AP Calculus BC 6.2: Riemann Sums

C Use a trapezoidal sum with the three subintervals indicated by the data in the table to approximate the

value of $\int_{0}^{10} R(t) d t$. Show the work that leads to your answer.

- Read "=" as " $\approx$ " for P5.

- The form of a trapezoidal sum includes three terms, each of which includes a product of two factors,

where one of the factors incorporates the $\frac{1}{2}$ as part of the product. To earn P5, at least five of the six

factors must be correct. If any of the six factors is incorrect, the response does not earn P6. Consider

the following examples:

○ $\frac{90+100}{2}(2-0)+\frac{100+150}{2}(8-2)+\frac{150+162}{2}(10-8)$ earns P5 and is sufficient to earn P6.

○ $\frac{190}{2}(2)+\frac{250}{2}(6)+\frac{312}{2}(2)$ earns P5 and is sufficient to earn P6.

$\circ \quad \frac{1}{2}((R(0)+R(2))(2)+(R(2)+R(8))(6)+(R(8)+R(10))(2))$ earns P5 and is eligible for P6.

○ $\frac{90+100}{2}(2)+\frac{100+150}{2}(2)+\frac{150+162}{2}(2)$ earns P5 but is not eligible for P6.

(Note that the factor of 2 in the second term of this expression is incorrect.)

- Special case: A response of (90+100)+(100+150) 3+(150+162) earns both P5 and P6.

- To be eligible for P6, a response must have earned P5.

Special case: A response of $95 \cdot 2+125 \cdot 6+156 \cdot 2$ earns P6 but does not earn P5.

- A response of $\frac{90+100}{2}(2-0)+\frac{100+150}{2}(8-2)+\frac{150+162}{2}(10-8)$ or equivalent banks P6

(i.e., subsequent errors in simplification will not be considered in scoring for P6).

- A response of $\frac{(90 \cdot 2+100 \cdot 6+150 \cdot 2)+(100 \cdot 2+150 \cdot 6+162 \cdot 2)}{2}$ or equivalent earns both P5

and P6. (Note that the average of the left Riemann sum and right Riemann sum is equivalent to the

trapezoidal sum.)

- A completely correct left Riemann sum (e.g., $90 \cdot 2+100 \cdot 6+150 \cdot 2=1080$ ) or a completely

correct right Riemann sum (e.g., $100 \cdot 2+150 \cdot 6+162 \cdot 2=1424$ ) earns P5 but does not earn P6.