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# Worked examples: Finding definite integrals using algebraic properties

Sal evaluates definite integrals of functions given their graphs. He does so using various properties of integrals.

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• But why is it negative? Since you are basically finding the area over that exact same interval. It is just going from the opposite direction. My intuition tells me it that should be 2 not -2. Can someone please explain this?
• This is a very tricky question, and the more I got into researching the definitions and history behind integrals, the further away I seemed to get from being able to answer. However, I then did a cleverly worded Google search and arrived at a math StackExchange question about this topic: https://math.stackexchange.com/questions/1316529/why-does-an-integral-change-signs-when-flipping-the-boundaries.

Essentially some of the answers are "here's a proof that it's true" (which is trivially easy if you know the definition of an integral), but I didn't find that satisfactory. After all, you're asking for the underlying reason rather than a mere demonstration of veracity.

For this underlying reason, I found one of the answers, written by David K, particularly insightful: "It seems to me that the notion that integration "is just taking the area underneath a curve" is what leads to confusion here.

Integration is really the measurement of the accumulated effect of something occurring at a particular rate with respect to something else. We could use it, for example, to figure out how much water is in a reservoir if we know the net rate of flow of water into the reservoir at each time during an interval of time. When the net rate of flow "in" is positive during an interval, the amount of water increases from the start to the end of that interval. When the net rate of flow is negative (it is actually flowing "out"), the amount of water decreases from the start to the end of that interval.

If you integrate "backward" (from the end of the time interval to the beginning, instead of from the beginning to the end), it is like playing back a video in reverse: whatever happened during that time interval is undone. The result is exactly opposite what happens when you integrate "forward.""

This says it all! It's all about how you conceptualize the integral; it's not just an "area under the curve", but rather an accumulated effect of a rate of change. The fact that changing upper and lower bounds changes the sign of the definite integral is then easily understood both conceptually and with an example.

If any of this was unclear, feel free to ask for clarification!
• Where do you find these Calculus problems? They seem challenging and fun to do
• Well, you could just buy a workbook or you can google it. Collegboard has a lot of Calculus problems including AP Calculus Exam practice tests.
• What about the area under the x-axis? If I were to say, integrate this function from x=4 to x=0, would that mean that the negative 7 becomes positive?
• Yes, the integral of f(x) from x=4 to x=0 would be positive 7.
• when we switch the bounds the value of definite integral remains the same but the sign changes from +ve to -ve and vice versa right?? is it because of the fact that we reverse the direction?
• Yes! I like to think of switching the bounds on the integral like negating the width of the rectangle. Our width changes from (b-a)/n to (a-b)/n. With b>a, the width then becomes negative switching the value of the integral.

Beware the switch for value from a graph when the graph is below the x-axis. The definite integral of a function below the x-axis will naturally by negative, but when you switch the bounds, it will become positive

:)
• The video makes the claim that the integral on the interval [n,n] is always 0. I understand how this can be true in most cases, but what if the function that is being integrated is not defined at n? Would, say, the integral on the interval [0,0] for 1/x be 0 or undefined? And, extrapolating the question a bit, if a function has a finite number of removable discontinuities over the range of which it is being integrated, will those be ignored as they are infinitely thin or will they make the integral undefined? Thanks!