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.

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:

[tex]f(t) = 348e^{0.22t}[/tex] [tex]10 \le t \le 16[/tex]

First, we calculate the integral

[tex]\int\limits^{16}_{10} {f(t)} \, dt[/tex]

This is calculated as:

[tex]\int\limits^{16}_{10} {f(t)} \, dt = \int\limits^{16}_{10} {348e^{0.22t}} \, dt[/tex]

Remove the constant

[tex]\int\limits^{16}_{10} {f(t)} \, dt = 348\int\limits^{16}_{10} {e^{0.22t}} \, dt[/tex]

Now, integrate

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

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

Expand

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

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

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

[tex]\int\limits^{16}_{10} {f(t)} \, dt = 39164.2[/tex]

From the question

[tex]t = 10[/tex] represents the start of 2010.

This means that

[tex]t = 16[/tex] 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