Let's solve each part step by step using the properties of the **Dirac delta function** (\(\delta(t)\)): ### **Delta Function Property** \[ \int_{-\infty}^{\infty} \delta(t-a) f(t)\,dt = f(a) \] If the limits include \(t=a\), the delta "picks out" \(f(a)\). --- ### **(a) \(\int_{-1}^{\infty} \delta(t) [3e^{-3t} + 6\sin(2\pi 10^6 t)]dt\)** **Step 1:** Identify the location of the delta: \(\delta(t)\) is nonzero only at \(t=0\). **Step 2:** Is \(t=0\) within integration limits? Yes, since \(-1 < 0 < \infty\). **Step 3:** Apply the property: \[ \int_{-1}^{\infty} \delta(t) f(t) dt = f(0) \] where \(f(t) = 3e^{-3t} + 6\sin(2\pi 10^6 t)\). **Step 4:** Compute \(f(0)\): \[ f(0) = 3e^{-3 \cdot 0} + 6\sin(2\pi 10^6 \cdot 0) = 3 \cdot 1 + 6 \cdot 0 = 3 \] **Final Answer (a):** \[ \boxed{3} \] --- ### **(b) \(\int_{-\infty}^{\infty} \delta(t + 3) e^{-1.5t} u(t) dt\)** **Step 1:** Delta is nonzero at \(t=-3\). **Step 2:** Plug \(t=-3\) into \(e^{-1.5t} u(t)\): - \(u(-3) = 0\) (since the unit step \(u(t)\) is 0 for \(t<0\)). \[ e^{-1.5 \cdot (-3)} u(-3) = e^{4.5} \cdot 0 = 0 \] **Final Answer (b):** \[ \boxed{0} \] --- ### **(c) \(\int_{0}^{\infty} \delta(t-2) e^{-2t} [u(t) - u(t+9)]dt\)** **Step 1:** Delta is nonzero at \(t=2\). **Step 2:** Is \(t=2\) within integration limits? Yes (\(0 < 2 < \infty\)). **Step 3:** Plug \(t=2\) into the rest: - \(u(2) = 1\) - \(u(2+9) = u(11) = 1\) So, \([u(2) - u(11)] = 1 - 1 = 0\) \[ e^{-2 \cdot 2} \times 0 = e^{-4} \times 0 = 0 \] **Final Answer (c):** \[ \boxed{0} \] --- ### **Summary of Answers** a) \(\boxed{3}\) b) \(\boxed{0}\) c) \(\boxed{0}\) If you need further explanation about any step, let me know!

Certainly! Here are the concise theoretical explanations for each problem in about 50 words: **(a)** The integral involves a delta function \(\delta(t)\), which "picks out" the value of the integrand at \(t=0\). Since the integration limits include zero, the integral equals the value of the remaining function evaluated at zero. **(b)** The delta function \(\delta(t+3)\) activates at \(t=-3\). But the integral limits are from \(-\infty\) to \(\infty\), so the delta "samples" the integrand at \(-3\). Since the step function \(u(t)\) is zero at \(-3\), the integral evaluates to zero. **(c)** The delta function \(\delta(t-2)\) is centered at \(t=2\), within the limits. The integrand includes step functions which evaluate to 1 or 0 at \(t=2\). Since \(u(2)=1\) and \(u(11)=1\), their difference is zero, resulting in the integral being zero.