Find the present value of the income stream 
from now until 5 years from now, given an interest rate 
(Note: Once you plug in the limits of integration, you are finished; you do not need to simplify our answer beyond that step.)
| Foundations:
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| The idea of an income stream is bit more complicated to set up than basic interest problems. We have two forces adjusting the balance of the account: the income stream, usually represented as Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle I(t)}
, which represents the desired income to be withdrawn from the account, and the interest rate paid to the account. In order to evaluate this result, we use the formula
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- present value = Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{0}^{T}I(t)e^{-rt}\,dt,}
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where  is the time when our stream will run out,  is the rate (compounded continuously) paid by the bank and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle I(t)}
is the desired continuous income stream.
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| Solution:
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| There isn't much to do here, except identify that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle T\,=\,5}
, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle I(t)\,=\,20+30x}
and the rate should be written as Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle r\,=\,10\%\,=\,0.10}
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Click to the final answer to see them in the formula!
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| Final Answer:
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- Present value = Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{0}^{T}I(t)e^{-rt}\,dt\,=\,\int _{0}^{5}(20+30x)e^{-0.10t}\,dt.}
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