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

Let 𝑓 be a function given by 𝑓(π‘₯) = ∫_(0) ^(π‘₯) (βˆ’π‘’^𝑑²) d𝑑. Which of the following must be true on the interval 0 < π‘₯ < 2? [A] 𝑓(π‘₯) > 0, decreasing, and concave up [B] 𝑓(π‘₯) < 0, decreasing, and concave down [C] 𝑓(π‘₯) < 0, decreasing, and concave up [D] 𝑓(π‘₯) > 0, decreasing, and concave down.

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Video Transcript

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

Let’s see if we can find any further information about 𝑓 of π‘₯ and negative 𝑒 to the 𝑑 squared. In order to do this, we will be using the fundamental theorem of calculus, which tells us that the integral between π‘Ž and 𝑏 of 𝑔 prime of 𝑑 with respect to 𝑑 is equal to 𝑔 of 𝑏 minus 𝑔 of π‘Ž. If we let negative 𝑒 to the 𝑑 squared be equal to 𝑔 prime of 𝑑 and π‘Ž equal to zero and 𝑏 equal to π‘₯, then we can say that the integral between zero and π‘₯ of 𝑔 prime of 𝑑 with respect to 𝑑 is equal to 𝑔 of π‘₯ minus 𝑔 of zero.

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

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

Now that we found the first derivative of 𝑓, let’s consider the second derivative of 𝑓 since this will tell us about the concavity of 𝑓. Second derivative of 𝑓, also known as 𝑓 double prime of π‘₯, it’s equal to the derivative of 𝑓 prime of π‘₯ with respect to π‘₯. And we know that 𝑓 prime of π‘₯ is equal to negative 𝑒 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„Ž by dπ‘₯ is equal to dβ„Ž by d𝑒 times d𝑒 by dπ‘₯. In our case, we can let β„Ž be equal to negative 𝑒 to the 𝑒 and 𝑒 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 𝑒 to the π‘₯ squared is equal to d by d𝑒 of negative 𝑒 to the 𝑒 times d by dπ‘₯ of π‘₯ squared. The first term here is exponential. So when we differentiate it, nothing will happen, giving us negative 𝑒 to the 𝑒.

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 𝑒 is equal to π‘₯ squared. And we found that 𝑓 double prime of π‘₯ is equal to negative two π‘₯𝑒 to the π‘₯ squared. And we’re ready to consider the concavity of 𝑓 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 𝑒 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 𝑓 double prime of π‘₯ must be negative over our integral. When 𝑓 double prime of π‘₯ is negative, this means that our function is concave down. So now, we’ve found that 𝑓 is decreasing and concave down. All we need to do is find if 𝑓 of π‘₯ is greater than or less than zero. In order to do this, let’s consider the graph of negative 𝑒 to the π‘₯ squared.

This is what our graph of negative 𝑒 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, let’s consider how this graph represents 𝑓 of π‘₯. We’re mainly concerned with the values of 𝑓 between the π‘₯-values of zero and two. Now, 𝑓 is the integral between zero and π‘₯ of negative 𝑒 to the 𝑑 squared with respect to 𝑑. 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 we’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 it’s below the π‘₯-axis, that means that our integral will evaluate to a negative number. And since our integral is equal to 𝑓 of π‘₯, this means we can say that 𝑓 of π‘₯ is less than zero. Therefore, we found all the information we need to know about 𝑓. 𝑓 is decreasing. 𝑓 is concave down. And 𝑓 of π‘₯ is less than zero.

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

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