Question Video: Finding the Definite Integration of an Odd Function Using the Properties of Definite Integration | Nagwa Question Video: Finding the Definite Integration of an Odd Function Using the Properties of Definite Integration | Nagwa

Question Video: Finding the Definite Integration of an Odd Function Using the Properties of Definite Integration Mathematics • Second Year of Secondary School

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The following figure represents the graph of the function 𝑓(π‘₯) = π‘₯Β². What does the graph suggest about the value of the lim_(π‘₯ β†’ 2) 𝑓(π‘₯).

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Video Transcript

The following figure represents the graph of the function 𝑓 of π‘₯ is equal to π‘₯ squared. What does the graph suggest about the value of the limit as π‘₯ approaches two of 𝑓 of π‘₯.

For this question, we have been given a function. And we’ve been asked to evaluate a limit of said function. To interpret this, let us recall the general form of a limit. The limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ is equal to 𝐿. What this statement tells us is that the value of 𝑓 of π‘₯ will approach 𝐿 as the value of π‘₯ approaches π‘Ž from both sides. But remember, we’re not concerned with the point where π‘₯ is actually equal to π‘Ž. Let us now apply this statement to our question.

The limit as π‘₯ approaches two of 𝑓 of π‘₯ is equal to some value. And we’ll call this value 𝐿 one. This 𝐿 one is the thing we need to find. What the statement is telling us is that the value of 𝑓 of π‘₯ approaches 𝐿 one as the value of π‘₯ approaches two from both sides. And remember, this doesn’t necessarily have to be the same as the value of our function when π‘₯ is equal to two. To find the value of 𝐿 one, we can see what happens to the value of our function as we get closer and closer to π‘₯ equals two. We’ll start by considering some value of π‘₯ which is slightly smaller than two.

Let’s say π‘₯ equals 1.5. In this case, 𝑓 of π‘₯ equals 2.25. Since we know that 𝑓 of π‘₯ equals π‘₯ squared, we could even verify our graphical findings by taking 1.5 squared. Let us now increase our value of π‘₯. So we’re approaching π‘₯ equals two from the left, so when π‘₯ is less than two. As we do this, we might begin to notice that the value of 𝑓 of π‘₯ appears to be approaching four. If we were to follow a similar process, starting with the value of π‘₯ which is slightly larger than two and then we were to approach π‘₯ equals two from the right. We would notice that the values of 𝑓 of π‘₯ also approach four.

Without really going into any calculations, we see that as π‘₯ gets closer and closer to two, 𝑓 of π‘₯ gets closer and closer to four. This is true if we approach from the left or from the right, so from both sides. Essentially, on our graph, we’re converging on the point two, four. Okay, so we have just shown that the value of 𝑓 of π‘₯ approaches four as the value of π‘₯ approaches two. Since this statement defines our limit, we can use our finding to say that the value of the limit is also four. In doing this, we have answered our question. Using our graph, we observed that the value of 𝑓 of π‘₯ approached four as the value of π‘₯ approached two from both the left and the right. We use this to say that the limit as π‘₯ approaches two of 𝑓 of π‘₯ is equal to four.

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