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### Course: Integral Calculus > Unit 1

Lesson 12: Definite integrals of common functions- Definite integrals: reverse power rule
- Definite integrals: reverse power rule
- Definite integral of rational function
- Definite integral of radical function
- Definite integral of trig function
- Definite integral involving natural log
- Definite integrals: common functions
- Definite integral of piecewise function
- Definite integral of absolute value function
- Definite integrals of piecewise functions

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# Definite integrals: reverse power rule

Examples of calculating definite integrals of polynomials using the fundamental theorem of calculus and the reverse power rule.

## Want to join the conversation?

- Why isn't "plus C" used in the anti derivatives? Is it because we are doing it for a definite integral?(24 votes)
- Yes this is because the integral is definite. For
**definite**integrals, because the antiderivative must be evaluated at the endpoints and the results must be subtracted, the "plus C" terms would cancel out in the subtraction. This is why "plus C" does not appear in the answer for a definite integral.(40 votes)

- At1:11I think Sal misspeaks, it is even so in the transcript. He says "So, four X to the first divided by four, well, that's just gonna be four X. " when he actually divided by one. He still did the problem correctly, but he said the wrong thing I believe.(4 votes)
- At2:09, I get the algebra, but I don't understand the mathematical logic behind it. I thought that with anti-derivatives, anything with a negative derivative would cancel that same amount of anything from the positive, so to speak. e.g. my intuition would be if your lower and upper bounds of [S4dx] were (for example) -1 and 1 (or -2 and 2, etc), then the resulting triangles calculated by the definite antiderivative (4x) would cancel each other out, equalling zero, rather than netting a positive value. Tell me where I'm going wrong, please!(1 vote)
- You have the terms
*negative derivative*and*negative antiderivative*confused. When the antiderivative is negative (in other words, the area is between the x-axis and the curve below the x-axis), then it can cancel out with a positive antiderivative (area is between x-axis and curve above x-axis). However, a part of the function with a negative derivative can still have a positive antiderivative.

And, as Howard Bradley implied, areas on the left of the y-axis and areas on the right of the y-axis do not cancel out. It is areas under the x-axis and areas above the x-axis that have the potential to cancel.(5 votes)

- I understand how to apply the technique taught in the video. But what exactly are we finding when we take the definite integral from -3 to 5 of 4dx ? What does the value 32 tells us?

Appreciate any replies!(1 vote)- If we integrate the function f(x) = 4 over the interval [-3,5] and find the answer is 32, this is telling us that the area under f(x) = 4 within the interval [-3,5] is equal to 32.(5 votes)

- Why is f(x)'s antiderivative called F(x)? Also, who was the one who came up with calculus's format, like f'(x), d/dx, etc.?(1 vote)
- Different mathematicians involved in the discovery and extension of calculus used different notations. For derivatives we have:

dy/dx : Leibniz notation

f'(x) : Lagrange notation

And newton notation is similar to lagrange but with a dot over the f instead.(5 votes)

- If you are asking why he didn't include the constant, its because it is canceled when you subtract the two evaluated antiderivatives.(3 votes)
- Why F(a)-F(b) gives us the area? Logically, different functions with the same a and b values will give the same area, but it is not true I think...(2 votes)
- This is because the 2nd Fundamental Theorum of Calculus says so. Sal has a video on the informal proof of the 1st and 2nd Fundamental Theorums here: https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-7/v/connecting-the-first-and-second-fundamental-theorems-of-calculus

If you would like a rigorous proof of them, the proof is here: https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-7/a/proof-of-fundamental-theorem-of-calculus(2 votes)

- I don't think -1^3 = +1. Am I right?(2 votes)
- You're right, but as per the video: -(-1)^3 = +1(1 vote)

- What grade level math is this?(2 votes)
- 1. at1:19, the reverse power rule was introduced with indefinite integral form, but it also applies to definite integrals, right?

2. in both examples above in the video, we don't need to consider the antiderivatives' constants?(1 vote)- 1. Yep! Remember that definite integrals are the same as indefinite integrals, but with an extra step of plugging in the bounds after integrating. So, any rule that applies to indefinite integrals has to apply to definite ones.

2. Definite integrals do not have a constant at the end. And there's a reason why. Here's a simple proof.

Suppose I need to find the definite integral of f(x) from a to b. So, if the antiderivative of f(x) is F(x), I get [F(x)+c] from a to b. If I plug in the bounds, I get F(b) + c - (F(a) + c). This simplifies to F(b) + c - F(a) - c. See that the constants cancel out and we get F(b) - F(a) (Also observe that this is the statement of the Fundamental Theorem of Calculus). So, even if you added a constant, it would get cancelled. So, we don't add it.(3 votes)

## Video transcript

- [Instructor] Let's evaluate
the definite integral from negative three to five of four dx. What is this going to be equal to? And I encourage you to pause the video and try to figure it out on your own. All right, so in order to evaluate this, we need to remember the
fundamental theorem of calculus, which connects the notion of a definite integral and antiderivative. So the fundamental theorem
of calculus tells us that our definite integral
from a to b of f of x dx is going to be equal to the antiderivative of our function f, which we
denote with the capital F, evaluated at the upper bound, minus our antiderivative, evaluated at the lower bound. So we just have to do
that right over here. So this is going to be equal to, well, what is the antiderivative of four? Well, you might immediately say, well, that's just going to be four x. You could even think of it in
terms of reverse power rule. Four is the same thing
as four x to the zero. So you increase zero by one. So it's going to be four x to the first, and then you divide by that new exponent. Four x to the first divided by one, well, that's just going to be four x. So the antiderivative is four x. This is, you could say,
our capital F of x, and we're going to evaluate that at five and at negative three. We're gonna find the
difference between these two. So what we have right over here, evaluating the antiderivative
at our upper bound, that is going to be four times five. And then from that,
we're going to subtract, evaluating our antiderivative
at the lower bound. So that's four times negative three. Four times negative three. And what is that going to be equal to? So this is 20 and then minus negative 12. So this is going to be plus 12, which is going to be equal to 32. Let's do another example where we're going to do
the reverse power rule. So let's say that we want to find the indefinite or we want to
find the definite integral going from negative one to three of seven x squared dx. What is this going to be equal to? Well, what we want to do is evaluate what is the
antiderivative of this? Or you could say what is,
if this is lowercase f of x, what is capital F of x? Well, the reverse power rule, we increase this exponent by one. So we're going to have
seven times x to the third, and then we divide by
that increased exponent. So seven x to the third divided by three, and we want to evaluate
that at our upper bound and then subtract from that and it evaluate it at our lower bound. So this is going to be equal to, so evaluating it at our upper bound, it's going to be seven
times three to the third. I'll just write that three
to the third over three. And then from that, we
are going to subtract this capital F of x, the antiderivative evaluated
at the lower bound. So that is going to be seven times negative one to the third, all of that over three. And so this first expression, let's see, this is going to be seven times three to the third over three. This is 27 over three. This is going to be the same
thing as seven times nine. So this is going to be 63. And this over here, negative one to the third
power is negative one. But then we were subtracting a negative, so this is just gonna be adding. And so this is just going
to be plus seven over three. Plus seven over three, if we wanted to express
this as a mixed number, seven over three is the
same thing as 2 1/3. So when we add everything together, we are going to get 65 1/3. And we are done.