Question

H(t)={ 0, t<0

1, t≥0
​1. Evaluate the following integrals involving H(t) : (a) ∫⁴₀H(s−3)ds.


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Answer 1

The value of the integral ∬⁴₀H(s-3)ds is 1, since H(t) equals 0 for t<0 and equals 1 for t≥0, and s ranges from 3 to 4 in the given integral.

Step-by-step explanation:

The student is asking to evaluate the integral of a step function H(t) from 0 to 4. The step function is defined as H(t)=0 for t<0 and H(t)=1 for t≥0. To evaluate the integral ∬⁴₀H(s-3)ds, we consider the effect of the shift in the step function. When s is less than 3, H(s-3) is 0, and when s is greater than or equal to 3, H(s-3) is 1. Thus, the integral from 0 to 3 is 0, and the integral from 3 to 4 is 1 times the width of the interval, which is 1. Therefore, the result of the integral is 1.