### Video Transcript

Determine the definite integral between the limits of five and seven of four π₯ times π₯ minus six to the sixth power with respect to π₯ rounded to one decimal place.

Now, the function four π₯ times π₯ minus six to the sixth power isnβt particularly nice to integrate. Not only is it the product of two functions, but one function is itself a composite function. Itβs a function of a function. So instead, we use a special strategy of introducing something extra, a new variable. This is called integration by substitution, and itβs sometimes referred to as the reverse chain rule. Now, the substitution rule for definite integrals says that if the derivative of π π prime is continuous on some close interval π to π and π is continues on the range of π’, which is equal to π of π₯, then the definite integral between π and π of π of π of π₯ times the derivative of π of π₯ with respect to π₯ is equal to the definite integral between π of π and π of π of π of π’ with respect to π’.

So how do we decide what π’ is going to be? Usually, we look for the function whose derivative also occurs. If thatβs not possible though, we try choosing π’ to be some more complicated parts of the integrand. This might be the inner function in a composite function or similar. Now in fact, π₯ minus six is the inner function of a composite function, but also its derivative is equal to one which technically does occur. So weβre going to let π’ be equal to π₯ minus six and this means the derivative of π’ with respect to π₯ is just one. Now, dπ’ by dπ₯ is absolutely not a fraction. But when dealing with integration by substitution, we sometimes treat it like one. And we can say that dπ’ must be equal to dπ₯.

Well, this is great. We can now replace π₯ minus six with π’ and dπ₯ with dπ’. But weβre not quite finish. What are we gonna do about the four π₯ and what are we going to do about the limits? Well, weβll rearrange our substitution by adding six to both sides. And then we find that π₯ is equal to π’ plus six. We now have an expression for π₯ in terms of π’. And we can use the substitution to change our limits. The upper limit is seven. And when π₯ is equal to seven, π’ is equal to seven minus six, which is of course one. Then when π₯ is equal to five, thatβs the lower limit, π’ is five minus six, which is negative one. And then before we do our substitutions, weβre going to recall that the integral of some constant multiplied by a function is equal to that constant multiplied by the integral of the function. So we can take out this constant factor of four and that leaves us to focus on integrating π₯ times π₯ minus six to the sixth power.

Letβs begin by replacing the limits. We want four times the definite integral between negative one and one. Then we replace π₯ with π’ plus six, π₯ minus six to the sixth power with π’ to the sixth power, and dπ₯ with dπ’. We distribute our parentheses and we see that we have an integrand solely in terms of π’ which weβre looking to integrate with respect to π’. And these are both polynomial terms. The integral of the sum of two functions is equal to the sum of the integrals of those functions. So we can just integrate π’ to the seventh power and six π’ to the sixth power individually. And when we do, we recall that the integral of ππ₯ to the πth power, where π and π are real constants and π is not equal to negative one, is ππ₯ to the power of π plus one over π plus one.

Essentially, we increase the exponent by one and then divide by that new number. This means π’ to the seventh power becomes π’ to the eighth power over eight. Six π’ to the sixth power becomes six π’ to the seventh power over seven. And weβre going to evaluate these between the limits of negative one and one by substituting each limit in and then finding the difference. Substituting one π and we get an eighth plus six-sevenths and substituting negative one and we get one-eighth minus six-sevenths. An eighth minus an eighth is zero. So we want four lots of six-sevenths minus negative six-sevenths or four lots of 12-sevenths. Popping these values into our calculator and we get 6.8571 and so on, which is 6.9 correct of one decimal place. And weβve determined our definite integral; itβs 6.9.