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Questions: Algebra BusinessCalculus
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Use \(\displaystyle \int\limits_{-8}^{-2} f{\left(x \right)}\, dx=-6\) and \(\displaystyle \int\limits_{-2}^{3} f{\left(x \right)}\, dx=-1\) to find the value of \(\displaystyle \int\limits_{-8}^{3} f{\left(x \right)}\, dx\).
Using the property \(\displaystyle \int_{a}^{b}f(x)\,dx = \int_{a}^{c}f(x)\,dx + \int_{c}^{b}f(x)\,dx\) gives:
\(\displaystyle \int\limits_{-8}^{3} f{\left(x \right)}\, dx = \int\limits_{-8}^{-2} f{\left(x \right)}\, dx + \int\limits_{-2}^{3} f{\left(x \right)}\, dx = -6+(-1)=-7\)
\begin{question}Use $\int\limits_{-8}^{-2} f{\left(x \right)}\, dx=-6$ and $\int\limits_{-2}^{3} f{\left(x \right)}\, dx=-1$ to find the value of $\int\limits_{-8}^{3} f{\left(x \right)}\, dx$.
\soln{9cm}{Using the property $\int_{a}^{b}f(x)\,dx = \int_{a}^{c}f(x)\,dx + \int_{c}^{b}f(x)\,dx$ gives:\newline
$\int\limits_{-8}^{3} f{\left(x \right)}\, dx = \int\limits_{-8}^{-2} f{\left(x \right)}\, dx + \int\limits_{-2}^{3} f{\left(x \right)}\, dx = -6+(-1)=-7$}
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Use <img class="equation_image" title=" \displaystyle \int\limits_{-8}^{-2} f{\left(x \right)}\, dx=-6 " src="/equation_images/%20%5Cdisplaystyle%20%5Cint%5Climits_%7B-8%7D%5E%7B-2%7D%20f%7B%5Cleft%28x%20%5Cright%29%7D%5C%2C%20dx%3D-6%20" alt="LaTeX: \displaystyle \int\limits_{-8}^{-2} f{\left(x \right)}\, dx=-6 " data-equation-content=" \displaystyle \int\limits_{-8}^{-2} f{\left(x \right)}\, dx=-6 " /> and <img class="equation_image" title=" \displaystyle \int\limits_{-2}^{3} f{\left(x \right)}\, dx=-1 " src="/equation_images/%20%5Cdisplaystyle%20%5Cint%5Climits_%7B-2%7D%5E%7B3%7D%20f%7B%5Cleft%28x%20%5Cright%29%7D%5C%2C%20dx%3D-1%20" alt="LaTeX: \displaystyle \int\limits_{-2}^{3} f{\left(x \right)}\, dx=-1 " data-equation-content=" \displaystyle \int\limits_{-2}^{3} f{\left(x \right)}\, dx=-1 " /> to find the value of <img class="equation_image" title=" \displaystyle \int\limits_{-8}^{3} f{\left(x \right)}\, dx " src="/equation_images/%20%5Cdisplaystyle%20%5Cint%5Climits_%7B-8%7D%5E%7B3%7D%20f%7B%5Cleft%28x%20%5Cright%29%7D%5C%2C%20dx%20" alt="LaTeX: \displaystyle \int\limits_{-8}^{3} f{\left(x \right)}\, dx " data-equation-content=" \displaystyle \int\limits_{-8}^{3} f{\left(x \right)}\, dx " /> . </p> </p><p> <p>Using the property <img class="equation_image" title=" \displaystyle \int_{a}^{b}f(x)\,dx = \int_{a}^{c}f(x)\,dx + \int_{c}^{b}f(x)\,dx " src="/equation_images/%20%5Cdisplaystyle%20%5Cint_%7Ba%7D%5E%7Bb%7Df%28x%29%5C%2Cdx%20%3D%20%5Cint_%7Ba%7D%5E%7Bc%7Df%28x%29%5C%2Cdx%20%2B%20%5Cint_%7Bc%7D%5E%7Bb%7Df%28x%29%5C%2Cdx%20" alt="LaTeX: \displaystyle \int_{a}^{b}f(x)\,dx = \int_{a}^{c}f(x)\,dx + \int_{c}^{b}f(x)\,dx " data-equation-content=" \displaystyle \int_{a}^{b}f(x)\,dx = \int_{a}^{c}f(x)\,dx + \int_{c}^{b}f(x)\,dx " /> gives:<br>
<img class="equation_image" title=" \displaystyle \int\limits_{-8}^{3} f{\left(x \right)}\, dx = \int\limits_{-8}^{-2} f{\left(x \right)}\, dx + \int\limits_{-2}^{3} f{\left(x \right)}\, dx = -6+(-1)=-7 " src="/equation_images/%20%5Cdisplaystyle%20%5Cint%5Climits_%7B-8%7D%5E%7B3%7D%20f%7B%5Cleft%28x%20%5Cright%29%7D%5C%2C%20dx%20%3D%20%5Cint%5Climits_%7B-8%7D%5E%7B-2%7D%20f%7B%5Cleft%28x%20%5Cright%29%7D%5C%2C%20dx%20%2B%20%5Cint%5Climits_%7B-2%7D%5E%7B3%7D%20f%7B%5Cleft%28x%20%5Cright%29%7D%5C%2C%20dx%20%3D%20-6%2B%28-1%29%3D-7%20" alt="LaTeX: \displaystyle \int\limits_{-8}^{3} f{\left(x \right)}\, dx = \int\limits_{-8}^{-2} f{\left(x \right)}\, dx + \int\limits_{-2}^{3} f{\left(x \right)}\, dx = -6+(-1)=-7 " data-equation-content=" \displaystyle \int\limits_{-8}^{3} f{\left(x \right)}\, dx = \int\limits_{-8}^{-2} f{\left(x \right)}\, dx + \int\limits_{-2}^{3} f{\left(x \right)}\, dx = -6+(-1)=-7 " /> </p> </p>