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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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  • piceratops ultimate style avatar for user Joshua Ogunmefun
    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?
    (14 votes)
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    • leaf red style avatar for user Ralph Schraven
      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!
      (67 votes)
  • blobby green style avatar for user Adam.Code3
    Where do you find these Calculus problems? They seem challenging and fun to do
    (3 votes)
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  • leaf green style avatar for user Jon Garcia
    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?
    (3 votes)
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  • marcimus orange style avatar for user Navneet
    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?
    (3 votes)
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    • duskpin ultimate style avatar for user Ms Demaray
      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

      :)
      (5 votes)
  • blobby green style avatar for user Lumbini
    Does the property Sal discussed here tell us that area here is a vector quantity?
    (1 vote)
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  • leaf green style avatar for user 石乐志大师
    What about below and backward? The same function but when the curve is under the x axis and start from 7 and end at 4. Will this give us a positive 2 since - - = + ?
    (1 vote)
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  • piceratops sapling style avatar for user Rice, Veniet
    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!
    (1 vote)
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Video transcript

- Do you want to evaluate the definite integral from three to three, of F of X, D X. And we're given the graph of F of X, and of Y equals F of X, and the area between F of X, and the X-axis over different intervals. Well when you look at this, you actually don't even have to look at this graph over here, because in general, if I have the definite integral of any function F of X, D X, from let's say, A to the same value, from one value to the same value, this is always going to be equal to zero. We're going from three to three. We could be going from negative pi to negative pi. It's always going to be zero. One way to think about it, we're starting and stopping here at three, so we're not capturing any area. Let's do another one. So here, we want to find the definite integral from seven to four, of F of X, D X. So we want to go from seven to four. So we want to go from seven to four. Now you might be tempted to say, okay well look, the area between F of X and X, is two, so maybe this thing is two. But the key realization is, this area only applies when you have the lower bound as the lower bound, and the higher value as the higher bound. So the integral from four to seven of F of X, D X, this thing. This thing is equal to two. This thing is depicting that area right over there. So what about this, where we've switched it? Instead of going from four to seven, we're going from seven to four. Well the key realization is, is if you switch the bounds, this is a key definite integral property, that's going to give you the negative value. So this is going to be equal to the negative of the integral from four to seven, of F of X, D X. And so this is going to be negative, and we just figured out the integral from four to seven of F of X, D X, well now that is this area. F of X is above the X-axis. It's a positive area. So it is going to be, so this thing right over here is going to evaluate to positive two, but we have that negative out front, so our original expression would evaluate to negative two.