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Thread: help with Integration word problem

  1. #1
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    help with Integration word problem

    Question as is

    Suppose the marginal cost is C'(x) = e-0.9x, where x is measured in units of 400 items and the cost is measured in units of 1,000 . Find the cost corresponding to the production interval [1,200, 1,600]. Round to the nearest dollar.

    Okay so the boundary is from 1.2 to 1.6 since its measured in 1000's of units.
    e-0.9x / -0.9

    I can factor out 1/-0.9 which simples to -1.1 and then I... plug in 1.2 into x and do the same for 1.6 and add the area of the two and multiply by 400?

    Last edited by XxXScorpionXxX; Apr 19th 2017 at 08:26 PM.
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  2. #2
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    Re: help with Integration word problem

    Quote Originally Posted by XxXScorpionXxX View Post
    Question as is

    Suppose the marginal cost is C'(x) = e-0.9x, where x is measured in units of 400 items and the cost is measured in units of 1,000 . Find the cost corresponding to the production interval [1,200, 1,600]. Round to the nearest dollar.

    Okay so the boundary is from 1.2 to 1.6 since its measured in 1000's of units.
    e-0.9x / -0.9

    I can factor out 1/-0.9 which simples to -1.1 and then I... plug in 1.2 into x and do the same for 1.6 and add the area of the two and multiply by 400?

    Isn't it just

    $\displaystyle \begin{align*} C &= \int_{1.2}^{1.6}{\mathrm{e}^{-0.9\,x}\,\mathrm{d}x} \\ &= \frac{1}{-0.9} \,\left[ \mathrm{e}^{-0.9\,x} \right] _{1.2}^{1.6} \end{align*}$

    BTW $\displaystyle \begin{align*} \frac{1}{-0.9} \end{align*}$ is NOT -1.1, it's $\displaystyle \begin{align*} -\frac{10}{9} = -1.11111111\dots \end{align*}$. Stay EXACT!
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