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7.5: Integral Applications

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In derivative story problems, often you’re given a function that describes the amount of something, and you’re asked to find the rate of change. With integral story problems, the reverse is often true. You’ll be given a function the describes the rate of change over time, and you’ll be asked to get the amount.

At time t=0 seconds, train starts with velocity 0 meters per second, and eventually gets up to a speed of 40 meters per second at time t=800 seconds. How far has it gone from t=0 to t=800?

Discussion: We know that distance is equal to velocity multiplied by time. So if we can figure out the train’s velocity, we can just multiply by 800 seconds and get an answer. So here are some potential answers:

Why can’t we just go with 40 meters per second? Because the train “eventually” gets up to 40 meters per second doesn’t mean it was going 40 meters per second the whole time — most of the trip it may have been going slower. So the answer of 32,000 meters (or 32 kilometers) is an upper bound, but isn’t the final answer. We need more information.

(Additional information) The speed of the train is given by the function f(t)=2t meters per second.

What this additional information is telling us is that there isn’t just a single velocity we can use and multiply by the 800 seconds — instead, the velocity is changing all the time so there are lots of velocities. Since the train goes from 0 to 40 meters per second, can we just use the “average” which is 20 meters per second? Well, let’s look at the speeds over time:

You can see the train spends much more time above 20 m/s than it does below, so that the train probably goes farther than 80020=16000 meters (16 km).

What about the speed at time t=400 which is 28.3 m/s? Can we use this value? Well, let’s see what that would mean visually. That’s basically treating the speed at the half-way point like it is a constant velocity:

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So if we use a velocity of 28.3 meters per second, that would give a final distance of 80028.322600 meters (22.6 km). But even this isn’t quite right. The train spends an equal amount of time slower than 28.3 and faster than 28.3, but the slow region (from time 0 to time 400) is slower than the fast region (time 400 to time 800) is fast. They don’t exactly cancel, so likely the train won’t quite reach 22.6 km. Somehow, instead, we need to take into account the speed of the train at every instant. Sound familiar? This is where an integral is going to come it. Basically if we take the speed of the train at a small time frame, and multiply by the time it is going that speed, you get a picture that looks like this:

train-screen.png

This is of course just the area under the curve! So we use the fundamental theorem of calculus to find the actual answer:

A quick check to see if this is reasonable: we’re saying the train went 21.3 km in those 13 minutes. Seems a bit fast for a train but not impossible. It’s less than our max of 32 km, but more than the guess of 16 km that we thought it would beat. Seems reasonable!


This page titled 7.5: Integral Applications is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Tyler Seacrest via source content that was edited to the style and standards of the LibreTexts platform.

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