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Calculus
Integrals
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Find the derivative of \(\displaystyle F(x) = \int\limits_{\cos{\left(x \right)}}^{9} \left(- 3 t^{2} + 9 t - 3\right)\, dt\).

Before using the Fundamental Theorem of Calculus part I the limits of integration must be reversed by using a property of integrals. This gives \(\displaystyle F(x) = - \int\limits_{9}^{\cos{\left(x \right)}} \left(- 3 t^{2} + 9 t - 3\right)\, dt\). Now using the Fundamental Theorem of Calculus part I and the chain rule with \(\displaystyle u=\cos{\left(x \right)}\) and \(\displaystyle F(u)=- \int\limits_{9}^{u} \left(- 3 t^{2} + 9 t - 3\right)\, dt\) gives: \(\displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(3 u^{2} - 9 u + 3\right)\left(- \sin{\left(x \right)}\right)=- \left(3 \cos^{2}{\left(x \right)} - 9 \cos{\left(x \right)} + 3\right) \sin{\left(x \right)}\)

Download \(\LaTeX\) 

\begin{question}Find the derivative of $F(x) = \int\limits_{\cos{\left(x \right)}}^{9} \left(- 3 t^{2} + 9 t - 3\right)\, dt$. 
    \soln{9cm}{Before using the Fundamental Theorem of Calculus part I the limits of integration must be reversed by using a property of integrals. This gives $F(x) = - \int\limits_{9}^{\cos{\left(x \right)}} \left(- 3 t^{2} + 9 t - 3\right)\, dt$. Now using the Fundamental Theorem of Calculus part I and the chain rule with $u=\cos{\left(x \right)}$  and $F(u)=- \int\limits_{9}^{u} \left(- 3 t^{2} + 9 t - 3\right)\, dt$ gives: $F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(3 u^{2} - 9 u + 3\right)\left(- \sin{\left(x \right)}\right)=- \left(3 \cos^{2}{\left(x \right)} - 9 \cos{\left(x \right)} + 3\right) \sin{\left(x \right)}$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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\begin{document}\begin{question}(10pts) The question goes here!
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HTML for Canvas 
<p> <p>Find the derivative of  <img class="equation_image" title=" \displaystyle F(x) = \int\limits_{\cos{\left(x \right)}}^{9} \left(- 3 t^{2} + 9 t - 3\right)\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28x%29%20%3D%20%5Cint%5Climits_%7B%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%5E%7B9%7D%20%5Cleft%28-%203%20t%5E%7B2%7D%20%2B%209%20t%20-%203%5Cright%29%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(x) = \int\limits_{\cos{\left(x \right)}}^{9} \left(- 3 t^{2} + 9 t - 3\right)\, dt " data-equation-content=" \displaystyle F(x) = \int\limits_{\cos{\left(x \right)}}^{9} \left(- 3 t^{2} + 9 t - 3\right)\, dt " /> . </p> </p>
HTML for Canvas
<p> <p>Before using the Fundamental Theorem of Calculus part I the limits of integration must be reversed by using a property of integrals. This gives  <img class="equation_image" title=" \displaystyle F(x) = - \int\limits_{9}^{\cos{\left(x \right)}} \left(- 3 t^{2} + 9 t - 3\right)\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28x%29%20%3D%20-%20%5Cint%5Climits_%7B9%7D%5E%7B%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%20%5Cleft%28-%203%20t%5E%7B2%7D%20%2B%209%20t%20-%203%5Cright%29%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(x) = - \int\limits_{9}^{\cos{\left(x \right)}} \left(- 3 t^{2} + 9 t - 3\right)\, dt " data-equation-content=" \displaystyle F(x) = - \int\limits_{9}^{\cos{\left(x \right)}} \left(- 3 t^{2} + 9 t - 3\right)\, dt " /> . Now using the Fundamental Theorem of Calculus part I and the chain rule with  <img class="equation_image" title=" \displaystyle u=\cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20u%3D%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle u=\cos{\left(x \right)} " data-equation-content=" \displaystyle u=\cos{\left(x \right)} " />   and  <img class="equation_image" title=" \displaystyle F(u)=- \int\limits_{9}^{u} \left(- 3 t^{2} + 9 t - 3\right)\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28u%29%3D-%20%5Cint%5Climits_%7B9%7D%5E%7Bu%7D%20%5Cleft%28-%203%20t%5E%7B2%7D%20%2B%209%20t%20-%203%5Cright%29%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(u)=- \int\limits_{9}^{u} \left(- 3 t^{2} + 9 t - 3\right)\, dt " data-equation-content=" \displaystyle F(u)=- \int\limits_{9}^{u} \left(- 3 t^{2} + 9 t - 3\right)\, dt " />  gives:  <img class="equation_image" title=" \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(3 u^{2} - 9 u + 3\right)\left(- \sin{\left(x \right)}\right)=- \left(3 \cos^{2}{\left(x \right)} - 9 \cos{\left(x \right)} + 3\right) \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20F%27%28x%29%3D%5Cfrac%7BdF%7D%7Bdu%7D%5Cfrac%7Bdu%7D%7Bdx%7D%20%3D%20%5Cleft%283%20u%5E%7B2%7D%20-%209%20u%20%2B%203%5Cright%29%5Cleft%28-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%5Cright%29%3D-%20%5Cleft%283%20%5Ccos%5E%7B2%7D%7B%5Cleft%28x%20%5Cright%29%7D%20-%209%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%203%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(3 u^{2} - 9 u + 3\right)\left(- \sin{\left(x \right)}\right)=- \left(3 \cos^{2}{\left(x \right)} - 9 \cos{\left(x \right)} + 3\right) \sin{\left(x \right)} " data-equation-content=" \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(3 u^{2} - 9 u + 3\right)\left(- \sin{\left(x \right)}\right)=- \left(3 \cos^{2}{\left(x \right)} - 9 \cos{\left(x \right)} + 3\right) \sin{\left(x \right)} " /> </p> </p>