Main content

### 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

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Definite integral of rational function

Sal finds the definite integral of (16-x³)/x³ between -1 and -2 using the reverse power rule.

## Want to join the conversation?

- Shouldn't it be - 7, as the interval for applying fund.th.of calculus should be [-2; -1] and not [-1; -2]?(34 votes)
- If you flip the bounds and add a negative sign to the entire thing (because you flipped the bounds), you should still get 7. The calculation is below:

- [[-8(-1)^(-2) - (-1)] - [-8(-2)^(-2) - (-2)]]

= - [[-7] - [0]]

= - [-7]

= 7

Intuitively, we are moving backwards from -1 to -2 (so the integral should be negative) but we are also below the x-axis (so the integral should be negative again). The two negatives cancel out and we are left with a positive answer.(7 votes)

- Why don't I have to account for a C constant when finding the antiderivative when it is a definite integral?(9 votes)
- If we have a function 𝒇(𝑥) and know its anti-derivative is 𝑭(𝑥) + C, then the definite integral from 𝑎 to 𝑏 is given by 𝑭(𝑏) + C - (𝑭(𝑎) + C). So we don't have to account for it because it cancels out.(32 votes)

- I learned that when we have integral from lower bound that's greater than the upper bound to that upper bound, which is less than the lower bound, we flip the bounds and have a negative integral. Why isn't that the case in this example?(7 votes)
- It turns out that your method is also correct, even though the way it was solved in the video was different. I also solved (and rewrote) the integral with the lower bound of -2 and an upper bound of -1. The solution is 7.(4 votes)

- Should we check if the function is continuous in the interval before we calculate the integration? If we calculate [-1,1], there is zero inside which is not defined, will the result be wrong?(5 votes)
- It doesn't matter. Point discontinuities don't affect the value of an integral, since the rectangle of area being lost has width 0.(7 votes)

- Shouldn't the answer be 7 instead of -7? When solving the problem myself, I got 7 and wasn't sure where I went wrong. So, I double checked with a graphing calculator which is also telling me that the answer is 7.(5 votes)
- You were correct, the answer was 7.(3 votes)

- I know that in this video, the simple algebraic manipulation of (a+b)/b can be used, but what if we encounter say b/(a+b) and is not the antiderivative of a trig function?(3 votes)
- You may need to use the u-substitution method for integration. The link below is about u-substitution if you need a review or study.

https://www.khanacademy.org/math/calculus-home/integration-techniques-calc/u-substitution-calc/v/u-substitution(5 votes)

- I'm looking at the graph of (16-x^3)/x^3 and I'm confused about what this integral means on the graph. The area of the given interval lies under the x axis, not above it. This suggests that the answer should be negative, however negative areas don't exist. Or do they? I remember from trigonometry that negative angles existed, so perhaps this is a similar concept i.e doing equations with numbers that can only be expressed in equations, but not real life.

Furthermore, the answer was positive anyway, but I'm not entirely sure why. Again, I understand the equation that Sal did in the video, but I don't see how to visualize it on the graph. Is the answer positive because the upperbound was lower than the lower bound, which gives an inverse area?

Can someone please confirm that I'm understanding this correctly.(3 votes)- Look at the order of the integral. he has it as -1 to -2, that's the same as 10 to 9, it's "backwards" you usually say lowest bound to upper bound. In cases when you have upper to lower, you make the result negative. so the integral from -2 to -1 woud be negative, but the bounds were swapped so it is a negative negative, or better known as a positive.

But to make sure this gets across, if it's below the x axis you will get a negative result under normal circumstances.(4 votes)

- In one of the practice exercises I had to find the def. integral from -1 to 1 of [(27/x^4)-3]dx and I got the result -24 wich was the correct answer for the exercise, however, when I graphed the function on desmos.com/calculator, the area under the curve looks like positive and either very large or infinite, so what is the truth?(3 votes)
- I would say that -24 is wrong. And that you should let KA know that they have an incorrect question/answer, if possible.

In order for the Fundamental Theorem of Calculus to hold, the integrand has to be continuous on the interval. This is not the case for 27/𝑥⁴ - 3, which has a discontinuity at 𝑥 = 0.(4 votes)

- Isn't (-2)^-2 supposed to be -1/4 instead of 1/4 since the base is negative?(3 votes)
- No, anything raised to an even power (x², x⁴, x⁻², etc.) will be a positive number.(3 votes)

- Is the notation for evaluating at bounds a3:09official? I've never seen it before but it seems useful(3 votes)
- It is part of the fundamental theorem of calculus.(2 votes)

## Video transcript

- [Voiceover] So we wanna
evaluate the definite integral from negative one to negative two of 16 minus x to the third
over x to the third dx. Now at first this might seem daunting, I have this rational expression,
I have xs in the numerators and xs in the denominators,
but we just have to remember, we just have to do some
algebraic manipulation, and this is going to seem
a lot more attractable. This is the same thing
as the definite integral from negative one to negative two of 16 over x to the third minus x to the third over x to the third, minus x to the third
over x to the third dx. And now what is that going to be equal to? That is going be equal
to the definite integral from negative one to negative two of, I could write this first
term right over here, let me do this in a different color. I could write, I could write this as 16x to the negative three,
x to the negative three, and this second one, we have minus x to the
third over x to the third. Well x to the third is just over, x to the third over x to the third is just going to be equal to one. So this is going to be minus one dx, so dx. And so what is this going to be equal to? Well, let's take the antiderivative
of each of these parts and then we're going to
have to evaluate them at the different bounds. So, let's see. The antiderivative of 16x
to the negative three, we're just gonna do the power rule for derivatives in reverse. You can view this as the
power rule of integration or the power rule of
taking the antiderivative where what you do is you're gonna increase our exponent by one, so you're gonna go from
negative three to negative two, and then you're gonna
divide by that amount, by negative two. So it's going to be 16
divided by negative two times x to the negative two. All I did is I increased the exponent and I divided by that amount. So that's the antiderivative here. And 16 divided by negative two, that is just negative eight. So we have negative eight,
x to the negative two. And then the antiderivative
of negative one, well, that's just negative
x, negative, negative x, negative x. And actually you might just know that, and then hey, if I take the
derivative of negative x I get negative one, or if you view this as negative x to the zero power,
because that's what one is, well, it's the same thing. You increase the exponent by one to get x to the first power. And then you divide by one and so, I mean, you can view it
as that right over there, but either way you get
to negative or minus x. And so now we want to evaluate that. We're going to evaluate that at the bounds and take the difference. So we're going to evaluate
that at negative two and then subtract from that
this evaluated at negative one. And let me do those in
two different colors so we can see what's going on. So we're gonna evaluate it at negative two and we're gonna evaluate
it at negative one. So let's first evaluate
it at negative two. So this is going to be equal to, this is going to be equal to, when you evaluate it at negative two, it's going to be negative
eight, negative eight times x to the negative two. So negative two the the negative two power minus negative two. And from that we're going to subtract it and evaluate it at negative one. So, it's going to be negative
eight times negative one to the negative two
power minus negative one. All right, so then what
is this going to be? So, negative two the negative two. So negative two to the negative two is equal to one over negative two squared, which is equal to 1/4. So this is equal to positive 1/4, but then negative eight times positive 1/4 is going to be equal to negative two. And then we have negative
two minus negative two, so that's negative two plus two. And so everything I've just
done in this purplish color that is just going to be zero. And then if we look at what's
going on in the orange, when you evaluate it at negative one, let's see, negative one
to the negative two power, well, that's one over
negative one squared. Well, this all is just going to be one. And so we're gonna have
negative eight plus one, which is equal to negative seven. So all of this evaluates
to negative seven. But remember, we're
subtracting negative seven, and so this is going to result, we deserve a little bit of a drum roll, this is going to be equal
to positive, positive seven. And obviously you don't have to write that positive out in front, I just wrote that just to emphasize that this is going to be a positive seven.