Video: AP Calculus AB Exam 1 β€’ Section II β€’ Part B β€’ Question 2

Let 𝑓 be a continuous function on the interval [βˆ’4, 8], where 𝑓 is twice differentiable with 𝑓″(π‘₯) > 0 on the open interval (βˆ’4, 8). Selected values of 𝑓 are shown in the table. i) Using the trapezoidal rule approximation with five trapezoids, find the approximate value of ∫_βˆ’4^8 (𝑓(π‘₯) dπ‘₯). ii) Using the result in part (i), find the average value of 𝑓 on [βˆ’4, 8]. iii) Find the average rate of change of 𝑓 over the interval [βˆ’4, 8]. iv) Examine 𝑓′(0) and 𝑓′(4) and determine if 𝑓′(0) is greater than, smaller than, or equal to 𝑓′(4). Explain why.

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Video Transcript

Let 𝑓 be a continuous function on the closed interval negative four to eight, where 𝑓 is twice differentiable with 𝑓 prime prime of π‘₯ greater than zero on the open interval negative four to eight. Selected values of 𝑓 are shown in the table. i) Using the trapezoidal rule approximation with five trapezoids, find the approximate value of the integral evaluated between negative four and eight of 𝑓 of π‘₯ with respect to π‘₯. ii) Using the result in part one, find the average value of 𝑓 on the closed interval negative four to eight. iii) Find the average rate of change of 𝑓 over the closed interval negative four to eight. And iv) Examine 𝑓 prime of zero and 𝑓 prime of four and determine if 𝑓 prime of zero is greater than, smaller than, or equal to 𝑓 prime of four. Explain why.

We’ll begin with part one. The trapezoidal rule approximation is the technique that allows us to approximate the value of definite integrals. We split the region between the curve and the π‘₯-axis into trapezoids. And we evaluate the area of each of these. Now, usually, the trapezoids are of equal width. By looking at the π‘₯-values in our table though, we can see that our trapezoids are going to have unequal width.

The first trapezoid is from negative four to negative two. So it will have a width of two units. The second trapezoid is from negative two to zero. So that also has a width of two units. But the third trapezoid will have a width of one unit, the fourth a width of three units, and the fifth a width of four units. So rather than use the trapezoidal rule approximation that we usually do, we have an unsimplified version. We take the difference between two consecutive π‘₯-values. Remember this gives us the width of the trapezoid. And we multiply this by a half of the sum of the corresponding 𝑓 of π‘₯ values. We’re going to do this for each of the trapezoids that we’re interested in and find their sum.

The width of the first trapezoid is negative two minus negative four. And we multiply that by the corresponding 𝑓 of π‘₯ values divided by two. That’s 16 plus 10 divided by two which is 26. The width of our second trapezoid is zero minus negative two. And we multiply that by half of the sum of the corresponding 𝑓 of π‘₯ values. That’s six plus 10 divided by two. That’s 16. The width of our third trapezoid is one minus zero. And we’re going to multiply that by five plus six divided by two, which is 11 over two. We add four minus one times three plus five over two which is 12. And finally, we add eight minus four times two plus three over two, which is equal to 10. The sum total is therefore 139 over two.

So we can say that an approximation for the total area of our five trapezoids is 139 over two square units. And this means the approximate value of the integral evaluated between negative four and eight of 𝑓 of π‘₯ with respect to π‘₯ is 139 over two. For part two, we’re going to use the result from part one to find the average value of 𝑓 over our closed interval. The formula for the average value of a function over a closed interval π‘Ž to 𝑏 is one over 𝑏 minus π‘Ž times the integral evaluated between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯.

We can let π‘Ž be equal to negative four and 𝑏 be equal to eight. And remember we already evaluated this integral between negative four and eight and we got 139 over two. One over eight minus negative four is one over 12. So the average value of 𝑓 is given as a 12 times 139 over two. That’s 139 over 24.

Now part three, we’re looking to find the average rate of change of 𝑓 over the closed interval negative four to eight. Once again, we have a formula we can use. The average rate of change of a function over a closed interval π‘Ž to 𝑏 is given by 𝑓 of 𝑏 minus 𝑓 of π‘Ž over 𝑏 minus π‘Ž. Once again, we let π‘Ž be equal to negative four and 𝑏 be equal to eight. So this becomes 𝑓 of eight minus 𝑓 of negative four over eight minus negative four. Eight minus negative four is 12.

And we refer to our table for the values for 𝑓 of eight and 𝑓 of negative four. They’re two and 16, respectively. Two minus 16 is negative 14. And then, we simplify our fraction by dividing both the numerator and the denominator by two. And we see the average rate of change of 𝑓 over the closed interval negative four to eight is negative seven-sixths.

For part four, we need to determine whether 𝑓 prime of zero is greater than, smaller than, or equal to 𝑓 prime of four. Now, we don’t actually have values for 𝑓 prime of zero and 𝑓 prime of four. But we’re told that 𝑓 is twice differentiable and that 𝑓 prime prime of π‘₯ is greater than zero on the open interval negative four to eight. Well, let’s recall what this actually means.

𝑓 prime prime of π‘₯ is the second derivative. It’s the derivative of 𝑓 prime of π‘₯. And in this case, it’s strictly greater than zero. This means that 𝑓 prime of π‘₯ the derivative of our function must be strictly increasing. In other words, as our value of π‘₯ increases, so does our value of 𝑓 prime of π‘₯. And this tells us that 𝑓 prime of four must be strictly greater than 𝑓 prime of zero. It cannot be equal to it. And we see then that 𝑓 prime of zero must be smaller than 𝑓 prime of four.

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