Video: Finding an Antiderivative Using Integration by Substitution

If 𝑓(π‘₯) is an antiderivative of 9π‘₯√(3π‘₯Β² βˆ’ 3) and 𝑓(2) = 25, what is the value of 𝑓(1)?

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Video Transcript

If 𝑓 of π‘₯ is an antiderivative of nine π‘₯ multiplied by the square root of three π‘₯ squared minus three and 𝑓 of two equals 25, what is the value of 𝑓 of one?

If 𝑓 of π‘₯ is an antiderivative of this function, then it means that 𝑓 of π‘₯ is its integral. We can say that 𝑓 of π‘₯ is equal to the integral of nine π‘₯ multiplied by the square root of three π‘₯ squared minus three with respect to π‘₯. Now, notice that this is an indefinite integral. It has no limits. Which means that when we perform this integration, we’d need to include a constant of integration. This is why we’re told that 𝑓 of π‘₯ is an antiderivative of the function rather than the antiderivative. Because there could be numerous different functions 𝑓 of π‘₯ depending on that value of that constant of integration.

We’ve been given some further information there. We’ve been told that 𝑓 of two is equal to 25. And so, we’ll be able to use this information later on to determine what our constant of integration is equal to. First, though, we need to perform this integral. Now, at first glance, this integral certainly doesn’t seem very straight forward because we have nine π‘₯ multiplied by the square root of a quadratic function. What we need to realize is that that factor of nine π‘₯ is actually a scalar multiple of the derivative of the function within the square root.

If we allow 𝑒 to represent that function three π‘₯ squared minus three, then we see that its derivative with respect to π‘₯, d𝑒 by dπ‘₯, is equal to six π‘₯, which is a scalar multiple of that factor of nine π‘₯. This tells us that we can use the method of integration by substitution in order to perform this integral. We’re going to make a substitution, 𝑒 equals three π‘₯ squared minus three, and convert our integral from an integral in terms of π‘₯ to an integral in terms of 𝑒. Let’s see what this looks like.

We already said that we’ll be letting 𝑒 equal three π‘₯ squared minus three. d𝑒 by dπ‘₯ is then equal to six π‘₯. And remember, d𝑒 by dπ‘₯ is not a fraction, but we can treat it a little like one. If d𝑒 by dπ‘₯ is equal to six π‘₯, then this means that d𝑒 is equivalent to six π‘₯ dπ‘₯. Or, dividing both sides by six, we can say that one-sixth d𝑒 is equal to π‘₯dπ‘₯. And therefore, we know how to replace π‘₯dπ‘₯ in our integral with something in terms of 𝑒. It’s equal to one-sixth d𝑒.

Making our substitution then, we keep the nine as it is. The square root of three π‘₯ squared minus three becomes the square root of 𝑒. And we replace π‘₯dπ‘₯ with one-sixth d𝑒. So we have that 𝑓 of π‘₯ is equal to the integral of nine root 𝑒 one-sixth d𝑒. We can bring the constant of nine-sixths out the front of the integral, giving nine-sixths the integral of root 𝑒 d𝑒. And, of course, nine over six will simplify to three over two. We then express root 𝑒 as 𝑒 to the power of one-half. So our integral becomes three over two the integral of 𝑒 to the power of one-half with respect to 𝑒.

In order to perform this integral, we recall that in order to integrate a power of 𝑒 not equal to negative one, we increase the power by one and divide by the new power. Increasing the power gives three over two and then dividing by three over two is equivalent to multiplying by its reciprocal, two-thirds. We have three over two multiplied by two-thirds 𝑒 to the power of three over two. But three over two multiplied by two-thirds is just equal to one. So we have just 𝑒 to the power of three over two. We mustn’t forget, though, that all important constant of integration as we’re performing an indefinite integral.

Finally, we need to reverse our substitution. Recall that 𝑒 was equal to three π‘₯ squared minus three. So replacing 𝑒 with three π‘₯ squared minus three gives 𝑓 of π‘₯ is equal to three π‘₯ squared minus three to the power of three over two plus c. So we have our expression for 𝑓 of π‘₯. But remember, we were asked to find the value of 𝑓 of one. And in order to do so, we need to know what our constant of integration, 𝑐, is equal to. In order to determine this, we use the other piece of information we were given, which is that 𝑓 of two is equal to 25. Substituting two for π‘₯ and 25 for 𝑓 of π‘₯ gives 25 equals three multiplied by two squared minus three to the power of three over two plus our constant of integration, 𝑐.

We can express a power of three over two as a square root cubed. So we have that 25 is equal to the square root of nine cubed plus 𝑐. The square root of nine is three and three cubed is 27. Finally, subtracting 27 from each side of this equation, we find that the value of our constant of integration, 𝑐, is negative two. Our function 𝑓 of π‘₯ then is equal to three π‘₯ squared minus three all to the power of three over two minus two.

And finally, to find the value of 𝑓 of one, we just need to substitute one into 𝑓 of π‘₯. We have three multiplied by one squared minus three all to the power of three over two minus two. Three multiplied by one squared is just three multiplied by one which is three. And three minus three is zero. Zero to the power of three over two is just zero. So we have zero minus two which simplifies to negative two. So by using the method of integration by substitution and then the given information to evaluate our constant of integration 𝑐, we’ve found that the value of 𝑓 of one is negative two.

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