### Video Transcript

Find the definite integral between
negative one and four of π₯ times the square root of π₯ plus five with respect to π₯
to the nearest thousandth.

In this question, our integrand is
the product of two functions, one of which is itself a composite function. It certainly is a tricky one to
integrate. And really, we have two
options. We could try the substitution
method or integration by parts. Notice though that the inner part
of our composite function has a nice simple derivative. And that tells us we might be able
to use integration by substitution.

The substitution rule for definite
integrals says that if π prime is continuous on the closed interval π to π, and
π is continuous over 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’. So, with this method, we try to
choose π’, which is π of π₯, to be some factor of the integrand whose differential
also occurs, albeit a scalar multiple of it. Here though, itβs not instantly
obvious what that might be.

So, instead, we choose π’ to be
some more complicated part of the function, here, the inner part in a composite
function that has a nice derivative. Weβll try π’ equals π₯ plus
five. The derivative of π₯ plus five with
respect to π₯ is simply one. And whilst we know that dπ’ by dπ₯
isnβt a fraction, we do treat it a little like one. We treat dπ’ and dπ₯ as
differentials. And we can say that dπ’ equals
dπ₯. Now, this may not seem instantly
helpful, as if we replace dπ₯ with dπ’ and π₯ plus five with π’, we still have a
part of our function thatβs in terms of π₯.

However, if we look back to our
substitution, we can rearrange this and say that π₯ must be equal to π’ minus
five. So, we can now replace each part of
our integrand and dπ₯ with dπ’. But what about our limits? Well, here we use our substitution
to redefine these. Our lower limit is when π₯ is equal
to negative one. And since π’ was equal to π₯ plus
five, π’ is equal to negative one plus five, which is equal to four. Then, when π₯ is equal to four,
thatβs our upper limit, π’ is equal to four plus five, which is nine.

So, we rewrite our definite
integral as the definite integral between four and nine, those are our new limits,
of π’ minus five β remember, we said π₯ is equal to π’ minus five β times the square
root of π’ with respect to π’. And actually, letβs rewrite the
square root of π’ as π’ to the power of one-half. And then, we can distribute the
parentheses. When we multiply π’ by π’ to the
power of one-half, we add their exponents. And we end up with π’ to the power
of three over two. So, our integrand is π’ to the
power of three over two minus five π’ to the power of one-half.

When integrating simple polynomial
terms whose exponent is not equal to negative one, we add one to the exponent and
divide by this new value. So, when we integrate π’ to the
power of three over two, we get π’ to the power of five over two divided by five
over two. And thatβs the same as two-fifths
times π’ to the power of five over two. Similarly, when we integrate
negative five π’ to the power of one-half, we get negative five π’ to the power of
three over two divided by three over two. And that simplifies to negative 10
over three π’ to the power of three over two.

Now, of course, we mustnβt forget
that weβre going to need to evaluate this between the limits of four and nine. Thatβs two-fifths of nine to the
power five over two minus ten-thirds of nine to the power of three over two minus
two-fifths times four to the power of five over two minus ten-thirds times four to
the power of three over two. Thatβs 316 over 15, which correct
to the nearest thousandth is 21.067.