Video: AP Calculus AB Exam 1 β€’ Section I β€’ Part A β€’ Question 27

If the integral from zero to π‘Ž 𝑔(π‘₯) dπ‘₯ = the integral of negative π‘Ž to zero 𝑔(π‘₯) dπ‘₯ for all positive values of π‘Ž, which of the following could be the graph of 𝑔?

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Video Transcript

If the integral from zero to π‘Ž of 𝑔 of π‘₯ dπ‘₯ is equal to negative the integral of negative π‘Ž to zero of 𝑔 of π‘₯ dπ‘₯ for all positive values of π‘Ž, which of the following could be the graph of 𝑔?

Let’s recall first of all what an integral means in terms of an area. Let’s suppose arbitrarily that our function 𝑔 looks something like this. The integral from zero to π‘Ž of 𝑔 of π‘₯ with respect to π‘₯ gives the area enclosed by the lines 𝑔 of π‘₯, the π‘₯-axis, π‘₯ equals zero, and π‘₯ equals π‘Ž. That’s the area shaded in orange.

The integral from negative π‘Ž to zero of 𝑔 of π‘₯ with respect to π‘₯ will give the area enclosed by the curve 𝑔 of π‘₯, the π‘₯-axis, and the lines π‘₯ equals negative π‘Ž and π‘₯ equals zero. That’s the area shaded in pink.

We’re told that the first area is equal to negative the second area, which means that the two areas have the same value. But they have different signs. This tells us that one area will be above the π‘₯-axis. And the other will be below the π‘₯-axis because areas above the π‘₯-axis give positive values. And areas below the π‘₯-axis, found using integration, give negative values.

As the statement is true for all positive values of π‘Ž, this means that any area to the right of the 𝑦-axis, that’s for positive values of π‘₯, must be the same as the corresponding area to the left of the 𝑦-axis. That’s for negative values of π‘₯. The size of the two areas will be the same. But one will be above the π‘₯-axis. And the other will be below the π‘₯-axis.

This tells us something important about the function 𝑔. It tells us that the function 𝑔 has rotational symmetry about the origin, which means if we were to rotate its graph 180 degrees about the origin, it would look exactly the same. This, in turn, tells us that 𝑔 is an odd function.

So to determine which of the four graphs could be the graph of 𝑔, we just need to determine which is an odd function. Well, graph A is not. We can see that it doesn’t have rotational symmetry about the origin. If we were to look at the areas represented by these two integrals, we’d see that they’re both below the π‘₯-axis. And therefore, they both have negative signs. So it wouldn’t be possible for the orange area to be the negative of the pink area. So graph A is out.

In the same way, we can rule out graphs C and D because, in each case, we can see that the two areas, although they have the same size, have the same sign. They’re on the same side of the π‘₯-axis. So again, it is impossible for the orange area to be the negative of the pink area.

If we look at our final graph then, graph B, we can see that this is indeed an odd function. It has rotational symmetry about the origin. This looks like it could potentially be a graph of a cubic such as 𝑦 equals negative π‘₯ cubed or 𝑦 equals negative four π‘₯ cubed. We can see that the pink area and the orange area are the same size. But they’re on opposite sides of the π‘₯-axis.

The pink area will have a positive value. And the orange area will have a negative value. But as they’re the same size, the orange area will be the negative of the pink area. Therefore, in graph B, it is true that the integral from zero to π‘Ž of 𝑔 of π‘₯ with respect to π‘₯ is equal to negative the integral from negative π‘Ž to zero of 𝑔 of π‘₯ with respect to π‘₯. So graph B is our answer.

Note that the question said β€œwhich of the following could be the graph of 𝑔?” because, of course, there are many other possibilities for the graph of this function. However, B is the only one of the four options that we’ve been given which is possible.

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