Consider the function \(f(t)=2 t^{4}\). Evaluate the following integrals: (a) \(\int_{0}^{10} f(t) \hat{\delta}(t-3) d t\) (b) \(\int_{0}^{10} f(t) \dot{\bar{\delta}}(t-3) d t\) (c) \(\int_{0}^{10} f(t) \ddot{\bar{\delta}}(t-3) d t\) (d) \(\int_{0}^{10} \ddot{f}(t) \hat{\delta}(t-3) d t\)

Short Answer

Expert verified
Question: Evaluate the following integrals containing the Dirac Delta function and its derivatives: a) Evaluate \(\int_{0}^{10} f(t) \hat{\delta}(t-3) dt\) b) Evaluate \(\int_{0}^{10} f(t) \dot{\hat{\delta}}(t-3) dt\) c) Evaluate \(\int_{0}^{10} f(t) \ddot{\hat{\delta}}(t-3) dt\) d) Evaluate \(\int_{0}^{10} \ddot{f}(t) \hat{\delta}(t-3) dt\) Given function: \(f(t) = 2t^4\) Answer: a) The integral evaluates to 162. b) The integral evaluates to -216. c) The integral evaluates to 216. d) The integral evaluates to 216.

Step by step solution

01

a) Evaluate \(\int_{0}^{10} f(t) \hat{\delta}(t-3) dt\)

To evaluate the integral, we can use the property of Dirac Delta function, which is \(\int_{-\infty}^{\infty} f(t) \hat{\delta}(t-a) dt = f(a)\). We have the integral \(\int_{0}^{10} 2t^4 \hat{\delta}(t-3) dt\), and using the property mentioned, the result is just \(f(3)\). So, we can calculate \(f(3) = 2(3^4) = 2(81) = 162\). The integral evaluates to 162.
02

b) Evaluate \(\int_{0}^{10} f(t) \dot{\hat{\delta}}(t-3) dt\)

The integral involves the derivative of the Dirac Delta function, which has the property: \(\int_{-\infty}^{\infty} f(t) \dot{\hat{\delta}}(t-a) dt = -f'(a)\). First, we need to find the derivative of \(f(t)\). \(f'(t) = \frac{d}{dt}(2t^4) = 8t^3\). So, the integral evaluates to \(-\int_{0}^{10} 8t^3 \dot{\hat{\delta}}(t-3) dt = -f'(3)\), which is \(-8(3^3) = -216\). The integral evaluates to -216.
03

c) Evaluate \(\int_{0}^{10} f(t) \ddot{\hat{\delta}}(t-3) dt\)

The integral involves the second derivative of the Dirac Delta function, which has the property: \(\int_{-\infty}^{\infty} f(t) \ddot{\hat{\delta}}(t-a) dt = f''(a)\). To calculate this integral, first, we need to find the second derivative of \(f(t)\). \(f''(t) = \frac{d^2}{dt^2}(2t^4) = 24t^2\). So, the integral evaluates to \(\int_{0}^{10} 24t^2 \ddot{\hat{\delta}}(t-3) dt = f''(3)\), which is \(24(3^2) = 216\). The integral evaluates to 216.
04

d) Evaluate \(\int_{0}^{10} \ddot{f}(t) \hat{\delta}(t-3) dt\)

In this integral, we have the second derivative of \(f(t)\) and the Dirac Delta function. We've already calculated the second derivative as: \(\ddot{f}(t) = 24t^2\). Now, we can apply the main property of the Dirac Delta function again, and the integral evaluates to \(\ddot{f}(3)\), which is \(24(3^2) = 216\). The integral evaluates to 216.

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