a] Evaluate each of the following integrals il int _(0)^(pi )/(6)24sin^5Theta cosTheta dTheta (3 marks) ii] int _(0)^2(3x)/(sqrt ((2x^2)+1))dx (Taking positive roots only) (3 marks) b] Evaluate int _(0)^15xe^4xdx , correct to 3 significant figures (4 marks) c] Express (11-3x)/(x^2)+2x-3 in terms of partial fractions hence evaluate int (11-3x)/(x^2)+2x-3dx (5 marks) d] Determine the area enclosed between the curves y=x^2+1 and y=7-x

a] <br />i) To evaluate the integral $\int_{0}^{\frac{\pi}{6}} 24\sin^5\Theta \cos\Theta d\Theta$, we can use the substitution method. Let $u = \sin\Theta$, then $du = \cos\Theta d\Theta$. The integral becomes $\int_{0}^{\frac{1}{2}} 24u^5 du$. Integrating, we get $\left[ 2u^6 \right]_{0}^{\frac{1}{2}} = 2 \left( \frac{1}{2} \right)^6 - 2(0)^6 = \frac{1}{8}$. Therefore, the correct answer is $\frac{1}{8}$.<br /><br />ii) To evaluate the integral $\int_{0}^{2} \frac{3x}{\sqrt{2x^2+1}} dx$, we can use the substitution method. Let $u = 2x^2 + 1$, then $du = 4x dx$. The integral becomes $\int_{1}^{5} \frac{3}{2\sqrt{u}} du$. Integrating, we get $\left[ 3\sqrt{u} \right]_{1}^{5} = 3\sqrt{5} - 3\sqrt{1} = 3\sqrt{5} - 3$. Therefore, the correct answer is $3\sqrt{5} - 3$.<br /><br />b) To evaluate the integral $\int_{0}^{1} 5xe^{4x} dx$, we can use integration by parts. Let $u = x$ and $dv = 5e^{4x} dx$. Then $du = dx$ and $v = \frac{5}{4}e^{4x}$. Applying the integration by parts formula, we get $\left[ uv \right]_{0}^{1} - \int_{0}^{1} v du = \left[ \frac{5}{4}xe^{4x} \right]_{0}^{1} - \int_{0}^{1} \frac{5}{4}e^{4x} dx$. Integrating, we get $\frac{5}{4}e^{4} - \frac{5}{4}e^{0} = \frac{5}{4}(e^{4} - 1)$. Therefore, the correct answer is $\frac{5}{4}(e^{4} - 1)$.<br /><br />c) To express $\frac{11-3x}{x^2+2x-3}$ in terms of partial fractions, we can factor the denominator as $(x+3)(x-1)$. Then, we can write $\frac{11-3x}{(x+3)(x-1)} = \frac{A}{x+3} + \frac{B}{x-1}$. Solving for $A$ and $B$, we get $A = 2$ and $B = -1$. Therefore, $\frac{11-3x}{x^2+2x-3} = \frac{2}{x+3} - \frac{1}{x-1}$. To evaluate $\int \frac{11-3x}{x^2+2x-3} dx$, we can integrate each term separately. The integral becomes $\int \frac{2}{x+3} dx - \int \frac{1}{x-1} dx$. Integrating, we get $2\ln|x+3| - \ln|x-1| + C$, where $C$ is the constant of integration.<br /><br />d) To determine the area enclosed between the curves $y=x^2+1$ and $y=7-x$, we need to find the points of intersection. Setting $x^2+1 = 7-x$, we get $x^2+2x-6=0$. Solving this quadratic equation, we get $x=-3$ and $x=2$. Therefore, the area enclosed between the curves is $\int_{-3}^{2} (7-x - (x^2+1)) dx = \int_{-3}^{2} (6-2x-x^2) dx$. Integrating, we get $\left[ 6x - \frac{2}{3}x^3 - \frac{1}{3}x^3 \right]_{-3}^{2} = \left[ 6x - x^3 \right]_{-3}^{2} = (12 - 8) - (-18 + 27) = 20 - 9 = 11$. Therefore, the area enclosed between the curves is 11.