### Video Transcript

Find the most general antiderivative capital πΉ of π₯ of the function lowercase π of π₯ is equal to three π₯ squared minus two π₯ minus one.

The question gives us the quadratic function lowercase π of π₯, and it wants us to find the most general antiderivative of this function lowercase π of π₯. Weβre told to call this most general antiderivative capital πΉ of π₯. Letβs start by recalling what an antiderivative is. We say that capital πΉ of π₯ is an antiderivative of lowercase π of π₯ if the derivative of capital πΉ of π₯ with respect to π₯ is equal to lowercase π of π₯.

But we know antiderivatives of a function are not unique. For example, the derivative of π₯ plus one with respect to π₯ is equal to one, so π₯ plus one is an antiderivative of one. However, if we were also to calculate the derivative of π₯ minus one with respect to π₯, we would also get one. So π₯ minus one is also an antiderivative of one. In fact, for any constant π, the derivative of π₯ plus π with respect to π₯ will be one. We call this the most general antiderivative because we can input any value of π and get an antiderivative. So now we know the form our most general antiderivative will take.

Letβs now discuss how weβll find the most general antiderivative of our quadratic function π of π₯. Letβs start by thinking about each time individually. Letβs start with three π₯ squared. We need the derivative of something to be equal to three π₯ squared. Recall the power rule for differentiation tells us for any constants π and π, the derivative of ππ₯ to the πth power with respect to π₯ is equal to π times π times π₯ to the power of π minus one. In other words, ππ₯ to the πth power is an antiderivative of πππ₯ to the power of π minus one. In our case, we want something which differentiates to give us an exponent of π₯ equal to two. So, weβll want our value of π equal to three. Letβs use π equal to three to try and find our value of π.

Letβs differentiate ππ₯ cubed with respect to π₯. To do this, we use the power rule for differentiation. We multiply by our exponent of π₯ and then reduce this exponent by one. This gives us three ππ₯ squared. Remember, we want this to be an antiderivative of three π₯ squared. So we need our coefficient of π₯ squared equal to three. Therefore, weβll choose our value of π equal to one. Therefore, what weβve shown so far is π₯ cubed is an antiderivative of three π₯ squared. And we could in fact check this by differentiating π₯ cubed with respect to π₯. If we did this by using the power rule for differentiation, we would indeed get three π₯ squared.

Letβs now do the same thing for our second term, negative two π₯. Again, we can try doing something very similar with the power rule for differentiation. This time, we need to notice that negative two π₯ is the same as negative two π₯ to the first power. So we want a function which differentiates to give us negative two π₯ to the first power. Therefore, we should set our value of π equal to two. So weβll set our value of π equal to two and try using the power rule for differentiation to differentiate ππ₯ squared with respect to π₯. Again, to do this, we multiply by our exponent of π₯ and then reduce this exponent by one. This gives us two ππ₯ to the first power. And of course, π₯ to the first power is just equal to π₯.

Remember, weβre looking for an antiderivative of negative two π₯, so we need our coefficient of π₯ to be equal to negative two. And we see that this is indeed the case if our value of π is equal the negative one. What weβve shown is negative π₯ squared is an antiderivative of negative two π₯. And once again, we could verify this by differentiating negative π₯ squared with respect to π₯. We would use the power rule for differentiation, and we would indeed get negative two π₯.

We could do exactly the same thing for our last term of negative one. But we already know the derivative of any linear function will just be the coefficient of π₯. So, just like we did in our earlier example, we can find an antiderivative of negative one by just having a linear function with the coefficient of π₯ equal to negative one. For example, negative π₯ will be an antiderivative of negative one. Once again, we could verify this by differentiating negative π₯ with respect to π₯. This is just the coefficient of π₯, which is negative one.

Now that we found an antiderivative for each of the three terms in our function lowercase π of π₯, we can add these together to find an antiderivative of our function lowercase π of π₯. Remember, the reason this is true is if we were to differentiate this function, we could in fact evaluate the derivative term by term. And each of these terms is just the antiderivative of one of the terms of our function lowercase π of π₯. Therefore, this derivative will evaluate to give us lowercase π of π₯. But remember, the question is asking us to find the most general antiderivative. And to do this, we need to add a constant to our function. Weβll represent this with a capital πΆ.

Therefore, we were able to show that the function lowercase π of π₯ is equal to three π₯ squared minus two π₯ minus one will have the most general antiderivative capital πΉ of π₯ is equal to π₯ cubed minus π₯ squared minus π₯ plus a constant πΆ.