Use the Trapezoidal Rule with 5 function values to obtain an estimate for:
${\int }_{-2}^6{\log }_e\sqrt{x+3} dx$
Simplify your answer as much as possible. 
 

$$\begin{gathered} Use Simpsons Rule with 5 function values to approximate \\ J0 {\int }_0^{\mathrm{\pi }} \mathcal{x}\sin \mathcal{x} d\mathcal{x} correct to 2 decimal places. \end{gathered}$$

$$\begin{gathered} Rationalize the denominator in the expression: \\ \frac{1}{\sqrt{n}+\sqrt{n+1}} where n is a positive integer. \end{gathered}$$

Using your result from part i or otherwise, find the value of the sum 
$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2 }+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+........+\frac{1}{\sqrt{99}+\sqrt{100}}$

$$\begin{gathered} Differentiate with respect to \mathcal{x}. \\ i. \frac{3}{{\mathcal{x}}^5} \\ ii. {(5-{\mathcal{x}}^2)}^7 \\ iii. {\mathcal{x}}^3{e}^{4\mathcal{x}} \\ Use Simpson\text{'}s Rule with five function value to approximate \\ {\int }_0^{-4}f\left(\mathcal{x}\right) d\mathcal{x} where f\left(\mathcal{x}\right)=\sqrt{\mathcal{x}} \\ Give your answer correct to 1 decimal place. \\ Find: \\ {\int }_0^{-1}(e^{5\mathcal{x}}-e^{-\mathcal{x}}) d\mathcal{x} \\ Find \\ \int \frac{{\mathcal{x}}^7-\mathcal{x}}{3{\mathcal{x}}^5} d\mathcal{x} \end{gathered}$$

$$\begin{gathered} Differentiate with respect to \mathcal{x}: \\ \left(i\right) \mathcal{y}=\frac{\mathcal{x}}{e^{\mathcal{x}}} \\ \left(ii\right) \mathcal{y}={(3-\cos \mathcal{x})}^{10} \end{gathered}$$

$$\begin{gathered} \left(i\right) Evaluate {\int }_0^4{e}^{\frac{\mathcal{x}}{4}} d\mathcal{x} \\ \left(ii\right) Find \int \frac{{\mathcal{x}}^2}{{\mathcal{x}}^3+2 }d\mathcal{x} \\ \left(iii\right) Evaluate {\int }_{\frac{\mathrm{\pi }}{3}}^{\mathrm{\pi }} Sin3\mathcal{x} d\mathcal{x} \end{gathered}$$

$Simplify \frac{3\mathcal{x}}{\mathcal{x}+2}-\frac{5\mathcal{x}+19}{{\mathcal{x}}^2+5\mathcal{x}+6}$

$$\begin{gathered} The diagram represents a function \mathcal{y}=f\left(\mathcal{x}\right) \\ Evaluate {\int }_0^{6}f\left(\mathcal{x}\right) d\mathcal{x} \end{gathered}$$