### Video Transcript

Determine the definite integral
between negative five and negative two of two over the square root of ๐ฅ plus six
d๐ฅ.

It may not be instantly obvious how
weโre 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
๐ฅ plus six multiplied by two is equal to the numerator. Thatโs two. And thatโs a hint to us that weโre
going to need to use integration by substitution to evaluate our integral.

Remember, in integration by
substitution, we introduce a new function. That is ๐ข, which we let ๐ equal
to ๐ of ๐ฅ. And we see that the definite
integral between ๐ and ๐ of ๐ of ๐ of ๐ฅ times ๐ prime of ๐ฅ with respect to ๐ฅ
is equal to the definite integral between ๐ of ๐ and ๐ of ๐ of ๐ of ๐ข with
respect to ๐ข. Weโre going to let ๐ข be equal to
the inner part of our composite function, thatโs ๐ฅ plus six, so that d๐ข by d๐ฅ is
equal to one. And whilst d๐ข by d๐ฅ isnโt a
fraction, we are allowed to treat it a little like one when weโre working with
integration by substitution. And we can say that d๐ข is equal to
d๐ฅ.

And so, we can replace d๐ฅ with d๐ข
and ๐ฅ plus six with ๐ข. But weโre going to need to do
something with our limits. We use our substitution to redefine
them. Our lower limit is when ๐ฅ is equal
to negative five. So, ๐ข then is equal to ๐ฅ plus
six, which is here negative five plus six, which is of course one. Then, our upper limit is when ๐ฅ is
equal to negative two. So, ๐ข 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 ๐ข with
respect to ๐ข.

Now, in fact, we can write one over
the square root of ๐ข as ๐ข to the power of negative one-half. And this next step isnโt 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 ๐ข
itself. So, this is equal to two times the
definite integral between one and four of ๐ข to the power of negative one-half. Now, when we integrate ๐ข to the
power of negative one-half, we add one to the exponent and then divide by that new
number. So, ๐ข to the power of negative
half becomes ๐ข to the power of one-half divided by one-half, which is the same as
two times ๐ข to the power of one-half.

We then replace ๐ข 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 weโre done. Since we changed our limits, we
donโt need to do anything more. Our definite integral is equal to
four.