AP®︎/College Calculus BC
Course: AP®︎/College Calculus BC > Unit 6Lesson 10: Finding antiderivatives and indefinite integrals: basic rules and notation: definite integrals
- 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
Definite integral of radical function
Sal finds the definite integral of 12∛x between -1 and 8 using the reverse power rule.
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- My calculator says -1^(4/3) is -1, giving me a final answer of 153 :/ any reason why?(4 votes)
- Make sure the - sign is associated with the 1, rather than negating the entire expression (calculators aren't too smart, they do what you say, not what you mean!).
-1^(4/3) = -1 because the calculator sees that expression as -(1^(4/3))
so associate the - with the 1 like so
(-1)^(4/3) = 1(15 votes)
- Why he didn't do anything with the 12 when taking the antiderivative of this function. Shouldn't that evaluate to 12x when taking the antiderivative?(4 votes)
- 12 is NOT a constant (The expression is not 12 alone, but 12x^1/3. The 12 would be a constant if it wasn't associated with any X, as in x^1/3 +12, for instance). Therefore Sal DID do something with the 12.
Taking x^1/3 alone and find its antiderivative will make you find :
3/4x^4/3(try taking the derivative of
3/4x^4/3and you'll get
But we dont want that ! We want the antiderivative of 12x^1/3
So now, put your 12 in the antiderivative you've found for x^1/3 :
12 . 3/4 . x^4/3
and the twelve becomes the 9 you can see in the rest of the video.(8 votes)
- Which exponent property makes it possible to break out 1/3 from 4/3?(1 vote)
- Differentiating it you would go from 4/3 to 1/3 because the power rule states you subtract 1 from any exponent. Since we're taking the antiderivative you would do the opposite: add 1 to the exponent, bringing it from 1/3 to 4/3.(3 votes)
- how would this work out if it was if there was more than just x under the radical like 12* sq-rt(4-x^2) ?(1 vote)
- To answer your question requires using trig substitutions:
That material depends on what you are learning here so I would suggest that you just keep working through the integration playlist without skipping ahead.(2 votes)
- why didnt sal take the 12 up front because it is a scalar(1 vote)
- I would assume it was because he knew, from experience (and prior preparation), that we was going to have to use it to deal with the 4/3 that was created from the Reverse Power Rule.(2 votes)
- definite integrals from6 to 2 of (8X^2+ 2)/X^(1/2)(1 vote)
- Can't you pull out the 12 as a coefficient(1 vote)
- [Voiceover] So, we want to evaluate the definite integral from negative one to eight of 12 times the cube root of x dx. Let's see, this is going to be the same thing as the definite integral from negative one to eight of 12 times, the cube root is the same thing as saying x to the 1/3 power dx and so now, if we want to take the antiderivative of the stuff on the inside, we're just going to do essentially the power rule, it conduces the power rule of integrals or it's the reverse of the power rule for derivatives, where we increase this exponent by one and then we divide by that increased exponent. So, this is going to be equal to 12 times x to the 1/3 plus one. Let me do that in another color, so we can keep track of it. X to the 1/3 plus one and then we're going to divide by 1/3 plus one and so, what's 1/3 plus one, well, that's 4/3, 1/3 plus 3/3, that's 4/3. So, I could write it this way. I could write this x to the 4/3 divided by 4/3 and this is going to be and I'm going to evaluate this at the bounds. So, I'm going to evaluate this at and I'll do this in different colors, I'm going to evaluate it at eight and I'm going to evaluate it at negative one and I'm going to subtract it and evaluate it at negative one from this expression evaluated at eight, and so what is this going to be equal to? Well, actually, let me simplify a little bit more. What is 12 divided by 4/3? So, 12, I'll do it right, well, I'll do it right over here, 12 over 4/3 is equal to 12 times 3/4, which will give you as 12/1 times 3/4, 12 divided by four is three, so this is going to be equal to nine, 3/4 of 12 is nine. So, this, we could rewrite this, we could write this as nine x to the 4/3 power. So, if we evaluate it at eight, this is going to be nine times eight to the 4/3 power and from that, we're going to subtract and evaluate it at negative one. So, this is going to be nine times negative one to the 4/3 power. So, what is eight to the 4/3 power? I'll do it over here. So, eight to the 4/3 is equal to eight to the 1/3 to the fourth power, these are just exponent properties here. Eight to the 1/3, the cube root of eight, or eight to the 1/3 power, that's two, 'cause two to the third power is eight and two to the fourth power, well two to the fourth power is equal to 16. So, eight to the 4/3 is 16 and what's a negative one to the 4/3? Well, same idea, negative one to the 4/3 is equal to negative one, there's several way you could do it. You could say negative one to the fourth and the cube root of that, or the cube root of negative one and then raise that to the fourth power, either way. So, let's do it the first way. Negative one to the fourth and then take the cube root of that. Well, negative one to the fourth is just one and then one to the 1/3 power, well that's just going to be equal to one. So, what we have here in blue, that's just equal to one. So, we have nine times 16 minus nine time one, well, that's just going to be nine times 15. We have 16/9 and then we're going take away a nine, so that's going to be nine times 15, so what is that? That is going to be equal to nine times 15 is 90 plus 45, which is equal to 135, 135 and we're done.