1. Problem Q6: Find the area between the curves $T_1(n) = n^2 + 3n$ and $T_2(n) = 2n^2$ from $n=0$ to $n=5$. 2. Formula: The area between two curves $y=f(x)$ and $y=g(x)$ over $[a,b]$ is given by $$\text{Area} = \int_a^b |f(x) - g(x)| \, dx.$$ Here, $f(n) = T_1(n)$ and $g(n) = T_2(n)$. 3. Calculate the difference: $$T_1(n) - T_2(n) = (n^2 + 3n) - 2n^2 = -n^2 + 3n.$$ 4. Find where the curves intersect by solving $-n^2 + 3n = 0$: $$n(-n + 3) = 0 \implies n=0 \text{ or } n=3.$$ 5. The sign of $T_1(n) - T_2(n)$ changes at $n=0$ and $n=3$. For $0 \le n \le 3$, $-n^2 + 3n \ge 0$, so $T_1(n) \ge T_2(n)$. For $3 < n \le 5$, $T_1(n) < T_2(n)$. 6. Compute the area as two integrals: $$\text{Area} = \int_0^3 (T_1(n) - T_2(n)) \, dn + \int_3^5 (T_2(n) - T_1(n)) \, dn.$$ 7. Calculate the first integral: $$\int_0^3 (-n^2 + 3n) \, dn = \left[-\frac{n^3}{3} + \frac{3n^2}{2}\right]_0^3 = \left(-\frac{27}{3} + \frac{27}{2}\right) - 0 = -9 + 13.5 = 4.5.$$ 8. Calculate the second integral: $$\int_3^5 (2n^2 - (n^2 + 3n)) \, dn = \int_3^5 (n^2 - 3n) \, dn = \left[\frac{n^3}{3} - \frac{3n^2}{2}\right]_3^5.$$ 9. Evaluate at bounds: At $n=5$: $\frac{125}{3} - \frac{75}{2} = 41.6667 - 37.5 = 4.1667$. At $n=3$: $\frac{27}{3} - \frac{27}{2} = 9 - 13.5 = -4.5$. 10. So second integral is $4.1667 - (-4.5) = 8.6667$. 11. Total area: $$4.5 + 8.6667 = 13.1667.$$ --- 12. Problem Q7: Find the enclosed area between parametric curves $x=t^2, y=t^3$ and $x=t, y=t^2$ for $t \in [0,1]$. 13. Formula: Area between parametric curves can be found by $$\text{Area} = \left| \int_a^b y_1(t) x_1'(t) \, dt - \int_a^b y_2(t) x_2'(t) \, dt \right|,$$ where $(x_1,y_1)$ and $(x_2,y_2)$ are the two curves. 14. Compute derivatives: $$x_1'(t) = \frac{d}{dt} t^2 = 2t,$$ $$x_2'(t) = \frac{d}{dt} t = 1.$$ 15. Compute integrals: $$I_1 = \int_0^1 y_1(t) x_1'(t) \, dt = \int_0^1 t^3 \cdot 2t \, dt = \int_0^1 2t^4 \, dt = 2 \cdot \frac{t^5}{5} \Big|_0^1 = \frac{2}{5}.$$ $$I_2 = \int_0^1 y_2(t) x_2'(t) \, dt = \int_0^1 t^2 \cdot 1 \, dt = \frac{t^3}{3} \Big|_0^1 = \frac{1}{3}.$$ 16. Area enclosed: $$|I_1 - I_2| = \left| \frac{2}{5} - \frac{1}{3} \right| = \left| \frac{6}{15} - \frac{5}{15} \right| = \frac{1}{15}.$$ --- 17. Problem Q8: Find area between $y=\sqrt{x}$ and $y=x^2$ using substitution $x=u^2$. 18. Find intersection points by solving $\sqrt{x} = x^2$: $$x^{1/2} = x^2 \implies x^{1/2 - 2} = 1 \implies x^{-3/2} = 1 \implies x=1.$$ 19. The curves intersect at $x=0$ and $x=1$. 20. Area between curves: $$\int_0^1 (\sqrt{x} - x^2) \, dx.$$ 21. Substitute $x = u^2$, so $dx = 2u \, du$, and when $x=0$, $u=0$; when $x=1$, $u=1$. 22. Rewrite integral: $$\int_0^1 (u - u^4) 2u \, du = \int_0^1 2u (u - u^4) \, du = \int_0^1 2u^2 - 2u^5 \, du.$$ 23. Integrate: $$\int_0^1 2u^2 \, du = 2 \cdot \frac{u^3}{3} \Big|_0^1 = \frac{2}{3},$$ $$\int_0^1 2u^5 \, du = 2 \cdot \frac{u^6}{6} \Big|_0^1 = \frac{1}{3}.$$ 24. Area: $$\frac{2}{3} - \frac{1}{3} = \frac{1}{3}.$$ --- 25. Problem Q9: Find area between piecewise functions $$f(x) = \frac{x^2}{2-x} \text{ for } 1 \le x \le 2,$$ $$g(x) = x \text{ for } 1 < x < 2.$$ 26. Area between curves: $$\int_1^2 \left| \frac{x^2}{2-x} - x \right| \, dx.$$ 27. Simplify the integrand: $$\frac{x^2}{2-x} - x = \frac{x^2 - x(2-x)}{2-x} = \frac{x^2 - 2x + x^2}{2-x} = \frac{2x^2 - 2x}{2-x} = \frac{2x(x-1)}{2-x}.$$ 28. For $x \in (1,2)$, numerator $2x(x-1) > 0$ and denominator $2-x > 0$ for $x<2$, so integrand is positive. 29. Compute integral: $$2 \int_1^2 \frac{x(x-1)}{2-x} \, dx.$$ 30. Use substitution $u = 2 - x$, so $du = -dx$, when $x=1$, $u=1$, when $x=2$, $u=0$. 31. Rewrite integral: $$2 \int_{u=1}^0 \frac{(2-u)(1-u)}{u} (-du) = 2 \int_0^1 \frac{(2-u)(1-u)}{u} \, du.$$ 32. Expand numerator: $$(2-u)(1-u) = 2 - 3u + u^2.$$ 33. Integral becomes: $$2 \int_0^1 \frac{2 - 3u + u^2}{u} \, du = 2 \int_0^1 \left( \frac{2}{u} - 3 + u \right) du.$$ 34. Integrate term by term: $$2 \left[ 2 \ln|u| - 3u + \frac{u^2}{2} \right]_0^1.$$ 35. Evaluate at $u=1$: $$2(2 \cdot 0 - 3 \cdot 1 + \frac{1}{2}) = 2(-3 + 0.5) = 2(-2.5) = -5.$$ 36. As $u \to 0^+$, $\ln u \to -\infty$, so integral diverges to $-\infty$. 37. The integral diverges, so the area is infinite due to the vertical asymptote at $x=2$. Final answers: - Q6 area: $13.1667$ - Q7 area: $\frac{1}{15}$ - Q8 area: $\frac{1}{3}$ - Q9 area: Diverges (infinite) due to vertical asymptote.