Video: AP Calculus AB Exam 1 β€’ Section I β€’ Part A β€’ Question 30

If 𝑓(π‘₯) is continuous, 𝑓(π‘₯) > 0 on [βˆ’2, 3], and 𝑓(π‘₯) is twice differentiable on (βˆ’2, 3) such that 𝑓′(π‘₯) < 0 and 𝑓″(π‘₯) > 0, which of the following has the greatest value? [A] Left Riemann sum approximation of ∫_βˆ’2^3 𝑓(π‘₯) dπ‘₯ with 5 subintervals of equal length. [B] Midpoint Riemann sum approximation of ∫_βˆ’2^3 𝑓(π‘₯) dπ‘₯ with 5 subintervals of equal length. [C] Right Riemann sum approximation of ∫_βˆ’2^3 𝑓(π‘₯) dπ‘₯ with 5 subintervals of equal length. [D] The trapezoidal sum approximation of ∫_βˆ’2^3 𝑓(π‘₯) dπ‘₯ with 5 subintervals of equal length.

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Video Transcript

If 𝑓 of π‘₯ is continuous, 𝑓 of π‘₯ is greater than zero on the interval between negative two and three, and 𝑓 of π‘₯ is twice differentiable on the interval between negative two and three such that 𝑓 prime of π‘₯ is less than zero and 𝑓 double prime of π‘₯ is greater than zero, which of the following has the greatest value? a) Left Riemann sum approximation of the integral from negative two to three of 𝑓 of π‘₯ with respect to π‘₯ with five subintervals of equal length. b) Midpoint Riemann sum approximation of the integral from negative two to three of 𝑓 of π‘₯ with respect to π‘₯ with five subintervals of equal length. c) Right Riemann sum approximation of the integral from negative two to three of 𝑓 of π‘₯ with respect to π‘₯ with five subintervals of equal length. Or d) The trapezoidal sum approximation of the integral from negative two to three of 𝑓 of π‘₯ with respect to π‘₯ with five subintervals of equal length.

Now, this is a lot of information to digest. Let’s break it down so we can make sense of the question. We’re told that 𝑓 of π‘₯ is continuous and is strictly greater than zero on the interval between negative two and three. This means that there are no discontinuities in 𝑓 of π‘₯. And for every π‘₯-value between negative two and three, which includes negative two and three, since we have square brackets here, 𝑓 of π‘₯ will be positive.

We’re told that 𝑓 of π‘₯ is twice differentiable on the interval between negative two and three with 𝑓 prime of π‘₯ being less than zero. 𝑓 prime of π‘₯ is the derivative of 𝑓 with respect to π‘₯. This also represents the gradient of 𝑓. Now, 𝑓 prime of π‘₯ is strictly less than zero. This means that 𝑓 of π‘₯ is a strictly decreasing function. Therefore, as π‘₯ increases, 𝑓 of π‘₯ decreases. We’re also told that 𝑓 double prime of π‘₯ is greater than zero. This means that the second differential of 𝑓 with respect to π‘₯ is always gonna be positive. This in fact tells us about the concavity of π‘₯. Since 𝑓 double prime of π‘₯ is greater than zero, this means that 𝑓 must be concave upwards.

Now that we have deduced a few things about 𝑓, let’s draw a quick sketch of what 𝑓 may look like. There are a few things which we must make sure that we have in our sketch. That is, that 𝑓 is continuous, 𝑓 is positive on all values between negative two and three, 𝑓 is strictly decreasing, and 𝑓 is concave upwards. Our curve clearly shows all of these properties. Let’s now consider what each of our options will look like on this graph of 𝑓. Let’s start by comparing the left Riemann sum approximation with the midpoint Riemann sum approximation.

Each of our four options uses five subintervals of equal length. Therefore, for each option, the intervals will be between negative two and negative one, between negative one and zero, between zero and one, between one and two, and finally between two and three. So these are the five subintervals we will be considering. For the left Riemann sum, we consider the left value on each interval. This gives us the height of each of the rectangles we need to find the area of. The width of each rectangle is given by that five equal subintervals which we’ve just found.

Therefore, the left Riemann sum approximation will look like this. Now, we can add in what the midpoint Riemann sum approximation will look like. This works in a similar way to the left Riemann sum. Except instead of using the left point of each interval as the height of each rectangle, it instead uses the value of the midpoint of the interval. So that’s these five points, as shown on the graph. Let’s now draw on what the midpoint Riemann sum approximation will look like.

We can clearly see from our sketch that the left Riemann sum is larger than the midpoint Riemann sum. We can see that each rectangle in the left Riemann sum is slightly larger than each rectangle in the midpoint Riemann sum. Therefore, we can deduce that option b the midpoint Riemann sum is not the correct answer here. Next, we’ll compare the right Riemann sum approximation with the left Riemann sum approximation.

Now, the right Riemann sum approximation is very similar to the left Riemann sum approximation. Except instead of using the value to the left of the interval in order to find the height of the rectangles, we instead use the value on the right of the interval. So that is these values here. Now, we can draw on the rectangles which we would use to find the right Riemann sum approximation. As we can see from our sketch, each of the rectangles in the left Riemann sum approximation is larger than their equivalent right Riemann sum approximations by each of these amounts, shown on the graph. So the left Riemann sum approximation must have a greater value than the right Riemann sum approximation. And therefore, our answer cannot be option c.

Next, let’s compare the left Riemann sum approximation with the trapezoidal sum approximation. In order to use the trapezoidal sum approximation, we form trapezoids where the vertices are the endpoint of each of the subintervals on the π‘₯-axis and the corresponding 𝑓 of π‘₯ values for each of these endpoints. This is what our trapezoids will look like. And thus, we can see from our sketch each of the rectangles in the left Riemann sum approximation are larger than their corresponding trapezoids from the trapezoidal sum approximation, which we can see from each of the shaded regions. Therefore, we can say that the value of the left Riemann sum approximation must be greater than the value of the trapezoidal sum approximation. And so, we can eliminate d as our final answer.

This leaves us with only one option. And that is that our answer must be a. That is that the left Riemann sum approximation of the integral from negative two to three of 𝑓 of π‘₯ with respect to π‘₯ with five subintervals of equal length has the greatest value of these four options.

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