### Video Transcript

The function π is even, continuous
on the closed interval negative eight to eight, and satisfies the definite integral
between negative eight and eight of π of π₯ with respect to π₯ is equal to 19 and
the definite integral between zero and four of π of π₯ with respect to π₯ is equal
to 13. Determine the definite integral
between negative eight and negative four of π of π₯ with respect to π₯.

We begin by recalling the property
of the integral of an even function That is, the definite integral between negative
π and π of that even function is equal to two times the definite integral between
zero and π of π of π₯ with respect to π₯. Now, in fact, weβre looking to find
the definite integral between negative eight and negative four of our even
function. So weβre going to do this in two
parts. Firstly, weβre going to split it up
and say that the definite integral must be equal to the integral between negative
eight and zero minus the integral between negative four and zero. Now, actually, weβll form an
equation using the first part of this integral and the fact that the function is
even.

The definite integral between
negative eight and eight of π of π₯ with respect to π₯ must be two times the
definite integral between zero and eight of πof π₯ with respect to π₯. Now, it also follows that this must
also be equal then to two times the definite integral between negative eight and
zero of π of π₯ with respect to π₯. Of course, in the question, we were
told that the definite integral between negative eight and eight of π of π₯ is
19. So we set 19 equal to two times the
definite integral that weβre looking for. And then, we divide both sides of
our equation by two. And we see that this is equal to 19
over two. The integral weβre looking for then
is equal to 19 over two minus the definite integral between negative four and zero
of π of π₯ with respect to π₯.

Now, once again, the function is
even. So this must, in turn, be equal to
19 over two minus the definite integral between zero and four of π of π₯ with
respect to π₯. Remember, this is because even
functions have reflection or symmetry about the π¦-axis. Now, weβre told in the question
that this definite integral is equal to 13. Then to evaluate 19 over two minus
13, we write 13 as 26 over two. So weβre looking to find 19 over
two minus 26 over two which is negative seven over two. And so, we found the definite
integral between negative eight and negative four of π of π₯ with respect to
π₯. Itβs negative seven over two.