Main content

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

Lesson 6: Applying properties of definite integrals- Negative definite integrals
- Finding definite integrals using area formulas
- Finding definite integrals using area formulas
- Definite integral over a single point
- Integrating scaled version of function
- Switching bounds of definite integral
- Integrating sums of functions
- Worked examples: Finding definite integrals using algebraic properties
- Finding definite integrals using algebraic properties
- Definite integrals on adjacent intervals
- Worked example: Breaking up the integral's interval
- Worked example: Merging definite integrals over adjacent intervals
- Definite integrals over adjacent intervals
- Functions defined by integrals: switched interval
- Finding derivative with fundamental theorem of calculus: x is on lower bound
- Finding derivative with fundamental theorem of calculus: x is on both bounds
- Definite integrals properties review

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 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.

## Want to join the conversation?

- 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)
- 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!(75 votes)

- Where do you find these Calculus problems? They seem challenging and fun to do(3 votes)
- 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.(7 votes)

- 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)
- Yes, the integral of f(x) from x=4 to x=0 would be positive 7.(6 votes)

- 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)
- 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)

- 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!(2 votes)
- Does the property Sal discussed here tell us that area here is a vector quantity?(1 vote)
- Area is never a vector quantity. A vector quantity has a magnitude and direction. Area is just the amount of 2 dimensional space a shape takes which has no direction. Hope this helps!(2 votes)

- 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)
- Yup, if the function was flipped across the x-axis, and the integral from 4 to 7 was -2 instead, the answer here would be positive 2. I think that's what you mean...let me know if you meant something else. :)(1 vote)

## 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.