Show that (а) Ifx() even, then S",x(1) dt = 2 f, x(t) dt II а

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**Theorem**: 

(a) If \( x(t) \) is an even function, then

\[
\int_{-a}^{a} x(t) \, dt = 2 \int_{0}^{a} x(t) \, dt
\]

**Explanation**: This theorem states that for even functions, the integral from \(-a\) to \(a\) can be simplified to twice the integral from \(0\) to \(a\). An even function satisfies \(x(t) = x(-t)\), meaning it is symmetrical around the y-axis. As a result, the area under the curve from \(-a\) to \(0\) mirrors the area from \(0\) to \(a\). Thus, the total integral is simply twice the integral from \(0\) to \(a\).
Transcribed Image Text:**Theorem**: (a) If \( x(t) \) is an even function, then \[ \int_{-a}^{a} x(t) \, dt = 2 \int_{0}^{a} x(t) \, dt \] **Explanation**: This theorem states that for even functions, the integral from \(-a\) to \(a\) can be simplified to twice the integral from \(0\) to \(a\). An even function satisfies \(x(t) = x(-t)\), meaning it is symmetrical around the y-axis. As a result, the area under the curve from \(-a\) to \(0\) mirrors the area from \(0\) to \(a\). Thus, the total integral is simply twice the integral from \(0\) to \(a\).
Expert Solution
Step 1

Since the function xt is even, the following relation is valid:

xt=x-t--->(1)

In this case, the given integral is 

I=-a+axtdt--->(2)

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