Differentiate both sides of the equation: \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x^{2}+y^{2}=9)\)

Simplify: \(\displaystyle 4x+\frac{\mathrm{d} }{\mathrm{d} x}y^{2}=0\)

Use implicit differentiation to differentiate the y term: \(\displaystyle 4x+2y\frac{dy}{dx}=0\)

Subtract 4x from both sides of the equation: \(\displaystyle 2y\frac{dy}{dx}=-4x\)

Divide both sides of the equation by 2y: \(\displaystyle \frac{dy}{dx}=\frac{-4x}{2y}=\frac{-2x}{y}\)

Plug in the appropriate values for x and y to find the slope of the tangent line: \(\displaystyle m=\frac{-2(2)}{(1)}=-4\)

Use slope-intercept form to solve for the equation of the tangent line: \(\displaystyle y=mx+b=y=-4x+b\)

Plug in the appropriate values of x and y into the equation, to find the equation of the tangent line: \(\displaystyle 1=(2)(-4)+b\)

Solve for b: \(\displaystyle 1=-8+b\)

\(\displaystyle b=9\)

Solution:\(\displaystyle y=-4x+9\)

To find \(\displaystyle \frac{\mathrm{d} z}{\mathrm{d} x}\) we must use implicit differentiation, which is an application of the chain rule.
Taking \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\) of both sides of the equation, we get

\(\displaystyle z+x\frac{\mathrm{d} z}{\mathrm{d} x}+2z\frac{\mathrm{d} z}{\mathrm{d} x}=3\)

and the derivatives were found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)
Note that for every derivative of a function with z, the additional term \(\displaystyle \frac{\mathrm{d} z}{\mathrm{d} x}\)appears; this is because of the chain rule, where z=g(x), so to speak, for the function it appears in. 

Using algebra to solve, we get

\(\displaystyle \frac{\mathrm{d} z}{\mathrm{d} x}=\frac{3-z}{x+2z}\)

\(\displaystyle \frac{x^{2}}{2}+y^{2}=4\)

Use implicit differentiation to calculate the slope of the tangent line:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(\frac{x^{2}}{2}+y^{2}=4)\)

\(\displaystyle \frac{2x}{2}+2y\frac{dy}{dx}=0\)

Simplify: \(\displaystyle x+2y\frac{dy}{dx}=0\)

Subtract x from both sides of the equation: \(\displaystyle 2y\frac{dy}{dx}=-x\)

Divide both sides of the equation by 2y: \(\displaystyle \frac{dy}{dx}=\frac{-x}{2y}\)

Plug in the x and y values from the point \(\displaystyle (1,\frac{\sqrt{14}}{2})\)into the equation, to calculate the slope of the tangent line: \(\displaystyle \frac{dy}{dx}=\frac{-1}{\frac{2\sqrt{14}}{2}}=\frac{-1}{\sqrt{14}}=\frac{-\sqrt{14}}{14}\)

Take the negative reciprocal of the slope of the tangent line to calculate the slope of the normal line: \(\displaystyle m=-\frac{1}{\frac{-\sqrt{14}}{14}}=\sqrt{14}\)

Plug the slope of the normal line into point slope form: \(\displaystyle y-\frac{\sqrt{14}}{2}=(x-1)\sqrt{14}\)

Add \(\displaystyle \frac{\sqrt{14}}{2}\) to both sides to isolate y: \(\displaystyle y=(x-1)\sqrt{14}+\frac{\sqrt{14}}{2}\)

Simplify the equation: \(\displaystyle y=x\sqrt{14}-\sqrt{14}+\frac{\sqrt{14}}{2}\)

Simplify further: \(\displaystyle y=x\sqrt{14}-\frac{\sqrt{14}}{2}\)

Solution: \(\displaystyle y=x\sqrt{14}-\frac{\sqrt{14}}{2}\)

\(\displaystyle \begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{-4}^{1.4}(11sin(17cos(x)))dx\\&\text{So the interval is }[-4,1.4]\text{ the subintervals have length }\frac{1.4-(-4)}{2}=\frac{27}{10}\\&\text{and the x-values are:}\\&[-4,-\frac{13}{10},\frac{7}{5}]\\&\int_{-4}^{1.4}(11sin(17cos(x)))dx\approx\frac{27}{20}[(11sin(17cos((-4))))+2(11sin(17cos((-\frac{13}{10}))))+(11sin(17cos((\frac{7}{5}))))]\\&\int_{-4}^{1.4}(11sin(17cos(x)))dx\approx-10.84\end{align*}\)

\(\displaystyle \begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{0}^{2}(6cos(20e^{(x)}))dx\\&\text{So the interval is }[0,2]\text{ the subintervals have length }\frac{2-(0)}{2}=1\\&\text{and the x-values are:}\\&[0,1,2]\\&\int_{0}^{2}(6cos(20e^{(x)}))dx\approx\frac{1}{2}[(6cos(20e^{((0))}))+2(6cos(20e^{((1))}))+(6cos(20e^{((2))}))]\\&\int_{0}^{2}(6cos(20e^{(x)}))dx\approx-5.20\end{align*}\)

\(\displaystyle \begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\\&\text{So the interval is }[-1,8.9]\text{ the subintervals have length }\frac{8.9-(-1)}{3}=\frac{33}{10}\\&\text{and the x-values are:}\\&[-1,\frac{23}{10},\frac{28}{5},\frac{89}{10}]\\&\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\approx\frac{33}{20}[(5sin(9e^{((-1))}))+2(5sin(9e^{((\frac{23}{10}))}))+2(5sin(9e^{((\frac{28}{5}))}))+(5sin(9e^{((\frac{89}{10}))}))]\\&\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\approx35.68\end{align*}\)

\(\displaystyle \begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{1}^{6.4}(-6cos(19x^{2}))dx\\&\text{So the interval is }[1,6.4]\text{ the subintervals have length }\frac{6.4-(1)}{3}=\frac{9}{5}\\&\text{and the x-values are:}\\&[1,\frac{14}{5},\frac{23}{5},\frac{32}{5}]\\&\int_{1}^{6.4}(-6cos(19x^{2}))dx\approx\frac{9}{10}[(-6cos(19(1)^{2}))+2(-6cos(19(\frac{14}{5})^{2}))+2(-6cos(19(\frac{23}{5})^{2}))+(-6cos(19(\frac{32}{5})^{2}))]\\&\int_{1}^{6.4}(-6cos(19x^{2}))dx\approx-16.73\end{align*}\)

\(\displaystyle \begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{2}^{6.5}(17cos(2sin(2x)))dx\\&\text{So the interval is }[2,6.5]\text{ the subintervals have length }\frac{6.5-(2)}{3}=\frac{3}{2}\\&\text{and the x-values are:}\\&[2,\frac{7}{2},5,\frac{13}{2}]\\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\approx\frac{3}{4}[(17cos(2sin(2(2))))+2(17cos(2sin(2(\frac{7}{2}))))+2(17cos(2sin(2(5))))+(17cos(2sin(2(\frac{13}{2}))))]\\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\approx27.55\end{align*}\)

\(\displaystyle \begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{0}^{4.4}(-196cos(3x)^{2})dx\\&\text{So the interval is }[0,4.4]\text{ the subintervals have length }\frac{4.4-(0)}{2}=\frac{11}{5}\\&\text{and the x-values are:}\\&[0,\frac{11}{5},\frac{22}{5}]\\&\int_{0}^{4.4}(-196cos(3x)^{2})dx\approx\frac{11}{10}[(-196cos(3(0))^{2})+2(-196cos(3(\frac{11}{5}))^{2})+(-196cos(3(\frac{22}{5}))^{2})]\\&\int_{0}^{4.4}(-196cos(3x)^{2})dx\approx-744.97\end{align*}\)

\(\displaystyle \begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\\&\text{So the interval is }[3,7.2]\text{ the subintervals have length }\frac{7.2-(3)}{3}=\frac{7}{5}\\&\text{and the x-values are:}\\&[3,\frac{22}{5},\frac{29}{5},\frac{36}{5}]\\&\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\approx\frac{7}{10}[(11cos(6e^{(4(3))}))+2(11cos(6e^{(4(\frac{22}{5}))}))+2(11cos(6e^{(4(\frac{29}{5}))}))+(11cos(6e^{(4(\frac{36}{5}))}))]\\&\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\approx-1.86\end{align*}\)