Video: Finding the General Antiderivative of a Given Function

Determine the most general antiderivative 𝐹(π‘₯) of the function 𝑓(π‘₯) = 4π‘₯(βˆ’π‘₯ + 5).

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Video Transcript

Determine the most general antiderivative capital 𝐹 of π‘₯ of the function lowercase 𝑓 of π‘₯ is equal to four π‘₯ multiplied by negative π‘₯ plus five.

The question gives us a function lowercase 𝑓 of π‘₯, and it wants us to find the most general antiderivative of our function lowercase 𝑓 of π‘₯. We’ll call this capital 𝐹 of π‘₯. Let’s start by recalling what we mean by capital 𝐹 of π‘₯ being an antiderivative of lowercase 𝑓 of π‘₯. We say that capital 𝐹 of π‘₯ is an antiderivative of lowercase 𝐹 of π‘₯ if capital 𝐹 prime of π‘₯ is equal to lowercase 𝑓 of π‘₯. In other words, the derivative of capital 𝐹 of π‘₯ with respect to π‘₯ is equal to lowercase 𝑓 of π‘₯. Normally, we’re used to being given a function and then asking to differentiate it. However, in this case, we’re given a function and then asked to find the function which differentiates to give it. We often call this process integration.

Let’s start by taking a closer look at our function lowercase 𝑓 of π‘₯. We can see it’s equal to four π‘₯ multiplied by negative π‘₯ plus five. This is a very complicated-looking function. We don’t know a function which differentiates to give something in this form. So instead, what we’ll do is we’ll simplify our expression for lowercase 𝑓 of π‘₯ by distributing four π‘₯ over our parentheses. Doing this, we get that lowercase 𝑓 of π‘₯ is equal to negative four π‘₯ squared plus 20π‘₯. Now, we’ve rewritten our function in the general form for a polynomial. And we know a lot about differentiating functions of this form. So to find our antiderivative, let’s start by recalling how we would differentiate π‘Ž times π‘₯ to the 𝑛th power.

The first thing we do is multiply our entire expression by the exponent of π‘₯. In this case, this is equal to 𝑛. The next thing we do is reduce our exponent by one. This gives us 𝑛 times π‘Ž multiplied by π‘₯ to the power of 𝑛 minus one. We call this the power rule for differentiation. So what we’ve described is how we would differentiate a polynomial term by term to get another polynomial. However, that’s not what we want to find out in this question. In this question, we’re given a polynomial, and we need to find the polynomial which differentiates to give us that function. In other words, we’re trying to do the reverse. So to do this, instead of using the power rule for differentiation, we’re going to try and do the reverse of this.

So let’s discuss our process for finding an antiderivative of functions in this form. When using the power rule for differentiation, the last thing we did is reduce our exponent by one. We need to do the reverse of this. So instead of reducing our exponent by one, we’re going to add one to our exponent. So our first step will be add one to the exponent of π‘₯. Next, we need to find the reverse of our first step in the power rule for differentiation. This was to multiply our coefficient by the exponent of π‘₯. The reverse of this will be to divide our coefficient by the exponent of π‘₯. But remember, we were multiplying the coefficient by the original exponents. We now need to divide the coefficient by the new exponent of π‘₯.

One way of seeing this is to consider what would happen if we applied this process to 𝑛 times π‘Ž multiplied by π‘₯ to the power of 𝑛 minus one. Remember, we want this process to give us π‘Ž times π‘₯ to the power of 𝑛. The first step tells us to add one to our exponent. In our case, our exponent of π‘₯ is 𝑛 minus one. So this gives us 𝑛 times π‘Ž times π‘₯ to the 𝑛th power. Next, we need to divide our coefficient by our new exponent. In our case, our new exponent is 𝑛. So we divide our coefficient by 𝑛. And of course, we can then cancel 𝑛 divided by 𝑛 to give us one. And we see this gives us π‘Ž times π‘₯ to the 𝑛th power. So this gives us a method of finding antiderivatives of functions in this form.

However, there’s one more thing we need to consider. And that is the derivative of any constant is always equal zero. Well, what does that mean in practice? Let’s consider the derivative of π‘₯ squared plus one. Well, we know the derivative of π‘₯ squared is equal to two π‘₯, and the derivative of one is equal to zero. So π‘₯ squared plus one is an antiderivative of two π‘₯. But now, consider the derivative of π‘₯ squared plus three. Again, we’ll differentiate this term by term. The derivative of π‘₯ squared with respect to π‘₯ is two π‘₯, and the derivative of the constant three is equal to zero. So π‘₯ squared plus three is also an antiderivative of two π‘₯. In fact, we can see π‘₯ squared plus any constant will be an antiderivative of two π‘₯.

One way we could represent this would be to say π‘₯ squared and then we can add any constant. We’ll call this constant 𝐢. Then, we know π‘₯ squared plus 𝐢 is an antiderivative of two π‘₯ for any value of 𝐢. And this is what we mean when we say the most general antiderivative of a function. So whenever we’re looking for a general antiderivative of a function, we add the last step which is to add a constant of integration which we usually call 𝐢. We’re now ready to find our general antiderivative of the function lowercase 𝑓 of π‘₯ given to us in the question.

Remember, since we can apply our derivative rules term by term, we can also use our antiderivative rules term by term. So let’s start with our first term of negative four π‘₯ squared. First, we want to add one to our exponent. In our case, the exponent of π‘₯ is two, so we add one to this to give three. Then, we need to do our second step which is to divide our coefficient by our new exponent. Our new exponent is equal to three. So we divide our entire expression by three. This gives us negative four π‘₯ cubed divided by three. And the third step is to add a constant of integration. But we can just do this at the end.

We now want to apply this process to our second term. It might be easier to consider this as 20 times π‘₯ to the first power. Again, we add one to our exponent of π‘₯ to get two and then divide by this new exponent of two. This gives us 20π‘₯ squared divided by two. And of course, 20 divided by two can simplify to give us 10. And remember, the last step we need to do is add our constant of integration which we will call 𝐢. And this is our final answer.

Therefore, we were able to find the most general antiderivative capital 𝐹 of π‘₯ of the function lowercase 𝑓 of π‘₯ is equal to four π‘₯ multiplied by negative π‘₯ plus five. We found that capital 𝐹 of π‘₯ is equal to negative four π‘₯ cubed divided by three plus 10π‘₯ squared plus a constant of integration 𝐢.

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