1. Evaluate $$\int_0^2 (3x^2 - 2x + 1) \, dx$$ - Use the power rule for integration: $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$ - Integrate term-by-term: $$\int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3$$ $$\int -2x \, dx = -2 \cdot \frac{x^2}{2} = -x^2$$ $$\int 1 \, dx = x$$ - Combine and evaluate from 0 to 2: $$\left[x^3 - x^2 + x\right]_0^2 = (8 - 4 + 2) - (0) = 6$$ 2. Evaluate $$\int (x^{1/2} + 4x^3) \, dx$$ - Integrate each term: $$\int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2}$$ $$\int 4x^3 \, dx = 4 \cdot \frac{x^4}{4} = x^4$$ - Final answer: $$\frac{2}{3} x^{3/2} + x^4 + C$$ 3. Evaluate $$\int_1^4 (\sqrt{x} + \frac{1}{\sqrt{x}}) \, dx$$ - Rewrite integrand: $$\sqrt{x} = x^{1/2}, \quad \frac{1}{\sqrt{x}} = x^{-1/2}$$ - Integrate: $$\int x^{1/2} \, dx = \frac{2}{3} x^{3/2}$$ $$\int x^{-1/2} \, dx = 2 x^{1/2}$$ - Evaluate from 1 to 4: $$\left[ \frac{2}{3} x^{3/2} + 2 x^{1/2} \right]_1^4 = \left( \frac{2}{3} \cdot 8 + 2 \cdot 2 \right) - \left( \frac{2}{3} \cdot 1 + 2 \cdot 1 \right) = \left( \frac{16}{3} + 4 \right) - \left( \frac{2}{3} + 2 \right) = \frac{16}{3} + 4 - \frac{2}{3} - 2 = \frac{14}{3} + 2 = \frac{20}{3}$$ 4. Evaluate $$\int_0^\pi \cos x \, dx$$ - Integral of cosine: $$\int \cos x \, dx = \sin x + C$$ - Evaluate from 0 to $$\pi$$: $$\sin \pi - \sin 0 = 0 - 0 = 0$$ 5. Evaluate $$\int (e^x + \sin x) \, dx$$ - Integrate term-by-term: $$\int e^x \, dx = e^x$$ $$\int \sin x \, dx = -\cos x$$ - Final answer: $$e^x - \cos x + C$$ 6. Evaluate $$\int_0^1 (2x + 1)^3 \, dx$$ - Use substitution: let $$u = 2x + 1$$, then $$du = 2 dx$$ or $$dx = \frac{du}{2}$$ - Change limits: when $$x=0$$, $$u=1$$; when $$x=1$$, $$u=3$$ - Integral becomes: $$\int_1^3 u^3 \cdot \frac{1}{2} \, du = \frac{1}{2} \int_1^3 u^3 \, du = \frac{1}{2} \cdot \frac{u^4}{4} \Big|_1^3 = \frac{1}{8} (81 - 1) = \frac{80}{8} = 10$$ 7. Evaluate $$\int \frac{\cos(\sqrt{x})}{\sqrt{x}} \, dx$$ - Use substitution: let $$t = \sqrt{x}$$, so $$x = t^2$$, $$dx = 2t \, dt$$ - Substitute: $$\int \frac{\cos t}{t} \cdot 2t \, dt = 2 \int \cos t \, dt = 2 \sin t + C = 2 \sin(\sqrt{x}) + C$$ 8. Evaluate $$\int_0^2 |x - 1| \, dx$$ - Split integral at $$x=1$$ where absolute value changes: $$\int_0^1 (1 - x) \, dx + \int_1^2 (x - 1) \, dx$$ - Compute each: $$\int_0^1 (1 - x) \, dx = \left[x - \frac{x^2}{2}\right]_0^1 = 1 - \frac{1}{2} = \frac{1}{2}$$ $$\int_1^2 (x - 1) \, dx = \left[ \frac{x^2}{2} - x \right]_1^2 = \left(2 - 2\right) - \left(\frac{1}{2} - 1\right) = 0 - (-\frac{1}{2}) = \frac{1}{2}$$ - Sum: $$\frac{1}{2} + \frac{1}{2} = 1$$ 9. Evaluate $$\int \frac{x}{x^2 + 1} \, dx$$ - Use substitution: let $$u = x^2 + 1$$, then $$du = 2x \, dx$$ - Rewrite integral: $$\int \frac{x}{x^2 + 1} \, dx = \frac{1}{2} \int \frac{2x}{u} \, dx = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln(x^2 + 1) + C$$ 10. Evaluate $$\int_0^{\pi/4} \sec^2 \theta \, d\theta$$ - Integral of $$\sec^2 \theta$$ is $$\tan \theta + C$$ - Evaluate: $$\tan \frac{\pi}{4} - \tan 0 = 1 - 0 = 1$$ 11. Evaluate $$\int_1^e \frac{\ln x}{x} \, dx$$ - Use substitution: let $$t = \ln x$$, then $$dt = \frac{1}{x} dx$$ - Integral becomes: $$\int_0^1 t \, dt = \frac{t^2}{2} \Big|_0^1 = \frac{1}{2}$$ 12. Evaluate $$\int \sin(3x) \, dx$$ - Use substitution: let $$u = 3x$$, then $$du = 3 dx$$ or $$dx = \frac{du}{3}$$ - Integral: $$\int \sin u \cdot \frac{1}{3} \, du = -\frac{1}{3} \cos u + C = -\frac{1}{3} \cos(3x) + C$$ 13. Evaluate $$\int_{-1}^2 (e^x + 2) \, dx$$ - Integrate term-by-term: $$\int e^x \, dx = e^x$$ $$\int 2 \, dx = 2x$$ - Evaluate: $$\left[e^x + 2x\right]_{-1}^2 = (e^2 + 4) - (e^{-1} - 2) = e^2 + 4 - e^{-1} + 2 = e^2 - e^{-1} + 6$$ 14. Evaluate $$\int x e^{x^2} \, dx$$ - Use substitution: let $$u = x^2$$, then $$du = 2x \, dx$$ or $$x \, dx = \frac{du}{2}$$ - Integral becomes: $$\int e^u \cdot \frac{1}{2} \, du = \frac{1}{2} e^u + C = \frac{1}{2} e^{x^2} + C$$ 15. Evaluate $$\int_0^1 \frac{dx}{\sqrt{1 - x^2}}$$ - Recognize integral as arcsine derivative: $$\int \frac{dx}{\sqrt{1 - x^2}} = \arcsin x + C$$ - Evaluate: $$\arcsin 1 - \arcsin 0 = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$ 16. Evaluate $$\int \frac{(\arctan x)^2}{1 + x^2} \, dx$$ - Use substitution: let $$t = \arctan x$$, then $$dt = \frac{1}{1 + x^2} dx$$ - Integral becomes: $$\int t^2 \, dt = \frac{t^3}{3} + C = \frac{(\arctan x)^3}{3} + C$$ 17. Evaluate $$\int_0^{\pi/2} \sin x \cos x \, dx$$ - Use identity: $$\sin x \cos x = \frac{1}{2} \sin 2x$$ - Integral: $$\int_0^{\pi/2} \frac{1}{2} \sin 2x \, dx = \frac{1}{2} \cdot \left[-\frac{\cos 2x}{2}\right]_0^{\pi/2} = -\frac{1}{4} (\cos \pi - \cos 0) = -\frac{1}{4} (-1 - 1) = \frac{1}{2}$$ 18. Evaluate $$\int \frac{2t + 3}{t^2 + 3t + 5} \, dt$$ - Let denominator be $$D = t^2 + 3t + 5$$, derivative $$D' = 2t + 3$$ - Integral becomes: $$\int \frac{D'}{D} \, dt = \ln|D| + C = \ln|t^2 + 3t + 5| + C$$ 19. Evaluate $$\int_0^4 (4 - t) \sqrt{t} \, dt$$ - Rewrite integrand: $$(4 - t) t^{1/2} = 4 t^{1/2} - t^{3/2}$$ - Integrate term-by-term: $$\int 4 t^{1/2} \, dt = 4 \cdot \frac{2}{3} t^{3/2} = \frac{8}{3} t^{3/2}$$ $$\int t^{3/2} \, dt = \frac{2}{5} t^{5/2}$$ - Evaluate from 0 to 4: $$\left[ \frac{8}{3} t^{3/2} - \frac{2}{5} t^{5/2} \right]_0^4 = \frac{8}{3} \cdot 8 - \frac{2}{5} \cdot 32 = \frac{64}{3} - \frac{64}{5} = \frac{320 - 192}{15} = \frac{128}{15}$$ 20. Evaluate $$\int_{-2}^2 f(x) \, dx$$ where $$f(x) = \begin{cases} 3 & -2 \leq x \leq 0 \\ 4 - x^2 & 0 < x \leq 2 \end{cases}$$ - Split integral: $$\int_{-2}^0 3 \, dx + \int_0^2 (4 - x^2) \, dx$$ - Compute each: $$3x \Big|_{-2}^0 = 0 - (-6) = 6$$ $$\left[4x - \frac{x^3}{3}\right]_0^2 = (8 - \frac{8}{3}) - 0 = \frac{24}{3} - \frac{8}{3} = \frac{16}{3}$$ - Sum: $$6 + \frac{16}{3} = \frac{18}{3} + \frac{16}{3} = \frac{34}{3}$$