Final answer:
To evaluate the given integral, we can use the properties of the Dirac delta function. The integral can be separated into two parts: one involving δ(t) and the other involving δ(t-1). Evaluating each part separately gives us a final result of 18.
Step-by-step explanation:
To evaluate the given integral, we will use the properties of the Dirac delta function. The Dirac delta function is zero everywhere except at t = 0 and t = 1, where it is undefined. It can be represented as δ(t) = ∞ at t = 0, and δ(t) = ∞ at t = 1.
Using the properties of the Dirac delta function, we can rewrite the integral as:
I = ∫₋₁ (t³ + 2) [δ(t) + 8δ(t−1)] dt = ∫₋₁ (t³ + 2)δ(t) + 8∫₋₁ (t³ + 2)δ(t−1) dt
Next, we evaluate each integral separately:
For the first integral, ∫₋₁ (t³ + 2)δ(t) dt, since δ(t) = ∞ at t = 0 and 0 everywhere else, the integral evaluates to (0³ + 2) = 2.
For the second integral, ∫₋₁ (t³ + 2)δ(t−1) dt, since δ(t−1) = ∞ at t = 1 and 0 everywhere else, we substitute t−1 = 0 into the integral limits, giving us ∫₀⁺ (t³ + 2)δ(t−1) dt. We can then evaluate the integral with the limits from 0 to a small positive value ε. The result will be (0³ + 2) = 2.
Finally, we add the two evaluations together: I = 2 + 8(2) = 2 + 16 = 18.