Video: Determining a Possible Graph of the Antiderivative π(π₯), Given the Graph of π(π₯) and Where g Is Defined as the Semi-Definite Integral of π

The graph of π is as shown. If π(π₯) = β«_(β2)^(π₯) π(π‘) dπ‘, π₯ > β2, which of the following is a possible graph of π? [A] Graph A [B] Graph B [C] Graph C [D] Graph D

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Video Transcript

The graph of π is as shown. If π of π₯ is the integral between negative two and π₯ of π of π‘ with respect to π‘ for π₯ greater than negative two, which of the following is a possible graph of π? (a), (b), (c), or (d).

Weβre asked to determine which of the four graphs (a), (b), (c), or (d) is a possible graph of the function π of π₯. And all we know about our function π of π₯ is that this is the integral between negative two and π₯ of the function π of π‘ for π₯ greater than negative two. And weβre given the graph of the function π, which is the integrand.

The first step to helping us decide which graph could represent π is to note the similarity between our integral for π and the version of the fundamental theorem of calculus. Which says that if π is continuous on the closed interval π, π and π of π₯ is defined as the integral between π and π₯ of π of π‘ with respect to π‘. Then the derivative of π with respect to π₯, thatβs π prime of π₯, is equal to π of π₯ for all π₯ in the open interval π, π.

Our function π of π₯ is almost identical to the π of π₯ in this formulation of the fundamental theorem of calculus, although notice that in our case the lower limit of integration is negative two. And a dummy variable π‘ helps us evaluate our limit. So by the fundamental theorem of calculus, the derivative of π with respect to π₯ β thatβs π prime of π₯, the slope of π β is the function π of π₯. And we have a graph of π of π₯. That is, we have a graph of the slope of π. So all we need to do is to match the slope of one of the possible graphs (a), (b), (c), or (d) with a graph of π, and then weβll have our solution.

The easiest way to compare is if we split our graph π into sections. We can define these sections by intervals of π₯. So letβs look first at how our graph π behaves between π₯ is negative two and π₯ is zero. Between π₯ is negative two and zero, the graph of π is all above the π₯-axis so that π is positive between these two π₯-values. This means that the slope or gradient of π is also positive between π₯ is negative two and zero. So letβs look at our four options and see which of these has a positive gradient between π₯ is negative two and zero.

In graph (a), the slope is positive between π₯ is negative two and π₯ is equal to zero. That is, as π₯ increases from left to right, the value of the function also increases. And this matches with our slope function π. And so far graph (a) is a possibility for our function π. If we look at graph (b) now between π₯ is negative two and zero, the slope of graph (b) also seems to be positive. So graph (b) could also be a possibility for our function π. If we look now at graph (c) between π₯ is negative two and zero, the slope is either negative or possibly zero. That is, as π₯ increases from left to right, the value of the function decreases. So we can eliminate graph (c) from our inquiries.

And now, if we look at graph (d) between π₯ is negative two and zero, we see again that the slope is negative. So we can discount graph (d) from our inquiries. So the gradient or slopes of graphs (c) and (d) are negative between π₯ is negative two and zero, which does not match our function π, which is the gradient of π. And weβre left with two graphs (a) and (b) as possible options for our function π.

So now, letβs look at our function π between π₯ is zero and positive two, remembering that π is the slope or gradient of the function π. Between π₯ is zero and positive two, our slope function π is below the π₯-axis. That is, all the values of π are negative between π₯ is zero and two. This means that our slope or gradient must be negative between π₯ is zero and two. So letβs look again at graph (a). Between π₯ is zero and π₯ is two, the slope or gradient of the function in graph (a) is negative. That is, as π₯ increases from zero to two, the function decreases. So the slope of graph (a) between π₯ is zero and π₯ is two does much our slope function π.

So now, letβs look again at graph (b). Between π₯ is zero and two, the slope or gradient of the function in graph (b) is positive. And this doesnβt match with our slope function π, which is negative between π₯ is zero and two. So now, we can eliminate graph (b) since this does not match with our slope function π between π₯ is zero and two. Only one possibility remains, and thatβs graph (a). So if π of π₯ is the integral between negative two and π₯ of π of π‘ with respect to π‘ where π is the function shown, the possible graph of the function π is graph (a).