# Video: Using Properties of Integration by Substitution

If π(π₯) is continuous for all real values of π₯, what is the value of β«_(π/2)^(π/2) π(2π₯) dπ₯? [A] 2 β«_(π)^(π) π(π₯) dπ₯ [B] (1/2) β«_(2π)^(2π) π(π₯) dπ₯ [C] β«_(π)^(π) π(π₯) dπ₯ [D] (1/2) β«_(π)^(π) π(π₯) dπ₯

04:17

### 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. Letβs consider both methods. Using integration by substitution, first of all then, weβre going to introduce a new variable. And weβre going to call it π’. Weβll 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 itβs 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, thereβs 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, letβs 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’. Theyβre all in terms of π₯. As this is a definite integral, the variable that weβre using makes no difference because weβre going to substitute values for this variable once weβve 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, thatβs 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 weβre 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. Thatβs 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₯, weβre now looking at this area here. Thatβs 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₯.