# Video: AP Calculus AB Exam 1 β’ Section I β’ Part A β’ Question 10

Let π be a function given by π(π₯) = β«_(0) ^(π₯) (βπ^π‘Β²) dπ‘. Which of the following must be true on the interval 0 < π₯ < 2? [A] π(π₯) > 0, decreasing, and concave up [B] π(π₯) < 0, decreasing, and concave down [C] π(π₯) < 0, decreasing, and concave up [D] π(π₯) > 0, decreasing, and concave down.

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### Video Transcript

Let π be a function given by π of π₯ is equal to the integral between zero and π₯ of negative π to the π‘ squared with respect to π‘. Which of the following must be true on the interval where π₯ is greater than zero but less than two. a) π of π₯ is greater than zero, decreasing, and concave up. b) π of π₯ is less than zero, decreasing, and concave down. c) π of π₯ is less than zero, decreasing, and concave up. And d) π of π₯ is greater than zero, decreasing, and concave down.

Letβs see if we can find any further information about π of π₯ and negative π to the π‘ squared. In order to do this, we will be using the fundamental theorem of calculus, which tells us that the integral between π and π of π prime of π‘ with respect to π‘ is equal to π of π minus π of π. If we let negative π to the π‘ squared be equal to π prime of π‘ and π equal to zero and π equal to π₯, then we can say that the integral between zero and π₯ of π prime of π‘ with respect to π‘ is equal to π of π₯ minus π of zero.

However, this must also be equal to our integral from the question since we defined π prime of π‘ as negative π to the π‘ squared. And we know that our integral in the question is also equal to π of π₯. And so here, we have formed a relationship between π of π₯ and π of π₯. We have that π of π₯ minus π of nought is equal to π of π₯. We can differentiate both sides of this equation with respect to π₯ since π of nought is a constant that will go to zero. And weβre left with π prime of π₯ is equal to π prime of π₯. However, weβve defined π prime of π‘ as negative π to the π‘ squared. Therefore, we can say that π prime of π₯ is equal to negative π to the π₯ squared.

Now π prime of π₯ is the derivative of π with respect to π₯, which tells us the slope of π. Now, π to the π₯ squared is an exponential. And we will have a positive value for any value of π₯. Therefore, negative π to the π₯ squared must be negative for all values of π₯. This means that the slope of π is negative for all values of π₯. So we can say that π is decreasing.

Now that we found the first derivative of π, letβs consider the second derivative of π since this will tell us about the concavity of π. Second derivative of π, also known as π double prime of π₯, itβs equal to the derivative of π prime of π₯ with respect to π₯. And we know that π prime of π₯ is equal to negative π to the π₯ squared. So we simply need to differentiate this function. Now, this is a compound function. We have a function of π₯ squared within an exponential. Therefore, we must use the chain rule in order to differentiate it.

The chain rule tells us that dβ by dπ₯ is equal to dβ by dπ’ times dπ’ by dπ₯. In our case, we can let β be equal to negative π to the π’ and π’ be equal to π₯ squared. Therefore, when we combine these two functions, we will arrive back at our original function. Applying the chain rule, we obtained that d by dπ₯ of negative π to the π₯ squared is equal to d by dπ’ of negative π to the π’ times d by dπ₯ of π₯ squared. The first term here is exponential. So when we differentiate it, nothing will happen, giving us negative π to the π’.

The second term here is π₯ squared. And we can find its derivative by using the power rule for differentiation. We multiply by the power and decrease the power by one, giving us two π₯. Now, we can simply substitute back in π’ is equal to π₯ squared. And we found that π double prime of π₯ is equal to negative two π₯π to the π₯ squared. And weβre ready to consider the concavity of π over the integral between zero and two. On this integral, since π₯ is strictly greater than zero and strictly less than two, π₯ will always have a positive value. And since π₯ is positive, this means that π to the π₯ squared will also be positive. However, negative two π₯ will be negative.

Therefore, we have a negative number multiplied by a positive number. And we can say that π double prime of π₯ must be negative over our integral. When π double prime of π₯ is negative, this means that our function is concave down. So now, weβve found that π is decreasing and concave down. All we need to do is find if π of π₯ is greater than or less than zero. In order to do this, letβs consider the graph of negative π to the π₯ squared.

This is what our graph of negative π to the π₯ squared will look like. We have an exponential curve, which has been flipped in the π₯-axis, due to the negative sign. Now, letβs consider how this graph represents π of π₯. Weβre mainly concerned with the values of π between the π₯-values of zero and two. Now, π is the integral between zero and π₯ of negative π to the π‘ squared with respect to π‘. Therefore, it is represented by the area between our curve and the π₯-axis between the π₯-values of zero and π₯.

The π₯-values which weβre interested in are such that π₯ is greater than zero and less than two. As we can see that between the values of zero and two, this area is always below the π₯-axis. And since itβs below the π₯-axis, that means that our integral will evaluate to a negative number. And since our integral is equal to π of π₯, this means we can say that π of π₯ is less than zero. Therefore, we found all the information we need to know about π. π is decreasing. π is concave down. And π of π₯ is less than zero.

And so our solution to this question must be option b.