What is the value of \(\int_{(n-1)T}^{nT} x(t)dt\) according to trapezoidal rule?(a) \([\frac{x(nT)-x[(n-1)T]}{2}]T\)(b) \([\frac{x(nT)+x[(n-1)T]}{2}]T\)(c) \([\frac{x(nT)-x[(n+1)T]}{2}]T\)(d) \([\frac{x(nT)+x[(n+1)T]}{2}]T\)I had been asked this question in an international level competition.My query is from Bilinear Transformations topic in section Digital Filters Design of Digital Signal Processing

What is the value of \(\int_{(n-1)T}^{nT} x(t)dt\) according to trapez

1.	What is the value of \(\int_{(n-1)T}^{nT} x(t)dt\) according to trapezoidal rule?(a) \([\frac{x(nT)-x[(n-1)T]}{2}]T\)(b) \([\frac{x(nT)+x[(n-1)T]}{2}]T\)(c) \([\frac{x(nT)-x[(n+1)T]}{2}]T\)(d) \([\frac{x(nT)+x[(n+1)T]}{2}]T\)I had been asked this question in an international level competition.My query is from Bilinear Transformations topic in section Digital Filters Design of Digital Signal Processing
Answer» Correct choice is (b) \([\frac{x(NT)+x[(n-1)T]}{2}]T\) The BEST EXPLANATION: The given integral is approximated by the trapezoidal rule. This rule states that if T is SMALL, the area (integral) can be approximated by the MEAN height of x(t) between the two limits and then multiplying by the width. That is \(\int_{(n-1)T}^{nT} x(t)dt=[\frac{x(nT)+x[(n-1)T]}{2}]T\)

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