# Video: AP Calculus AB Exam 1 • Section I • Part A • Question 10

Let 𝑓 be a function given by 𝑓(𝑥) = ∫_(0) ^(𝑥) (−𝑒^𝑡²) d𝑡. Which of the following must be true on the interval 0 < 𝑥 < 2? [A] 𝑓(𝑥) > 0, decreasing, and concave up [B] 𝑓(𝑥) < 0, decreasing, and concave down [C] 𝑓(𝑥) < 0, decreasing, and concave up [D] 𝑓(𝑥) > 0, decreasing, and concave down.

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### Video Transcript

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

Let’s see if we can find any further information about 𝑓 of 𝑥 and negative 𝑒 to the 𝑡 squared. In order to do this, we will be using the fundamental theorem of calculus, which tells us that the integral between 𝑎 and 𝑏 of 𝑔 prime of 𝑡 with respect to 𝑡 is equal to 𝑔 of 𝑏 minus 𝑔 of 𝑎. If we let negative 𝑒 to the 𝑡 squared be equal to 𝑔 prime of 𝑡 and 𝑎 equal to zero and 𝑏 equal to 𝑥, then we can say that the integral between zero and 𝑥 of 𝑔 prime of 𝑡 with respect to 𝑡 is equal to 𝑔 of 𝑥 minus 𝑔 of zero.

However, this must also be equal to our integral from the question since we defined 𝑔 prime of 𝑡 as negative 𝑒 to the 𝑡 squared. And we know that our integral in the question is also equal to 𝑓 of 𝑥. And so here, we have formed a relationship between 𝑔 of 𝑥 and 𝑓 of 𝑥. We have that 𝑔 of 𝑥 minus 𝑔 of nought is equal to 𝑓 of 𝑥. We can differentiate both sides of this equation with respect to 𝑥 since 𝑔 of nought is a constant that will go to zero. And we’re left with 𝑔 prime of 𝑥 is equal to 𝑓 prime of 𝑥. However, we’ve defined 𝑔 prime of 𝑡 as negative 𝑒 to the 𝑡 squared. Therefore, we can say that 𝑓 prime of 𝑥 is equal to negative 𝑒 to the 𝑥 squared.

Now 𝑓 prime of 𝑥 is the derivative of 𝑓 with respect to 𝑥, which tells us the slope of 𝑓. Now, 𝑒 to the 𝑥 squared is an exponential. And we will have a positive value for any value of 𝑥. Therefore, negative 𝑒 to the 𝑥 squared must be negative for all values of 𝑥. This means that the slope of 𝑓 is negative for all values of 𝑥. 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 𝑥, it’s equal to the derivative of 𝑓 prime of 𝑥 with respect to 𝑥. And we know that 𝑓 prime of 𝑥 is equal to negative 𝑒 to the 𝑥 squared. So we simply need to differentiate this function. Now, this is a compound function. We have a function of 𝑥 squared within an exponential. Therefore, we must use the chain rule in order to differentiate it.

The chain rule tells us that dℎ by d𝑥 is equal to dℎ by d𝑢 times d𝑢 by d𝑥. In our case, we can let ℎ be equal to negative 𝑒 to the 𝑢 and 𝑢 be equal to 𝑥 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 d𝑥 of negative 𝑒 to the 𝑥 squared is equal to d by d𝑢 of negative 𝑒 to the 𝑢 times d by d𝑥 of 𝑥 squared. The first term here is exponential. So when we differentiate it, nothing will happen, giving us negative 𝑒 to the 𝑢.

The second term here is 𝑥 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 𝑥. Now, we can simply substitute back in 𝑢 is equal to 𝑥 squared. And we found that 𝑓 double prime of 𝑥 is equal to negative two 𝑥𝑒 to the 𝑥 squared. And we’re ready to consider the concavity of 𝑓 over the integral between zero and two. On this integral, since 𝑥 is strictly greater than zero and strictly less than two, 𝑥 will always have a positive value. And since 𝑥 is positive, this means that 𝑒 to the 𝑥 squared will also be positive. However, negative two 𝑥 will be negative.

Therefore, we have a negative number multiplied by a positive number. And we can say that 𝑓 double prime of 𝑥 must be negative over our integral. When 𝑓 double prime of 𝑥 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 𝑥 is greater than or less than zero. In order to do this, let’s consider the graph of negative 𝑒 to the 𝑥 squared.

This is what our graph of negative 𝑒 to the 𝑥 squared will look like. We have an exponential curve, which has been flipped in the 𝑥-axis, due to the negative sign. Now, let’s consider how this graph represents 𝑓 of 𝑥. We’re mainly concerned with the values of 𝑓 between the 𝑥-values of zero and two. Now, 𝑓 is the integral between zero and 𝑥 of negative 𝑒 to the 𝑡 squared with respect to 𝑡. Therefore, it is represented by the area between our curve and the 𝑥-axis between the 𝑥-values of zero and 𝑥.

The 𝑥-values which we’re interested in are such that 𝑥 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 𝑥-axis. And since it’s below the 𝑥-axis, that means that our integral will evaluate to a negative number. And since our integral is equal to 𝑓 of 𝑥, this means we can say that 𝑓 of 𝑥 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 𝑥 is less than zero.

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