### Video Transcript

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

It may not be instantly obvious how
weβre going to evaluate this integral. However, if we look carefully, we
can see that the numerator is a scalar multiple of the derivative of the inner
function on our denominator. In other words, the derivative of
π₯ plus six multiplied by two is equal to the numerator. Thatβs two. And thatβs a hint to us that weβre
going to need to use integration by substitution to evaluate our integral.

Remember, in integration by
substitution, we introduce a new function. That is π’, which we let π equal
to π of π₯. And we see that 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’. Weβre going to let π’ be equal to
the inner part of our composite function, thatβs π₯ plus six, so that dπ’ by dπ₯ is
equal to one. And whilst dπ’ by dπ₯ isnβt a
fraction, we are allowed to treat it a little like one when weβre working with
integration by substitution. And we can say that dπ’ is equal to
dπ₯.

And so, we can replace dπ₯ with dπ’
and π₯ plus six with π’. But weβre going to need to do
something with our limits. We use our substitution to redefine
them. Our lower limit is when π₯ is equal
to negative five. So, π’ then is equal to π₯ plus
six, which is here negative five plus six, which is of course one. Then, our upper limit is when π₯ is
equal to negative two. So, π’ is equal to negative two
plus six, which is four. And so, our integral is equal to
the definite integral between one and four of two over the square root of π’ with
respect to π’.

Now, in fact, we can write one over
the square root of π’ as π’ to the power of negative one-half. And this next step isnβt entirely
necessary, but it can make the process simpler. We recall that we can take any
constant factors outside of the integral and focus on integrating the function in π’
itself. So, this is equal to two times the
definite integral between one and four of π’ to the power of negative one-half. Now, when we integrate π’ to the
power of negative one-half, we add one to the exponent and then divide by that new
number. So, π’ to the power of negative
half becomes π’ to the power of one-half divided by one-half, which is the same as
two times π’ to the power of one-half.

We then replace π’ with four and
one and find the difference. We get two times two lots of four
to the power of one-half minus two lots of one to the power of one-half. Well, four to the power of one-half
is two. And one to the power of one-half is
one. So, we have two times two times two
minus two times one, which is simply equal to four. And weβre done. Since we changed our limits, we
donβt need to do anything more. Our definite integral is equal to
four.