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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 6

Lesson 1: Exploring accumulations of change

# Worked example: accumulation of change

An example relating rates of change with a leaky bathtub. Created by Sal Khan.

## Want to join the conversation?

• What if the graph was a curve? why is the integral of the function equal to the area function of the curve? like why is the area function of y=2x be x^2? i don't get how integrating a function can give you the area of the function :( • Really Great Questions. It is just as great that you are thinking this way! Your questions will be answered in the very next section of videos called "Riemann Sums", which will guide your intuition of how when dx, which can be thought of as length along the x axis, times f(x), which can be thought of as the height of the graph above the x axis at x, are multiplied together, they form a width x height = area argument the represents the area of the region under the curve. There is more to it than that, but the Riemann Sum videos will explain it in more detail.
Keep Studying!
• How can you find the area under R(t) when it is discontinuous? • Why did you multiply all of those numbers by 1/2? And why did you use 40 minutes? • The question doesn't indicate what happened to the leak after 20 minutes, by the time the drain valve was opened, the leak was contributing 2 gals / minute as a minimum, but the rate of change on that leak had been linearly increasing at a rate of 1 gal/minute for each 10 minutes leading up to this point.

As the plot does not continue beyond 20 minutes, I calculated both options, with the actual amount of water being in between the two points.

How did the leak stop after the drain valve opened, but before the tub was empty, and how are we to know that this should not be projected going forwards? • What would the area under the curve represent if the Y axis was NOT a rate?
ie. a temperature vs time graph or a position vs time graph.
(1 vote) • The integer always represents the accumulation of the quantity represented by the y-axis over the quantity represented by the x-axis. There are some that are very easy to understand, and other that apparently makes no sense.

In your examples, if the x-axis is time and the y-axis is temperature, the integral would represent the accumulation of temperature that happen in a certain time, this could be useful if you are dealing with a material that absorbs heat and want to know how much heat it has absorbed over a period of time.

Your second example is more complex to find a useful representation, if the x-axis represents time and the y-axis represents position, then the integral represents the total accumulated "position" that something accumulated over time. I really can't think of a useful interpretation of that one.
• i tried solving problem myself using Riemann sums (left rule) using rectangles of width 5 for both the curves and i got the answer to be close to 600 gallons.. i realized that the difference should be 90 gallons because the value of the function at t=20 is 2 and not 20. where am i going wrong? • First off, remember that using a Riemann sum with intervals of width 5 is going to lead to an approximation which may be fairly far off from the actual area under the curve. If I write a left-hand approximation using 12 rectangles of width 5, here's what I get:

L_12 = 5(0 + 0.5 + 1 + 1.5 + 2 + 18.75 + 17.5 + 16.25 + 15 + 13.75 + 12.5 + 11.25) = 550

So it does underestimate the exact value, in large part because you are leaving off that highest point. Does that help?
• How would you model the level of water in the bathtub if water was also being put in at different rates? • At , Sal says that the area of the trapezium or trapezoid is average height times the base, but from I recall that the area of a trapezium or trapezoid is (upper bound+lower bound)*height*0.5? Where does the area formula for the trapezium Sal had stated come from? • Sal chose to divide the trapezoid into a square and a triangle - both easy to compute the area in your head.
He could have used your formula, but then we would have to imagine the trapezoid laying on its side since what you call the upper and lower bound need to be parallel to each other. In that case the upper bound is 20, the lower bound is 10 and the height is 40 giving
(20+10)(40)(1/2) = (30)(40)(1/2)=(30)(20)=600, the same result.
• The slope of the line is positive between t=0 and t=20. Meaning the rate of gal/min is positive. My understanding is that the tank is being filled over the first 20 minutes.

Then, between t=20 to t=60 the slope is negative. My understanding is that this is when the draining of water begins. So, to answer the problem, I would calculate the area under the curve between t=20 and t=60.

In the video (), to find the total water drained the areas under both curves were added.

My question: please explain where is the flaw in my understanding of the problem.  