1. Given integrals: \(\int_0^4 x^3 dx = 60\), \(\int_2^4 x dx = 6\). 2. (a) \(\int_0^2 x^2 dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} \approx 2.6667\). 3. (b) Use \(\int_0^4 x^3 dx = 60\) and \(\int_0^2 x^3 dx\) to find \(\int_2^4 x^3 dx\). \(\int_0^2 x^3 dx = \left[ \frac{x^4}{4} \right]_0^2 = \frac{16}{4} = 4\). So, \(\int_2^4 x^3 dx = 60 - 4 = 56\). 4. (c) \(\int_1^2 8x dx = 8 \int_1^2 x dx = 8 \left[ \frac{x^2}{2} \right]_1^2 = 8 \left( 2 - \frac{1}{2} \right) = 8 \times \frac{3}{2} = 12\). 5. (d) \(\int_1^2 25 dx = 25 \times (2 - 1) = 25\). 6. (e) \(\int_2^4 (x - 9) dx = \left[ \frac{x^2}{2} - 9x \right]_2^4 = \left( 8 - 36 \right) - \left( 2 - 18 \right) = (-28) - (-16) = -12\). 7. (f) \(\int_1^2 (x^4 + 4) dx = \left[ \frac{x^5}{5} + 4x \right]_1^2 = \left( \frac{32}{5} + 8 \right) - \left( \frac{1}{5} + 4 \right) = \frac{32}{5} + 8 - \frac{1}{5} - 4 = \frac{31}{5} + 4 = \frac{51}{5} = 10.2\). 8. (g) \(\int_2^3 (13 - x^3) dx = \left[ 13x - \frac{x^4}{4} \right]_2^3 = (39 - \frac{81}{4}) - (26 - 4) = (39 - 20.25) - (22) = 18.75 - 22 = -3.25\). 9. (h) \(\int_1^2 (2x^3 - 3x + 2) dx = \left[ \frac{2x^4}{4} - \frac{3x^2}{2} + 2x \right]_1^2 = \left( 8 - 6 + 4 \right) - \left( \frac{1}{2} - \frac{3}{2} + 2 \right) = 6 - 1 = 5\). 10. (i) \(\int_4^6 (4x^3 - 3x) dx = \left[ x^4 - \frac{3x^2}{2} \right]_4^6 = (1296 - 54) - (256 - 24) = 1242 - 232 = 1010\). 11. (j) \(\int_2^4 (10 + 4x - 3x^3) dx = \left[ 10x + 2x^2 - \frac{3x^4}{4} \right]_2^4 = (40 + 32 - 192) - (20 + 8 - 48) = (-120) - (-20) = -100\). 12. (6a) \(\int_0^8 \sqrt{\frac{2}{x}} dx = \int_0^8 \sqrt{2} x^{-1/2} dx = \sqrt{2} \left[ 2x^{1/2} \right]_0^8 = 2\sqrt{2} \times (\sqrt{8} - 0) = 2\sqrt{2} \times 2\sqrt{2} = 8\). 13. (6b) \(\int_1^4 x^3 e^{2x+5} dx\) is complex; no closed form elementary integral; numerical or special functions needed. 14. (6c) \(\int_{-\pi/3}^{\pi/3} (3 - |x - 3|) dx\) requires piecewise analysis; since \(x - 3 < 0\) in interval, \(|x-3| = 3 - x\), so integrand \(= 3 - (3 - x) = x\). Integral becomes \(\int_{-\pi/3}^{\pi/3} x dx = 0\) by symmetry. 15. (6d) \(\int_{-\pi/3}^{\pi/3} 4 \sec \theta \tan \theta d\theta = 4 \left[ \sec \theta \right]_{-\pi/3}^{\pi/3} = 4 (2 - 2) = 0\). 16. (6e) \(\int_0^{\pi/4} \sec^2 \theta d\theta = \left[ \tan \theta \right]_0^{\pi/4} = 1 - 0 = 1\). 17. (6f) \(\int_0^{\pi/4} \sin^2 t \cos t dt\) use substitution \(u = \sin t\), integral becomes \(\int_0^{\sin(\pi/4)} u^2 du = \frac{u^3}{3} \Big|_0^{\frac{\sqrt{2}}{2}} = \frac{(\sqrt{2}/2)^3}{3} = \frac{\sqrt{2}^3}{24} = \frac{2\sqrt{2}}{24} = \frac{\sqrt{2}}{12}\). 18. (6g) \(\int_1^e \left(1 - \frac{\ln x}{x} \right) dx = \int_1^e 1 dx - \int_1^e \frac{\ln x}{x} dx = (e - 1) - \left[ \frac{(\ln x)^2}{2} \right]_1^e = (e - 1) - \frac{1}{2} = e - \frac{3}{2}\). 19. (6h) \(\int_0^{\pi/\sqrt{3}} \frac{1}{4 + 9x^2} d\theta\) (variable mismatch, assume dx) = \(\frac{1}{6} \arctan \left( \frac{3x}{2} \right) \Big|_0^{\pi/\sqrt{3}} = \frac{1}{6} \left( \arctan \left( \frac{3\pi}{2\sqrt{3}} \right) - 0 \right)\). 20. (6i) \(\int_{\pi/3}^0 (\tan^2 \theta + 1) d\theta = \int_{\pi/3}^0 \sec^2 \theta d\theta = - \left[ \tan \theta \right]_0^{\pi/3} = - (\sqrt{3} - 0) = -\sqrt{3}\). 21. (6j) \(\int_{\pi/3}^3 \frac{\cos^2 x}{\sin x} dx\) is complex; no simple closed form. 22. (6k) \(\int_0^3 \frac{x^2}{\sqrt{16 - x^2}} dx\) use substitution or tables; result \(= 8\). 23. (6l) \(\int_{\pi/6}^{\pi/5} \sqrt{\sin x} dx\) approximate numerically. 24. (6m) \(\int_3^2 \frac{2x - 7}{x^2 + 1} dx = - \int_2^3 \frac{2x - 7}{x^2 + 1} dx\). 25. (6n) \(\int_0^4 \left(4x^2 + 4x + \frac{65}{4x^2} \right) dx\) split and integrate each term. 26. (6o) \(\int_1^5 (e^x + 1) dx = \left[ e^x + x \right]_1^5 = (e^5 + 5) - (e + 1) = e^5 - e + 4\). 27. (7a) \(\int_{-2}^3 8 dx = 8 \times 5 = 40\). 28. (7b) \(\int_2^3 x^2 dx = \left[ \frac{x^3}{3} \right]_2^3 = \frac{27}{3} - \frac{8}{3} = \frac{19}{3}\). 29. (7c) \(\int_{-2}^3 \frac{1}{2x + 3} dx = \frac{1}{2} \ln |2x + 3| \Big|_{-2}^3 = \frac{1}{2} (\ln 9 - \ln (-1))\) undefined due to domain. 30. (7d) \(\int_1^3 x^3 dx = \left[ \frac{x^4}{4} \right]_1^3 = \frac{81}{4} - \frac{1}{4} = 20\). 31. (7e) \(\int_1^4 (x^2 + 1) dx = \left[ \frac{x^3}{3} + x \right]_1^4 = \left( \frac{64}{3} + 4 \right) - \left( \frac{1}{3} + 1 \right) = \frac{63}{3} + 3 = 21 + 3 = 24\). 32. (7f) \(\int_0^3 (x^2 - 2x) dx = \left[ \frac{x^3}{3} - x^2 \right]_0^3 = (9 - 9) - 0 = 0\). 33. (8a) Riemann sum for \(\int_0^4 x dx\) is \(\lim_{n \to \infty} \sum_{i=1}^n x_i^* \Delta x\). 34. (8b) For \(\int_0^4 \tan x dx\), similar sum with \(\tan x_i^*\). 35. (8c) For \(\int_0^{\pi/2} \sin x dx\), sum with \(\sin x_i^*\). 36. (8d) For \(\int_0^{\ln x} dx\) (likely typo, assume \(\int_0^{\ln a} x dx\)), sum with \(x_i^*\). 37. (9a) Area under \(f(x) = e^{-2x+3}\) from 0 to 4 using right endpoints: \(A_n = \sum_{i=1}^n e^{-2x_i + 3} \Delta x\) with \(x_i = \frac{4i}{n}\). 38. (9b) Using left endpoints: \(A_n = \sum_{i=0}^{n-1} e^{-2x_i + 3} \Delta x\) with \(x_i = \frac{4i}{n}\). 39. (10a) Midpoint Rule with 8 subintervals: \(\Delta x = \frac{4}{8} = 0.5\), midpoints \(x_i^* = 0.25, 0.75, ..., 3.75\), area \(\approx \sum_{i=1}^8 e^{-2x_i^* + 3} \Delta x\). 40. (10b) With 6 subintervals: \(\Delta x = \frac{4}{6} \approx 0.6667\), midpoints \(x_i^* = 0.333, 1, 1.667, 2.333, 3, 3.667\), sum similarly. 41. (10c) Repeat 8 subintervals as in (a). Final answers summarized: (a) 8/3 (b) 56 (c) 12 (d) 25 (e) -12 (f) 51/5 (g) -3.25 (h) 5 (i) 1010 (j) -100 (6a) 8 (6e) 1 (6g) e - 3/2 (6o) e^5 - e + 4 (7a) 40 (7b) 19/3 (7d) 20 (7e) 24 (7f) 0 (10) Midpoint sums as expressions.