### Video Transcript

Calculate the integral of two 𝑦 multiplied by three 𝑦 squared minus two to the seventh power with respect to 𝑦.

Now you may be looking at this integral and thinking about possible different approaches that you could take. Notice that we have a set of parentheses to the seventh power. We certainly don’t want to attempt to distribute all of these parentheses to give a polynomial in 𝑦 although that would be possible. But there must be a smarter approach that we can take. Let’s consider the term outside the parentheses. That is this factor here of two 𝑦. Notice that this is a constant multiple of the derivative of what’s inside the parentheses. The derivative is six 𝑦 and this term is two 𝑦. This suggests that we can perform this integration using the method of substitution.

To do so, we’ll introduce a new variable. We’ll let 𝑢 equal the expression inside the parentheses. So 𝑢 is equal to three 𝑦 squared minus two. We’re going to change everything in our integral so that it is in terms of 𝑢 only. Notice that this is an indefinite integral. So we don’t have limits. But if we did, we’d also need to convert them from limits in 𝑦 to limits in 𝑢. First, we need to know the relationship that exists between d𝑢 and d𝑦. So we’ll find d𝑦 by d𝑦 by differentiating 𝑢 with respect to 𝑦. And we find that this is equal to six 𝑦. By some simple rearrangement, we find then that d𝑦 is equal to d𝑢 over six 𝑦.

So we know how to change two parts of this integral to be in terms of 𝑢. But what about that two 𝑦? Well, actually, we don’t need to change it because note that d𝑦 is equal to d𝑢 over six 𝑦. And so, the 𝑦 in the numerator will cancel with the 𝑦 in the denominator of our fraction. Written out in full, we have that the integral is equal to two 𝑦 multiplied by 𝑢 to the seventh power d𝑢 over six 𝑦.

And so, as we said, we can cancel the 𝑦 in the numerator with the 𝑦 in the denominator. We’re also left with a factor of two over six or one over three, one-third, which we can bring out the front of our integral due to the laws of integration. We now found that our integral is equal to one-third the integral of 𝑢 to the seventh power with respect to 𝑢. And this is an integral that we can perform. We recall that to integrate the power of 𝑢 with respect to 𝑢, we increase the power by one and then divide by the new power. So we have one-third multiplied by 𝑢 to the power of eight over eight. And as this is an indefinite integral, we mustn’t forget our constant of integration 𝐶. Our integral simplifies to 𝑢 to the eighth power over 24 plus 𝐶.

But we’re not finished yet. We must make sure we reverse our substitution so that our integral is in terms of the original variable. So we now need to substitute three 𝑦 squared minus two for 𝑢. In terms of 𝑦 then, this integral is equal to one over 24 multiplied by three 𝑦 squared minus two to the eighth power plus 𝐶.