therefore, the integral ∫₁² t²(2t - 1)dt can be evaluated using power rule and the formula for the integral of a polynomial. This gives:

1. ∫₁² t²(2t - 1)dt = 2∫₁² t³ - t²dt = 2[(1/4)t⁴ - (1/3)t³] evaluated from 1 to 2.
2. Substituting the upper and lower limits of integration, we get: 2[(1/4)(2)⁴ - (1/3)(2)³] - 2[(1/4)(1)⁴ - (1/3)(1)³].
3. Simplifying the expression, we get: 2[(1/4)(16) - (1/3)(8)] - 2[(1/4)(1) - (1/3)(1)].
4. Further simplifying, we find that ∫₁² t²(2t - 1)dt = 2(5/6) - 2(1/12) = 10/6 - 2/12 = 14/6 - 2/12 = 26/12 - 2/12 = 24/12 = 2.

Part 3: From 2 to 3, g(t) = 1. Therefore, the integral ∫₂³ t²(1)dt can be evaluated using power rule and the formula for the integral of a polynomial. This gives:

1. ∫₂³ t²(1)dt = ∫₂³ t²dt = [(1/3)t³] evaluated from 2 to 3.
2. Substituting the upper and lower limits of integration, we get: (1/3)(3)³ - (1/3)(2)³.
3. Simplifying the expression, we get: (1/3)(27) - (1/3)(8) = 27/3 - 8/3 = 19/3.

Finally, we add up the results from the three parts:

• ∫₀³ t²dg(t) = 0 + 2 + 19/3 = 2 + 19/3 = 6/3 + 19/3 = 25/3.