### Video Transcript

Find the antiderivative of the function π of π₯ is equal to two π₯ squared plus three π₯ plus three.

First letβs recall how to find the derivative of a general function of π₯ with a coefficient π and a power of π. The first thing we need to do is to multiply by the power of π₯, which is π in this case. The second thing we need to do is to lower the power of π₯ by one, which is π minus one in this case. Now we have recalled how to find the derivative of a general function. In order to find the antiderivative, we would need to reverse these steps.

Reversing the last step of our process, we find that in order to find the antiderivative we must first raise the power of π₯ by one. We must then divide by the power of π₯. Itβs worth noting that when finding the derivative, we multiply it by the original power of π₯ before it was lowered by one. In order to find the reverse of this process, we must therefore divide by the new power of π₯ after it has already been raised by one. Letβs look at a couple of examples to make sure we arenβt missing any steps.

Imagine we had a function π of π₯ equals π₯ cubed. Using our rules, the derivative of this function would be three π₯ squared. Now letβs imagine another function: π of π₯ equals π₯ cubed plus one. Taking the derivative of this function, we find the answer is also three π₯ squared. The same will be true for adding or subtracting any constant. We can represent this constant as π. Since constants do not affect the derivative of a function, in some sense we can say they disappear.

We must be careful when moving in the reverse direction and finding the antiderivative since by this logic constants can also appear. We should therefore include a final step to our process for finding an antiderivative where we add the constant π. Let us now formalize the steps that we have for finding an antiderivative. We start with the derivative of a function of π₯ equals ππ₯ to the power of π. The antiderivative to this is therefore π of π₯.

As our first step, we raise the power of π₯ by one to give π plus one. As our second step, we divide by this new power π plus one. Finally, we add our constant π. We have now formalized the process, and we can use this on the function given in the question. The notation here, which you may not be familiar with, simply denotes that we are taking the antiderivative with respect to π₯. Letβs apply our steps turn-by-turn, looking first at two π₯ squared.

We first raise our power of π₯ by one to give two π₯ to the power of two plus one. This is of course two π₯ cubed. We then divide by our new power to give two π₯ cubed over three. Our final step, adding the constant π, we will come back to later. For now, letβs move on to the second term in our equation: three π₯. We first raise our power of π₯ by one to give three π₯ to the power of one plus one. This is three π₯ squared. We then divide by our new power, two, to give three π₯ squared over two.

When looking at our final term three, it may be useful to see that three can be written as three π₯ to the power of zero. This is because π₯ to the power of zero is equal to one. And therefore, weβre simply times-ing three by one, which is three. Using this, we can see that π₯ to the power of zero can have its power raised by one to give three π₯ to the power of zero plus one. This is of course three π₯ to the power of one. Dividing by the new power one gives us the same term. We can therefore rewrite this as three π₯.

Finally, you may have noticed that for each of our terms, we skipped step three and we did not add a constant. We could have done this throughout the process, labelling each constant as one, two, and three, one for each term. Since these constants are not set, we can instead take another approach to tidy things up by defining big πΆ to equal π one add π two add π three. We can simply add one constant to the end of our equation.

Now we have completed our steps. We have found that the antiderivative of the given function is two π₯ cubed over three add three π₯ squared over two add three π₯ add πΆ.