Question

Evaluate the integral: (−1,3) ∫(t⁵+2) [δ(t)+3δ(t−1)] dt

(-2,2) ∫ t⁶ [ δ(t) + δ(t+1.1) + δ(t-3)] dt

Answer 1

To evaluate the given integral, we simplify the integrals using the properties of the Dirac delta function and evaluate them separately.

Step-by-step explanation:

To evaluate the given integral:

(−1,3) ∫(t⁵+2) [δ(t)+3δ(t−1)] dt(-2,2) ∫ t⁶ [ δ(t) + δ(t+1.1) + δ(t-3)] dt

We can first simplify the integrals by using the properties of the Dirac delta function and then evaluate them separately.

First, we simplify the first integral [δ(t)+3δ(t−1)]. Since δ(t) is zero for all t except when t = 0 and δ(t−1) is zero for all t except when t = 1, the integral becomes:

(-1,3) ∫(t⁵+2) [δ(t)+3δ(t−1)] dt = (0,1) ∫(t⁵+2) [1] dt + (1,3) ∫(t⁵+2) [3] dt

We can now evaluate each integral separately to get the final answer.