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Signal and systems miscellaneous
  1. The solution of integral 2(t3 + 3) δ(t + 2)dt
    – 1
    1. – 5
    2. – 2
    3. Zero
    4. 11
Correct Option: A

Given 2(t3 + 3) δ(t + 2)dt
– 1

Since, 2x(t) δ(t + t0)dt = x(– t0)
– 1

Therefore, 2(t3 + 3) δ(t + 2)dt = t3 + 3|at t = –2
– 1

= (– 2)3 + 3
= – 8 + 3 = – 5
Hence, alternative (A) is the correct choice.


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