### Video Transcript

If π of π₯ is continuous for
all real values of π₯, what is the value of the integral from π over two to π
over two of π of two π₯ with respect to π₯? a) Is it two multiplied by the
integral from π to π of π of π₯ with respect to π₯. b) A half multiplied by
the integral from two π to two π of π of π₯ with respect to π₯. c) The
integral from π to π of π of π₯ with respect π₯. Or d, a half the integral from
π to π of π of π₯ with respect to π₯.

Now, there are two ways that we
can approach this question. One is a formal method using
integration by substitution. And the second is a geometric
approach. Letβs consider both
methods. Using integration by
substitution, first of all then, weβre going to introduce a new variable. And weβre going to call it
π’. Weβll let π’ be equal to two
π₯. This means that in our
integrand, π of two π₯ will be equal to π of π’. We also know, using
differentiation, that dπ’ by dπ₯ is equal to two. And although dπ’ by dπ₯ is
absolutely not a fraction, we can treat it a little like one. So itβs equivalent to say that
a half dπ’ is equal to dπ₯.

This means that we can replace
dπ₯ in our integral with a half dπ’. Now, thereβs one more thing we
must remember to do. This is a definite
integral. So we must remember to change
our limits, which are currently limits in terms of π₯ to limits in terms of
π’. Using the relation π’ equals
two π₯, we know that when π₯ equals π over two, π’ is equal to π. And when π₯ equals π over two,
π’ is equal to π. So the limits for our integral
in terms of π’ are π and π.

Now, letβs rewrite this
integral fully in terms of π’. The lower limit is π. The upper limit is π. π of two π₯ becomes π of π’,
and dπ₯ becomes a half dπ’. Although we can write this in
any order. The factor of a half is just a
multiplicative constant. So we can bring it out the
front of our integral. And we have a half multiplied
by the integral from π to π of π of π’ with respect to π’.

Now, none of the answers given
are in terms of π’. Theyβre all in terms of π₯. As this is a definite integral,
the variable that weβre using makes no difference because weβre going to
substitute values for this variable once weβve integrated. We can therefore change the
variable to anything we like. It could be π‘ or π¦ or, more
usefully, we could change it to π₯. And we have that the integral
is equal to a half the integral from π to π of π of π₯ with respect to
π₯. Which we see is option d in the
list we were given.

Now, thatβs the formal method,
using integration by substitution to show why these two integrals are
equivalent. The other method we could use
is to consider a geometric approach. Suppose we have the function π¦
equals π of π₯. And weβre integrating between
the values of π and π. Now, this will give the area
under the curve, π¦ equals π of π₯, between the vertical lines π₯ equals π and
π₯ equals π. Thatβs all this area shaded in
orange.

Using our knowledge of
transformations of graphs, we know that π¦ equals π of two π₯ will be a
horizontal stretch with a scale factor of one-half. When we find the integral from
π over two to π over two of π of two π₯ with respect to π₯, weβre now looking
at this area here. Thatβs the area below the
transformed graph between the limits of π over two and π over two. Because the graph has been
stretched by a scale factor of one-half. Or we can think of this as
being compressed by a factor of two. The second area will be half
the area we began with.

So we can say that the integral
from π over two to π over two of π of two π₯ with respect to π₯ is a half the
integral from π to π of π of π₯ with respect π₯. Which, again, is option d. So there are the two ways that
we could approach this problem. A more formal approach using
integration by substitution or a geometric explanation using transformation of
graphs. The correct answer is option
d. The integral is equal to
one-half the integral from π to π of π π₯ with respect to π₯.