Let's solve each problem step by step: --- ## **4. Solve the Integral Equation using a Transform** Given: \[ f(t) - \int_{-\infty}^{\infty} f(t-\tau) \tau e^{-2\tau} H(\tau) d\tau = e^{-4t} H(t) \] ### **Step 1: Recognize the Convolution** The integral is a **convolution**: \[ \int_{-\infty}^{\infty} f(t-\tau)g(\tau) d\tau = (f * g)(t) \] Here, \(g(\tau) = \tau e^{-2\tau} H(\tau)\). ### **Step 2: Take Laplace Transform** Recall: - \(\mathcal{L}[f(t)] = F(s)\) - \(\mathcal{L}[f * g] = F(s) G(s)\) - \(\mathcal{L}[e^{-at}H(t)] = \frac{1}{s + a}\) Take Laplace transforms of both sides: \[ \mathcal{L}[f(t)] - \mathcal{L}[f * g] = \mathcal{L}[e^{-4t}H(t)] \] \[ F(s) - F(s) G(s) = \frac{1}{s + 4} \] \[ F(s)[1 - G(s)] = \frac{1}{s + 4} \] ### **Step 3: Find \(G(s)\)** Compute Laplace transform of \(g(\tau) = \tau e^{-2\tau} H(\tau)\): \[ G(s) = \mathcal{L}[\tau e^{-2\tau} H(\tau)] = \int_0^\infty \tau e^{-2\tau} e^{-s\tau} d\tau = \int_0^\infty \tau e^{-(s+2)\tau} d\tau \] Recall Laplace: \(\mathcal{L}[t e^{-a t}] = \frac{1}{(s + a)^2}\): So, \[ G(s) = \frac{1}{(s + 2)^2} \] ### **Step 4: Solve for \(F(s)\)** \[ F(s) \left[1 - \frac{1}{(s + 2)^2}\right] = \frac{1}{s + 4} \] \[ F(s) \left(\frac{(s + 2)^2 - 1}{(s + 2)^2}\right) = \frac{1}{s + 4} \] \[ F(s) = \frac{1}{s + 4} \frac{(s + 2)^2}{(s + 2)^2 - 1} \] \[ (s + 2)^2 - 1 = s^2 + 4s + 4 - 1 = s^2 + 4s + 3 = (s + 1)(s + 3) \] So, \[ F(s) = \frac{(s + 2)^2}{(s + 1)(s + 3)(s + 4)} \] --- ### **Step 5: Inverse Laplace Transform** Decompose: \[ \frac{(s + 2)^2}{(s + 1)(s + 3)(s + 4)} \] Let’s write \((s + 2)^2 = s^2 + 4s + 4\). Let’s express as partial fractions: \[ \frac{(s + 2)^2}{(s + 1)(s + 3)(s + 4)} = \frac{A}{s + 1} + \frac{B}{s + 3} + \frac{C}{s + 4} \] Multiply both sides by \((s+1)(s+3)(s+4)\): \[ (s + 2)^2 = A(s + 3)(s + 4) + B(s + 1)(s + 4) + C(s + 1)(s + 3) \] Expand and solve for A, B, C: Let \(s = -1\): \[ (-1 + 2)^2 = A(2)(3) + 0 + 0 \implies 1 = 6A \implies A = \frac{1}{6} \] Let \(s = -3\): \[ (-3 + 2)^2 = 0 + B(-2)(1) + 0 \implies 1 = B(-2) \implies B = -\frac{1}{2} \] Let \(s = -4\): \[ (-4 + 2)^2 = 0 + 0 + C(-3)(-1) \implies 4 = 3C \implies C = \frac{4}{3} \] So, \[ F(s) = \frac{1}{6}\frac{1}{s+1} - \frac{1}{2}\frac{1}{s+3} + \frac{4}{3}\frac{1}{s+4} \] --- ### **Step 6: Inverse Laplace** \[ f(t) = \frac{1}{6}e^{-t} - \frac{1}{2}e^{-3t} + \frac{4}{3}e^{-4t}, \quad t \geq 0 \] --- ### **Final Answer:** \[ \boxed{ f(t) = \frac{1}{6}e^{-t} - \frac{1}{2}e^{-3t} + \frac{4}{3}e^{-4t} \quad \text{for } t \geq 0 } \] --- ## **5. Evaluate the Line Integral** Given: \[ \vec{F} = 3y\vec{i} + (x^2 - y)\vec{j} \] Curve \(C\): upper half of unit circle (\(x^2 + y^2 = 1\), \(y \geq 0\)), from \((-1,0)\) to \((1,0)\), then along \(y = x^2 - 1\) from \((1,0)\) to \((-1,0)\). ### **Step 1: Recognize the Path is a Closed Curve** Since the curve is closed, use **Green's Theorem**: \[ \oint_{C} \vec{F} \cdot d\vec{r} = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \] where \(P = 3y\), \(Q = x^2 - y\). ### **Step 2: Compute the Curl** \[ \frac{\partial Q}{\partial x} = 2x \] \[ \frac{\partial P}{\partial y} = 3 \] \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - 3 \] ### **Step 3: Set Up the Double Integral** Region \(R\): bounded above by \(y = \sqrt{1 - x^2}\), below by \(y = x^2 - 1\), for \(x \in [-1, 1]\). \[ \iint_{R} (2x - 3) dA = \int_{x=-1}^{1} \int_{y = x^2 - 1}^{y = \sqrt{1 - x^2}} (2x - 3) dy dx \] ### **Step 4: Integrate with Respect to \(y\)** \[ \int_{x=-1}^{1} (2x - 3) \left[ y \right]_{y = x^2 - 1}^{y = \sqrt{1 - x^2}} dx \] \[ = \int_{x=-1}^{1} (2x - 3) \left( \sqrt{1 - x^2} - (x^2 - 1) \right) dx \] \[ = \int_{x=-1}^{1} (2x - 3)\sqrt{1 - x^2} dx - \int_{x=-1}^{1} (2x - 3)(x^2 - 1) dx \] --- #### **First Integral:** \[ I_1 = \int_{-1}^{1} (2x - 3)\sqrt{1 - x^2} dx \] Split: \[ I_1 = 2\int_{-1}^{1} x\sqrt{1 - x^2} dx - 3\int_{-1}^{1} \sqrt{1 - x^2} dx \] - \(\int_{-1}^{1} x\sqrt{1 - x^2} dx = 0\) (odd function over symmetric interval). - \(\int_{-1}^{1} \sqrt{1 - x^2} dx = \pi/2\) (area of upper half unit circle). So, \[ I_1 = 2 \times 0 - 3 \times \frac{\pi}{2} = -\frac{3\pi}{2} \] --- #### **Second Integral:** \[ I_2 = \int_{-1}^{1} (2x - 3)(x^2 - 1) dx \] Expand: \[ (2x - 3)(x^2 - 1) = 2x^3 - 2x - 3x^2 + 3 \] So, \[ I_2 = \int_{-1}^{1} [2x^3 - 2x - 3x^2 + 3] dx \] Integrate term by term: - \(\int_{-1}^{1} 2x^3 dx = 0\) (odd) - \(\int_{-1}^{1} -2x dx = 0\) (odd) - \(\int_{-1}^{1} -3x^2 dx = -3 \left[ \frac{x^3}{3} \right]_{-1}^{1} = -3 \left( \frac{1}{3} - \left(-\frac{1}{3}\right) \right) = -3 \times \frac{2}{3} = -2\) - \(\int_{-1}^{1} 3 dx = 3 \times (1 - (-1)) = 3 \times 2 = 6\) So, \[ I_2 = 0 + 0 - 2 + 6 = 4 \] --- ### **Step 5: Combine** \[ \text{Total} = I_1 - I_2 = -\frac{3\pi}{2} - 4 \] --- ### **Final Answer:** \[ \boxed{ \int_{C} \vec{F} \cdot d\vec{r} = -\frac{3\pi}{2} - 4 } \] --- **If you need clarifications or more details on any step, feel free to ask!**

**Final Answers:** **4. Solution to the Integral Equation:** \[ \boxed{ f(t) = \frac{1}{6} e^{-t} - \frac{1}{2} e^{-3t} + \frac{4}{3} e^{-4t} \quad \text{for } t \geq 0 } \] --- **5. Value of the Line Integral:** \[ \boxed{ \oint_{C} \vec{F} \cdot d\vec{r} = -\frac{3\pi}{2} - 4 } \]