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Questions: Algebra BusinessCalculus

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Calculus
Integrals
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Find the derivative of \(\displaystyle F(x) = \int\limits_{\ln{\left(x \right)}}^{0} \frac{1}{t^{2}}\, dt\).

Before using the Fundamental Theorem of Calculus part I the limits of integration must be reversed by using a property of integrals. This gives \(\displaystyle F(x) = - \int\limits_{0}^{\log{\left(x \right)}} \frac{1}{t^{2}}\, dt\). Now using the Fundamental Theorem of Calculus part I and the chain rule with \(\displaystyle u=\ln{\left(x \right)}\) and \(\displaystyle F(u)=- \int\limits_{0}^{u} \frac{1}{t^{2}}\, dt\) gives: \(\displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(- \frac{1}{u^{2}}\right)\left(\frac{1}{x}\right)=- \frac{1}{x \log{\left(x \right)}^{2}}\)

Download \(\LaTeX\) 

\begin{question}Find the derivative of $F(x) = \int\limits_{\ln{\left(x \right)}}^{0} \frac{1}{t^{2}}\, dt$. 
    \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 $F(x) = - \int\limits_{0}^{\log{\left(x \right)}} \frac{1}{t^{2}}\, dt$. Now using the Fundamental Theorem of Calculus part I and the chain rule with $u=\ln{\left(x \right)}$  and $F(u)=- \int\limits_{0}^{u} \frac{1}{t^{2}}\, dt$ gives: $F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(- \frac{1}{u^{2}}\right)\left(\frac{1}{x}\right)=- \frac{1}{x \log{\left(x \right)}^{2}}$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution 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_{\ln{\left(x \right)}}^{0} \frac{1}{t^{2}}\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28x%29%20%3D%20%5Cint%5Climits_%7B%5Cln%7B%5Cleft%28x%20%5Cright%29%7D%7D%5E%7B0%7D%20%5Cfrac%7B1%7D%7Bt%5E%7B2%7D%7D%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(x) = \int\limits_{\ln{\left(x \right)}}^{0} \frac{1}{t^{2}}\, dt " data-equation-content=" \displaystyle F(x) = \int\limits_{\ln{\left(x \right)}}^{0} \frac{1}{t^{2}}\, 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_{0}^{\log{\left(x \right)}} \frac{1}{t^{2}}\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28x%29%20%3D%20-%20%5Cint%5Climits_%7B0%7D%5E%7B%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%7D%20%5Cfrac%7B1%7D%7Bt%5E%7B2%7D%7D%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(x) = - \int\limits_{0}^{\log{\left(x \right)}} \frac{1}{t^{2}}\, dt " data-equation-content=" \displaystyle F(x) = - \int\limits_{0}^{\log{\left(x \right)}} \frac{1}{t^{2}}\, dt " /> . Now using the Fundamental Theorem of Calculus part I and the chain rule with  <img class="equation_image" title=" \displaystyle u=\ln{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20u%3D%5Cln%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle u=\ln{\left(x \right)} " data-equation-content=" \displaystyle u=\ln{\left(x \right)} " />   and  <img class="equation_image" title=" \displaystyle F(u)=- \int\limits_{0}^{u} \frac{1}{t^{2}}\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28u%29%3D-%20%5Cint%5Climits_%7B0%7D%5E%7Bu%7D%20%5Cfrac%7B1%7D%7Bt%5E%7B2%7D%7D%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(u)=- \int\limits_{0}^{u} \frac{1}{t^{2}}\, dt " data-equation-content=" \displaystyle F(u)=- \int\limits_{0}^{u} \frac{1}{t^{2}}\, dt " />  gives:  <img class="equation_image" title=" \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(- \frac{1}{u^{2}}\right)\left(\frac{1}{x}\right)=- \frac{1}{x \log{\left(x \right)}^{2}} " src="/equation_images/%20%5Cdisplaystyle%20F%27%28x%29%3D%5Cfrac%7BdF%7D%7Bdu%7D%5Cfrac%7Bdu%7D%7Bdx%7D%20%3D%20%5Cleft%28-%20%5Cfrac%7B1%7D%7Bu%5E%7B2%7D%7D%5Cright%29%5Cleft%28%5Cfrac%7B1%7D%7Bx%7D%5Cright%29%3D-%20%5Cfrac%7B1%7D%7Bx%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%5E%7B2%7D%7D%20" alt="LaTeX:  \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(- \frac{1}{u^{2}}\right)\left(\frac{1}{x}\right)=- \frac{1}{x \log{\left(x \right)}^{2}} " data-equation-content=" \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(- \frac{1}{u^{2}}\right)\left(\frac{1}{x}\right)=- \frac{1}{x \log{\left(x \right)}^{2}} " /> </p> </p>