# Video: APCALC02AB-P1A-Q29-534128470452

If π’ = π₯Β³, which of the following integrals is equivalent to β«_(0) ^(3) π₯Β²π^π₯Β³ dπ₯? [A]β«_(0) ^(27) π^π’ dπ’ [B] 1/3 β«_(0) ^(3) π’π^π’ dπ’ [C] 1/3 β«_(0) ^(27) π^π’ dπ’ [D] β«_(0) ^(3) π^π’ dπ’

03:30

### Video Transcript

If π’ is equal to π₯ cubed, which of the following integrals is equivalent to the integral from zero to three of π₯ squared π to the power of π₯ cubed with respect to π₯?

We can see that there are four possible options given, all in terms of the variable π’. Weβre told in the question that π’ is equal to π₯ cubed. So weβre going to be changing the variable in the original integral from π₯ to π’. This is called the method of substitution. When we perform the method of substitution, we must make sure that we change all parts of the integral from the first variable to the variable weβre substituting, including the limits of the integral. Letβs see how this works.

If π’ is equal to π₯ cubed, then π to the power of π₯ cubed is equivalent to π to the power of π’. Weβll see how to deal with that π₯ squared factor in a moment. But if weβre changing our integral to be in terms of π’, then we need to be integrating with respect to π’ instead of with respect to π₯. So we need to know the relationship that exists between dπ’ and dπ₯.

First, we find dπ’ by dπ₯ by differentiating π’ with respect to π₯, using the power rule of differentiation. And we find that dπ’ by dπ₯ is equal to three π₯ squared. By a simple rearrangement, we find that dπ₯ is equal to one over three π₯ squared dπ’. So weβve found the relationship between dπ₯ and dπ’.

Now, we donβt actually need to change that π₯ squared term to be in terms of π’ because we know that we also have an π₯ squared term in the denominator here. So the two π₯ squareds are going to cancel one another out. So what weβre left with is the integral from zero to three of π to the power of π’ one-third dπ’. But actually, we arenβt finished yet. Those limits of zero and three are in terms of our variable π₯, not the variable π’. So we need to replace these limits with limits in terms of π’.

For our lower limit, we substitute the lower limit of π₯ equals zero into our equation relating π’ and π₯. And we find that π’ is equal to zero cubed, which is equal to zero. So in fact, the lower limit for our integral is actually the same as the original lower limit for the integral in terms of π₯.

However, when π₯ is equal to three, π’ will be equal to three cubed or three to the power of three, which is 27. So the upper limit for the integral is not three, but 27. That factor of one-third is just a constant. So we can bring it out at the front of our integral. And we now have that our original integral is equivalent to one-third the integral from zero to 27 of π to the power of π’ with respect to π’.

We can see that, in the other options, a variety of mistakes have been made. For example, in answer a, the limits have been converted to limits in π’. But we donβt have the factor of one-third. So we havenβt taken account of the relationship between dπ’ and dπ₯ correctly. In options b and d, the limits havenβt been changed from limits in π₯ to limits in π’. And other mistakes have been made, such as incorrectly dealing with the π₯ squared term and incorrectly finding the relationship between dπ’ and dπ₯.

The correct answer is option c. The integral is equivalent to one-third the integral from zero to 27 of π to the power of π’ with respect to π’.