### Video Transcript

Determine the definite integral between two and zero of five over the square root of negative π₯ plus seven dπ₯.

This is not an expression thatβs nice to integrate using our standard rules for finding the antiderivative. So instead, we use the substitution rule for definite integrals. This says that if π prime is continuous on the close interval π to π and π is continuous 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 π prime 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’.

Now, letβs look carefully at the expression negative π₯ plus seven. We know that its derivative is equal to negative one, and the numerator of our fraction is a scalar multiple of this derivative. That means we can let π’, which is our function π of π₯, be equal to negative π₯ plus seven. We saw that its derivative is equal to negative one. And the substitution rule also says it is permissible to operate with dπ₯ and dπ’ after integral signs as if they were differential. In practical terms, it means weβre allowed to treat dπ’ by dπ₯, in this case only, a little bit like a fraction.

So, we say that dπ’ is equal to negative dπ₯ or, alternatively, negative dπ’ equals dπ₯. And this is great because we can now replace negative π₯ plus seven with π’. And we can replace dπ₯ with negative dπ’. Our integral becomes five over the square root of π’ negative dπ’. But of course, convention dictates that we would actually write this as the integral of negative five over the square root of π’ dπ’. Before we do anything here, we are going to need to change our limits. We can see that π is equal to two and π is equal to zero.

Now, according to the substitution rule, we replace π₯ with π and π₯ with π in our substitution. And so, we find that when π₯ is equal to two, π’ is negative two plus seven, which is simply five. Similarly, when π₯ is equal to zero, π’ is zero plus seven, which is seven. And we now see that weβre looking to find the definite integral between five and seven of negative five over the square root of π’ dπ’.

Letβs take out the constant factor of negative five and write one over the square root of π’ as π’ to the power of negative one-half. Then, we recall that to integrate a power term whose exponent is not equal to negative one, we add one to the exponent and then divide by that new value. This means the integral of π’ to the power of negative one-half is π’ to the power of a half divided by one-half. But of course, dividing by one-half is the same as multiplying by two. So, we have two π’ to the power of one-half. We need to evaluate this between the limits of five and seven.

When we do, we get two times seven to the power of one-half minus two times five to the power of one-half. Now, of course, seven to the power of one-half is root seven and five to the power of one-half is root five. We can also distribute this negative five over both terms inside our parentheses. And when we do, we find that the definite integral between two and zero of five over the square root of negative π₯ plus seven with respect to π₯ is negative 10 root seven plus 10 root five.