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A car moving at a speed of v mph achieves 25 + 0.1v mpg (miles per gallon) for v between 20 and...

Question:

A car moving at a speed of {eq}v \ mph {/eq} achieves {eq}25 + 0.1v \ mpg {/eq} (miles per gallon) for {eq}v {/eq} between {eq}20 {/eq} and {eq}60 mph. {/eq} Your speed as a function of time, {eq}t {/eq}, in hours, is given by

{eq}v= 50 \frac {t}{t+1} {/eq}

How many gallons of gas do you consume between {eq}t = 2 {/eq} and {eq}t = 3 {/eq}?

Area under a curve:

To get the total amount of fuel consumed we will find the area under the fuel function with respect to time, and then integrate it from t = 2 and t = 3.

Answer and Explanation:

The function of fuel consumed given to us is 25 +0.1v, where v is the function of speed, which is given by

{eq}v= 50 \frac {t}{t+1} {/eq}

We just substitute v in the fuel function to get the fuel function in terms of time

fuel consumed = {eq}25 + 0.1*50 \frac {t}{t+1} {/eq}

Now we integrate this function from t = 2 to t = 3

{eq}\int_{2}^{3} 25 + 0.1*50 \frac {t}{t+1} dt {/eq}

{eq}25t + 5(t - ln(t+1)) \Big]_{2}^{3} {/eq}

= 28.56 gallons


Learn more about this topic:

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How to Find Area Between Functions With Integration

from Math 104: Calculus

Chapter 14 / Lesson 3
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