Question

The rate at which revenue was generated​ (in millions of dollars per​ year) for a certain company for the years 2010 through 2016 can be approximated by ​f(t)=348e0.22t ​(10≤t≤​16), where t=10 corresponds to the start of the year 2010. Find and interpret ∫1016f(t)dt.

Answer 1

The revenue of a company represents the income generated by the company within a time frame. The total revenue generated by the company from the start of 2010 to the start of 2016 is $39.16 billion

Given that:

$f(t) = 348e^{0.22t}$  $10 \le t \le 16$

First, we calculate the integral

$\int\limits^{16}_{10} {f(t)} \, dt$

This is calculated as:

$\int\limits^{16}_{10} {f(t)} \, dt = \int\limits^{16}_{10} {348e^{0.22t}} \, dt$

Remove the constant

$\int\limits^{16}_{10} {f(t)} \, dt = 348\int\limits^{16}_{10} {e^{0.22t}} \, dt$

Now, integrate

$\int\limits^{16}_{10} {f(t)} \, dt = 348 \times \frac{1}{0.22} (e^{0.22t})|\limits^{16}_{10}$

$\int\limits^{16}_{10} {f(t)} \, dt = \frac{ 348}{0.22} (e^{0.22t})|\limits^{16}_{10}$

Expand

$\int\limits^{16}_{10} {f(t)} \, dt = \frac{ 348}{0.22} (e^{0.22\times 16} -e^{0.22\times 10} )$

$\int\limits^{16}_{10} {f(t)} \, dt = \frac{ 348}{0.22} (e^{3.52} -e^{2.2} )$

$\int\limits^{16}_{10} {f(t)} \, dt = \frac{ 348}{0.22} \times 24.759$

$\int\limits^{16}_{10} {f(t)} \, dt = 39164.2$

From the question

$t = 10$ represents the start of 2010.

This means that

$t = 16$ represents the start of 2016.

So, the interpretation is:

The total revenue from the start of 2010 to the start of 2016 is $39.16 billion

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brainly.com/question/376674