Video: Using Properties of Integration by Substitution

If 𝑓(π‘₯) is continuous for all real values of π‘₯, what is the value of ∫_(π‘Ž/2)^(𝑏/2) 𝑓(2π‘₯) dπ‘₯? [A] 2 ∫_(π‘Ž)^(𝑏) 𝑓(π‘₯) dπ‘₯ [B] (1/2) ∫_(2π‘Ž)^(2𝑏) 𝑓(π‘₯) dπ‘₯ [C] ∫_(π‘Ž)^(𝑏) 𝑓(π‘₯) dπ‘₯ [D] (1/2) ∫_(π‘Ž)^(𝑏) 𝑓(π‘₯) dπ‘₯

04:17

Video Transcript

If 𝑓 of π‘₯ is continuous for all real values of π‘₯, what is the value of the integral from π‘Ž over two to 𝑏 over two of 𝑓 of two π‘₯ with respect to π‘₯? a) Is it two multiplied by the integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ with respect to π‘₯. b) A half multiplied by the integral from two π‘Ž to two 𝑏 of 𝑓 of π‘₯ with respect to π‘₯. c) The integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ with respect π‘₯. Or d, a half the integral from π‘Ž to 𝑏 of 𝑓 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. Let’s consider both methods. Using integration by substitution, first of all then, we’re going to introduce a new variable. And we’re going to call it 𝑒. We’ll let 𝑒 be equal to two π‘₯. This means that in our integrand, 𝑓 of two π‘₯ will be equal to 𝑓 of 𝑒. We also know, using differentiation, that d𝑒 by dπ‘₯ is equal to two. And although d𝑒 by dπ‘₯ is absolutely not a fraction, we can treat it a little like one. So it’s equivalent to say that a half d𝑒 is equal to dπ‘₯.

This means that we can replace dπ‘₯ in our integral with a half d𝑒. Now, there’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 𝑒. Using the relation 𝑒 equals two π‘₯, we know that when π‘₯ equals π‘Ž over two, 𝑒 is equal to π‘Ž. And when π‘₯ equals 𝑏 over two, 𝑒 is equal to 𝑏. So the limits for our integral in terms of 𝑒 are π‘Ž and 𝑏.

Now, let’s rewrite this integral fully in terms of 𝑒. The lower limit is π‘Ž. The upper limit is 𝑏. 𝑓 of two π‘₯ becomes 𝑓 of 𝑒, and dπ‘₯ becomes a half d𝑒. 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‘Ž to 𝑏 of 𝑓 of 𝑒 with respect to 𝑒.

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

Now, that’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 𝑦 equals 𝑓 of π‘₯. And we’re integrating between the values of π‘Ž and 𝑏. Now, this will give the area under the curve, 𝑦 equals 𝑓 of π‘₯, between the vertical lines π‘₯ equals π‘Ž and π‘₯ equals 𝑏. That’s all this area shaded in orange.

Using our knowledge of transformations of graphs, we know that 𝑦 equals 𝑓 of two π‘₯ will be a horizontal stretch with a scale factor of one-half. When we find the integral from π‘Ž over two to 𝑏 over two of 𝑓 of two π‘₯ with respect to π‘₯, we’re now looking at this area here. That’s the area below the transformed graph between the limits of π‘Ž over two and 𝑏 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‘Ž over two to 𝑏 over two of 𝑓 of two π‘₯ with respect to π‘₯ is a half the integral from π‘Ž to 𝑏 of 𝑓 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‘Ž to 𝑏 of 𝑓 π‘₯ with respect to π‘₯.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.