1. Problema 1: Calcular las integrales indicadas. 2. Integral a) $$\int x^4 (2x+5)^6 \, dx$$ 3. Usamos sustitución: sea $$u = 2x+5$$, entonces $$du = 2 dx$$ o $$dx = \frac{du}{2}$$. 4. Expresamos $$x$$ en función de $$u$$: $$x = \frac{u-5}{2}$$. 5. Reescribimos $$x^4$$: $$x^4 = \left(\frac{u-5}{2}\right)^4 = \frac{(u-5)^4}{2^4} = \frac{(u-5)^4}{16}$$. 6. La integral queda: $$\int x^4 (2x+5)^6 dx = \int \frac{(u-5)^4}{16} u^6 \frac{du}{2} = \int \frac{(u-5)^4 u^6}{32} du = \frac{1}{32} \int (u-5)^4 u^6 du$$. 7. Expandimos $$ (u-5)^4 $$ usando binomio de Newton: $$(u-5)^4 = u^4 - 4 \cdot 5 u^3 + 6 \cdot 25 u^2 - 4 \cdot 125 u + 625 = u^4 - 20 u^3 + 150 u^2 - 500 u + 625$$. 8. Multiplicamos por $$u^6$$: $$u^6 (u-5)^4 = u^{10} - 20 u^9 + 150 u^8 - 500 u^7 + 625 u^6$$. 9. La integral es: $$\frac{1}{32} \int (u^{10} - 20 u^9 + 150 u^8 - 500 u^7 + 625 u^6) du$$. 10. Integramos término a término: $$\frac{1}{32} \left( \frac{u^{11}}{11} - 20 \frac{u^{10}}{10} + 150 \frac{u^9}{9} - 500 \frac{u^8}{8} + 625 \frac{u^7}{7} \right) + C$$. 11. Simplificamos coeficientes: $$\frac{1}{32} \left( \frac{u^{11}}{11} - 2 u^{10} + \frac{150}{9} u^9 - 62.5 u^8 + \frac{625}{7} u^7 \right) + C$$. 12. Reemplazamos $$u = 2x+5$$ para la respuesta final. --- 13. Problema 2a: Dada la aceleración $$a(t) = \sqrt{t} - \frac{1}{\sqrt{t}}$$, hallar la velocidad $$v(t)$$ con condición $$v(0) = 2$$. 14. Sabemos que $$v(t) = \int a(t) dt + C$$. 15. Calculamos $$\int \left( t^{1/2} - t^{-1/2} \right) dt = \int t^{1/2} dt - \int t^{-1/2} dt$$. 16. Integrales: $$\int t^{1/2} dt = \frac{t^{3/2}}{3/2} = \frac{2}{3} t^{3/2}$$ $$\int t^{-1/2} dt = \frac{t^{1/2}}{1/2} = 2 t^{1/2}$$ 17. Entonces: $$v(t) = \frac{2}{3} t^{3/2} - 2 t^{1/2} + C$$. 18. Usamos condición inicial $$v(0) = 2$$: $$v(0) = 0 - 0 + C = 2 \Rightarrow C = 2$$. 19. Por tanto: $$v(t) = \frac{2}{3} t^{3/2} - 2 t^{1/2} + 2$$. --- 20. Problema 2b: Hallar el espacio $$s(t)$$ con $$s(0) = 5$$. 21. Sabemos que $$s(t) = \int v(t) dt + D$$. 22. Integramos: $$\int \left( \frac{2}{3} t^{3/2} - 2 t^{1/2} + 2 \right) dt = \frac{2}{3} \int t^{3/2} dt - 2 \int t^{1/2} dt + 2 \int dt$$. 23. Calculamos cada integral: $$\int t^{3/2} dt = \frac{t^{5/2}}{5/2} = \frac{2}{5} t^{5/2}$$ $$\int t^{1/2} dt = \frac{2}{3} t^{3/2}$$ $$\int dt = t$$ 24. Entonces: $$s(t) = \frac{2}{3} \cdot \frac{2}{5} t^{5/2} - 2 \cdot \frac{2}{3} t^{3/2} + 2 t + D = \frac{4}{15} t^{5/2} - \frac{4}{3} t^{3/2} + 2 t + D$$. 25. Usamos condición inicial $$s(0) = 5$$: $$s(0) = 0 - 0 + 0 + D = 5 \Rightarrow D = 5$$. 26. Por tanto: $$s(t) = \frac{4}{15} t^{5/2} - \frac{4}{3} t^{3/2} + 2 t + 5$$. --- Respuesta final: Integral 1a: $$\int x^4 (2x+5)^6 dx = \frac{1}{32} \left( \frac{(2x+5)^{11}}{11} - 2 (2x+5)^{10} + \frac{150}{9} (2x+5)^9 - 62.5 (2x+5)^8 + \frac{625}{7} (2x+5)^7 \right) + C$$ Velocidad: $$v(t) = \frac{2}{3} t^{3/2} - 2 t^{1/2} + 2$$ Espacio: $$s(t) = \frac{4}{15} t^{5/2} - \frac{4}{3} t^{3/2} + 2 t + 5$$