Video: Antiderivatives

In this video, we will learn how to find the antiderivative of a function. The antiderivative of a function 𝑓(π‘₯) is the function 𝐹(π‘₯) where 𝐹′(π‘₯) = 𝑓(π‘₯).

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Video Transcript

Antiderivatives

At this point, you know how the calculate derivatives of many functions. And have been introduced to a variety of their applications. In this video, we want to ask a question that turns that process around. Given a function 𝑓, how do we find a function with the derivative 𝑓? And why would we be interested in such a function?

The answer to the first part of that question is the antiderivative. That question being, given a function 𝑓, how do we find a function with the derivative 𝑓? The antiderivative of a function 𝑓 is a function with the derivative 𝑓. In other words, it’s a function that reverses what the derivative does. Let’s consider the function 𝑓 of π‘₯ equals two π‘₯. To take the antiderivative of this function, we need a function whose derivative is two π‘₯. We know that when we take the derivative of π‘₯ squared, we get two π‘₯. And that means the antiderivative of two π‘₯ would be π‘₯ squared. However, π‘₯ squared is not the only antiderivative of two π‘₯. π‘₯ squared plus one is also an antiderivative of two π‘₯. And that’s because the derivative of a constant is zero. To account for this, we say that the antiderivative of two π‘₯ is π‘₯ squared plus 𝑐, where 𝑐 is any constant value.

Now let’s look at why we might be interested in doing something like this. Let’s consider what we know about motion. If we start with the position of an object and the function 𝑠 of 𝑑, we can find the velocity of that object, the 𝑣 of 𝑑, by taking the derivative of our position function. And if we take the derivative of our velocity function, we can calculate the acceleration of our object, the π‘Ž of 𝑑. But what if we wanted to move in the other direction? If we knew our acceleration and we wanted to calculate the position, then we would need the antiderivative. This is just one of the many cases when we have a need for an antiderivative. Let’s move on and look at some examples of finding the antiderivative.

Find the most general antiderivative capital 𝐹 of π‘₯ of the function lower case 𝑓 of π‘₯ equals two π‘₯ to the seventh power minus three π‘₯ to the fifth power minus π‘₯ squared.

To do this, we’ll take the antiderivative of each of these terms separately. We need the antiderivative of two times π‘₯ to the seventh power. We need to know what function, when we take the derivative, equals two times π‘₯ to the seventh power. And for π‘₯ to the π‘Ž power, we can find the antiderivative by taking π‘₯ to the π‘Ž plus one power over π‘Ž plus one. And then, in the general form, we add 𝑐 to represent any constant. Let’s apply this to two times π‘₯ to the seventh power. We’ll leave two to the side, and we’ll take π‘₯ to the seven plus one power and divide by seven plus one.

The antiderivative of two times π‘₯ to the seventh power is two times π‘₯ to the eighth power over eight. And we can reduce this to one over four. Two times π‘₯ to the seventh power has an antiderivative of π‘₯ to the eighth power over four. And we’ll repeat this process with negative three π‘₯ to the fifth. We can keep the negative three and we have π‘₯ to the five plus one power all over five plus one. Negative three times π‘₯ to the sixth power over six, which will reduce to π‘₯ to the sixth power over two. And we’ll make sure that we keep that negative.

We’ll repeat the process one final time. We’re dealing with negative π‘₯ squared, so I’ll pull out a negative one. Then we’ll have negative one times π‘₯ to the two plus one power over two plus one, negative π‘₯ cubed over three. Because we’re looking for the most general form, we can’t forget this constant at the end. Which makes our antiderivative 𝐹 of π‘₯ equals π‘₯ to the eighth power over four minus π‘₯ to the sixth power over two minus π‘₯ cubed over three plus 𝑐.

Now let’s look at a case where we don’t want the most general form.

Determine the antiderivative capital 𝐹 of the function lower case 𝑓 of π‘₯ equals five π‘₯ to the fourth plus four π‘₯ cubed where capital 𝐹 of one equals negative two.

Before we do anything else, we’ll calculate the general antiderivative. And that means we’ll follow the same process from the previous example. We’ll pull out the constant, add one to our exponent, and then divide by the value of the new exponent. In this case, we’ll have five times π‘₯ to the fifth power divided by five. And we’ll reduce that to π‘₯ to the fifth. Now, for the second term, take out that four, we’ll raise π‘₯ cubed to π‘₯ to the fourth power, and then divide by four. Which reduces to π‘₯ to the fourth power. The four in the numerator and the denominator cancel out.

If we were finding the general form, we would add a constant 𝑐. And we say that capital 𝐹 of π‘₯ equals π‘₯ to the fifth power plus π‘₯ to the fourth power plus 𝑐. And we wanna plug in 𝐹 of one to help us find the value of 𝑐. 𝐹 of one equals negative two. One to the fifth power plus one to the fourth power. One plus one equals two. So two plus 𝑐 has to equal negative two. Subtract two from both sides. And we see that the constant value is negative four. We’ll take that information and plug it in to what we found for the general antiderivative. An antiderivative under these conditions is π‘₯ to the fifth plus π‘₯ to the fourth minus four.

If the second derivative of 𝑓 of π‘₯ equals three π‘₯ to the fifth plus three π‘₯ cubed plus five π‘₯ squared plus two, determine 𝑓 of π‘₯.

If we’re given a second derivative, we can take the antiderivative which will give us the first derivative of 𝑓 of π‘₯. And then we’ll take the antiderivative of that value which will give us our original 𝑓 of π‘₯. The process won’t be any different than our previous examples. We’ll just have to do that twice. Let’s take the antiderivative of three π‘₯ to the fifth which would become three times π‘₯ to the sixth power over six. And the second term, three times π‘₯ cubed, has an antiderivative of three times π‘₯ to the fourth over four. Five π‘₯ squared becomes five π‘₯ cubed over three. And the antiderivative of two is two π‘₯. Make sure you add your constant term.

Now, at this point, we have the first derivative of this function. And we could simplify some of the coefficients here. But because we’re going to take the antiderivative again, I’ll wait and simplify in the last step. Now, we need the antiderivative of three times π‘₯ to the sixth power over six. Three-sixths times π‘₯ to the seventh power over seven. Three-fourths π‘₯ to the fourth is three-fourths times π‘₯ to the fifth over five. Five-thirds times π‘₯ cubed. Five-thirds times π‘₯ to the fourth over four. Two π‘₯ becomes two times π‘₯ squared over two. The antiderivative of a constant is that constant times π‘₯. And we’ll need an additional constant on the end, which we can call 𝐷.

We’ve found our 𝑓 of π‘₯ value, but we want to simplify. We can reduce this three over six to one-half. And then, we’ll have π‘₯ to the seventh power over 14. We multiplied the denominators in our second term. And we get three times π‘₯ to the fifth over 20. Our third term is five times π‘₯ to the fourth over 12. In our fourth term, the twos cancel out and we have π‘₯ squared. Our final two terms, the 𝑐π‘₯ plus 𝐷 can’t be simplified any further. Which means the general antiderivative of the function we were given is π‘₯ to the seventh power over 14 plus three times π‘₯ to the fifth power over 20 plus five times π‘₯ to the fourth power over 12 plus π‘₯ squared plus 𝑐π‘₯ plus 𝐷.

For our last two examples, we’re going to consider what at first might seem like an irregular form.

Determine the most general antiderivative capital 𝐹 of π‘₯ of the function lower case 𝑓, given that lower case 𝑓 of π‘₯ equals five over two plus four over π‘₯.

This first term, five over two, acts just like a constant. And we take its antiderivative by saying five over two times π‘₯. At first, it might not seem very clear what we can do with four over π‘₯. But what if we wrote it like this, four times of one over π‘₯? Now we’re saying, what function has a derivative of one over π‘₯? The natural log of π‘₯ has a derivative of one over π‘₯. This means the antiderivative of four times one over π‘₯ is going to be four times the natural log of π‘₯. Because four times the natural log of π‘₯ has a derivative of four times one over π‘₯. So we’ll bring down four times the natural log of π‘₯. And since we’re considering a general form, we’ll still need to add a constant value 𝑐. Capital 𝐹 of π‘₯ equals five over two π‘₯ plus four times the natural log of π‘₯ plus 𝑐.

By considering the product rule, find the function 𝑓 so that 𝑓 prime of π‘₯ equals 𝑒 to the π‘₯ power over the square root of π‘₯ plus two times 𝑒 to the π‘₯ power times the square root of π‘₯.

We’ll first need to remember the product rule for derivatives. That product rule tells us the derivative of the function 𝑓 of π‘₯ times the function 𝑔 of π‘₯ equals 𝑓 of π‘₯ times the derivative of 𝑔 of π‘₯ plus 𝑔 of π‘₯ times the derivative of 𝑓 of π‘₯. Before we try to find an 𝑓 of π‘₯ and a 𝑔 of π‘₯, let’s rewrite this function. We have 𝑓 prime of π‘₯ equals 𝑒 to the π‘₯ power. And we know that it’s being multiplied by one over the square root of π‘₯. We can write that as π‘₯ to the negative one-half power. We’re multiplying 𝑒 to the π‘₯ power times π‘₯ to the negative one-half power plus two times 𝑒 to the π‘₯ power times π‘₯ to the one-half power.

Something that we know is that the derivative of 𝑒 to the π‘₯ power equals 𝑒 to the π‘₯ power. If we say that 𝑓 of π‘₯ equals 𝑒 to the π‘₯ power, then 𝑓 prime of π‘₯ also equals 𝑒 to the π‘₯ power. This means that π‘₯ to the negative one-half power equals 𝑔 prime of π‘₯. And it means that 𝑔 of π‘₯ equals two times π‘₯ to the one-half power. 𝑔 of π‘₯ equals two times π‘₯ to the one-half power. If we check that derivative, we get two times one-half times π‘₯ to the one-half minus one power, which is in fact π‘₯ to the negative one-half power. But what does this mean for us? Well, in the product rule, this value is the derivative of 𝑓 of π‘₯ times 𝑔 of π‘₯. And that means the antiderivative is going to be 𝑓 of π‘₯ times 𝑔 of π‘₯. We know 𝑓 of π‘₯ and we know 𝑔 of π‘₯, which means the antiderivative equals two times π‘₯ to the one-half power times 𝑒 to the π‘₯ power. And we can put that back in the form it was given to us in. Two times the square root of π‘₯ times 𝑒 to the π‘₯ power.

Let’s briefly recap this key point about antiderivatives. If you’re given some function 𝑓 prime of π‘₯, its antiderivative will be the function 𝑓 of π‘₯ which derivative would then be 𝑓 prime of π‘₯. Our very first example was when 𝑓 prime of π‘₯ equals two π‘₯, we can take the antiderivative which gives us π‘₯ squared plus any constant 𝑐. And π‘₯ squared plus any constant 𝑐, if you take that derivative, would then give you two π‘₯.

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