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 $f(n)$ and $g(n)$ over $[a,b]$ is given by $$\text{Area} = \int_a^b |f(n) - g(n)| \, dn.$$ Since $T_1(n)$ and $T_2(n)$ are continuous and we can check which is greater over $[0,5]$, we can remove the absolute value accordingly. 3. Compare $T_1(n)$ and $T_2(n)$: $$T_1(n) - T_2(n) = (n^2 + 3n) - 2n^2 = -n^2 + 3n = n(3 - n).$$ For $n$ in $[0,3]$, $T_1(n) > T_2(n)$; for $n$ in $(3,5]$, $T_2(n) > T_1(n)$. 4. Compute area as sum of two integrals: $$\text{Area} = \int_0^3 (T_1(n) - T_2(n)) \, dn + \int_3^5 (T_2(n) - T_1(n)) \, dn.$$ 5. Calculate each integral: $$\int_0^3 (-n^2 + 3n) \, dn = \left[-\frac{n^3}{3} + \frac{3n^2}{2}\right]_0^3 = \left(-\frac{27}{3} + \frac{27}{2}\right) - 0 = -9 + 13.5 = 4.5.$$ $$\int_3^5 (2n^2 - (n^2 + 3n)) \, dn = \int_3^5 (n^2 - 3n) \, dn = \left[\frac{n^3}{3} - \frac{3n^2}{2}\right]_3^5 = \left(\frac{125}{3} - \frac{75}{2}\right) - \left(\frac{27}{3} - \frac{27}{2}\right) = \left(41.6667 - 37.5\right) - \left(9 - 13.5\right) = 4.1667 + 4.5 = 8.6667.$$ 6. Total area: $$4.5 + 8.6667 = 13.1667.$$ --- 7. Problem Q7: Find the enclosed area between parametric curves $x = t^2, y = t^3$ and $x = t, y = t^2$ for $t \in [0,1]$. 8. Formula: Area between parametric curves can be found by $$\text{Area} = \int (y_1 \, dx_1 - y_2 \, dx_2)$$ or equivalently by integrating over $t$ the difference in $y$ times derivative of $x$. 9. Parametric derivatives: For curve 1: $x_1 = t^2$, $y_1 = t^3$, so $dx_1/dt = 2t$. For curve 2: $x_2 = t$, $y_2 = t^2$, so $dx_2/dt = 1$. 10. Area formula: $$\text{Area} = \int_0^1 y_1 \frac{dx_1}{dt} dt - \int_0^1 y_2 \frac{dx_2}{dt} dt = \int_0^1 t^3 (2t) dt - \int_0^1 t^2 (1) dt = \int_0^1 2t^4 dt - \int_0^1 t^2 dt.$$ 11. Compute integrals: $$\int_0^1 2t^4 dt = 2 \cdot \frac{t^5}{5} \Big|_0^1 = \frac{2}{5}.$$ $$\int_0^1 t^2 dt = \frac{t^3}{3} \Big|_0^1 = \frac{1}{3}.$$ 12. Area: $$\frac{2}{5} - \frac{1}{3} = \frac{6}{15} - \frac{5}{15} = \frac{1}{15}.$$ --- 13. Problem Q8: Compute area between $y = \sqrt{x}$ and $y = x^2$ by substituting $x = u^2$. 14. Original area: $$\text{Area} = \int_0^1 (\sqrt{x} - x^2) dx.$$ 15. Substitute $x = u^2$, so $dx = 2u du$, and limits $x=0 \to u=0$, $x=1 \to u=1$. 16. Rewrite integrand: $$\sqrt{x} = \sqrt{u^2} = u,$$ $$x^2 = (u^2)^2 = u^4.$$ 17. Area integral becomes: $$\int_0^1 (u - u^4) 2u du = \int_0^1 2u^2 - 2u^5 du = 2 \int_0^1 u^2 du - 2 \int_0^1 u^5 du.$$ 18. Compute integrals: $$2 \cdot \frac{u^3}{3} \Big|_0^1 - 2 \cdot \frac{u^6}{6} \Big|_0^1 = \frac{2}{3} - \frac{2}{6} = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}.$$ --- Final answers: - Q6 area = $13.1667$ - Q7 area = $\frac{1}{15}$ - Q8 area = $\frac{1}{3}$