1. Evaluate the integral $$\int e^x \cos 2x \, dx$$ Step 1. State the problem: We want to find $$\int e^x \cos 2x \, dx$$. Step 2. Use integration by parts or the formula for integrals of the form $$\int e^{ax} \cos(bx) \, dx$$ which is: $$\int e^{ax} \cos(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx) + C$$ Step 3. Here, $$a=1$$ and $$b=2$$, so: $$\int e^x \cos 2x \, dx = \frac{e^x}{1^2 + 2^2} (1 \cdot \cos 2x + 2 \cdot \sin 2x) + C = \frac{e^x}{5} (\cos 2x + 2 \sin 2x) + C$$ 2. Evaluate the integral $$\int \frac{x^2}{\sqrt{9 - 25x^2}} \, dx$$ Step 1. State the problem: Find $$\int \frac{x^2}{\sqrt{9 - 25x^2}} \, dx$$. Step 2. Use substitution: Let $$u = 25x^2$$, then $$du = 50x \, dx$$, but this is complicated. Instead, use trigonometric substitution: Let $$x = \frac{3}{5} \sin \theta$$, then: $$dx = \frac{3}{5} \cos \theta \, d\theta$$ Step 3. Substitute into the integral: $$x^2 = \left(\frac{3}{5} \sin \theta\right)^2 = \frac{9}{25} \sin^2 \theta$$ $$\sqrt{9 - 25x^2} = \sqrt{9 - 25 \cdot \frac{9}{25} \sin^2 \theta} = \sqrt{9 - 9 \sin^2 \theta} = 3 \cos \theta$$ Step 4. The integral becomes: $$\int \frac{\frac{9}{25} \sin^2 \theta}{3 \cos \theta} \cdot \frac{3}{5} \cos \theta \, d\theta = \int \frac{9}{25} \sin^2 \theta \cdot \frac{3}{5} \, d\theta = \int \frac{27}{125} \sin^2 \theta \, d\theta$$ Step 5. Use the identity $$\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$$: $$\int \frac{27}{125} \sin^2 \theta \, d\theta = \frac{27}{125} \int \frac{1 - \cos 2\theta}{2} \, d\theta = \frac{27}{250} \int (1 - \cos 2\theta) \, d\theta$$ Step 6. Integrate: $$\frac{27}{250} \left( \theta - \frac{\sin 2\theta}{2} \right) + C$$ Step 7. Back-substitute: $$\theta = \arcsin \left( \frac{5x}{3} \right)$$ $$\sin 2\theta = 2 \sin \theta \cos \theta = 2 \cdot \frac{5x}{3} \cdot \frac{\sqrt{9 - 25x^2}}{3} = \frac{10x \sqrt{9 - 25x^2}}{9}$$ Step 8. Final answer: $$\frac{27}{250} \arcsin \left( \frac{5x}{3} \right) - \frac{27}{500} \cdot \frac{10x \sqrt{9 - 25x^2}}{9} + C = \frac{27}{250} \arcsin \left( \frac{5x}{3} \right) - \frac{3x \sqrt{9 - 25x^2}}{50} + C$$ 3. Find the value of $$c$$ such that the area bounded by $$y = x^2 - c^2$$ and $$y = c^2 - x^2$$ is 500. Step 1. State the problem: Find $$c$$ such that the area between the parabolas is 500. Step 2. The curves intersect where: $$x^2 - c^2 = c^2 - x^2 \implies 2x^2 = 2c^2 \implies x^2 = c^2 \implies x = \pm c$$ Step 3. The area between curves from $$-c$$ to $$c$$ is: $$A = \int_{-c}^c \left[ (c^2 - x^2) - (x^2 - c^2) \right] dx = \int_{-c}^c 2(c^2 - x^2) \, dx$$ Step 4. Calculate the integral: $$A = 2 \int_{-c}^c (c^2 - x^2) \, dx = 2 \left[ c^2 x - \frac{x^3}{3} \right]_{-c}^c = 2 \left( c^3 - \frac{c^3}{3} - (-c^3 + \frac{c^3}{3}) \right)$$ Step 5. Simplify: $$= 2 \left( c^3 - \frac{c^3}{3} + c^3 - \frac{c^3}{3} \right) = 2 \left( \frac{2c^3}{3} + \frac{2c^3}{3} \right) = 2 \cdot \frac{4c^3}{3} = \frac{8c^3}{3}$$ Step 6. Set area equal to 500: $$\frac{8c^3}{3} = 500 \implies c^3 = \frac{500 \cdot 3}{8} = 187.5$$ Step 7. Solve for $$c$$: $$c = \sqrt[3]{187.5} = 5.73$$ (approx) 4. Find the volume of the solid generated by revolving the region between $$f(x) = x$$ and $$g(x) = 1 + x^2$$ over $$[0,2]$$ about the x-axis. Step 1. State the problem: Find volume of solid formed by revolving area between $$y = x$$ and $$y = 1 + x^2$$ from 0 to 2 about x-axis. Step 2. The volume by washer method is: $$V = \pi \int_0^2 \left[ (1 + x^2)^2 - (x)^2 \right] dx$$ Step 3. Expand: $$(1 + x^2)^2 = 1 + 2x^2 + x^4$$ Step 4. So integrand is: $$1 + 2x^2 + x^4 - x^2 = 1 + x^2 + x^4$$ Step 5. Integrate: $$V = \pi \int_0^2 (1 + x^2 + x^4) dx = \pi \left[ x + \frac{x^3}{3} + \frac{x^5}{5} \right]_0^2 = \pi \left( 2 + \frac{8}{3} + \frac{32}{5} \right)$$ Step 6. Calculate sum inside parentheses: $$2 = \frac{30}{15}, \quad \frac{8}{3} = \frac{40}{15}, \quad \frac{32}{5} = \frac{96}{15}$$ Sum = $$\frac{30 + 40 + 96}{15} = \frac{166}{15}$$ Step 7. Final volume: $$V = \pi \cdot \frac{166}{15} = \frac{166 \pi}{15}$$ 5. Find the surface area generated by rotating $$y = x^3$$, $$0 < x < 2$$ about the x-axis. Step 1. State the problem: Find surface area of curve $$y = x^3$$ rotated about x-axis from 0 to 2. Step 2. Surface area formula: $$S = 2\pi \int_0^2 y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx$$ Step 3. Compute derivative: $$\frac{dy}{dx} = 3x^2$$ Step 4. Substitute: $$S = 2\pi \int_0^2 x^3 \sqrt{1 + (3x^2)^2} \, dx = 2\pi \int_0^2 x^3 \sqrt{1 + 9x^4} \, dx$$ Step 5. Use substitution: Let $$u = 1 + 9x^4$$, then: $$du = 36x^3 dx$$ Step 6. Rewrite integral: $$S = 2\pi \int_0^2 x^3 \sqrt{u} \frac{du}{36x^3} = \frac{2\pi}{36} \int_{u(0)}^{u(2)} \sqrt{u} \, du = \frac{\pi}{18} \int_1^{145} u^{1/2} \, du$$ Step 7. Integrate: $$\frac{\pi}{18} \left[ \frac{2}{3} u^{3/2} \right]_1^{145} = \frac{\pi}{27} \left( 145^{3/2} - 1 \right)$$ Step 8. Final answer: $$S = \frac{\pi}{27} \left( 145^{3/2} - 1 \right)$$