AP Calculus AB — Riemann Sums, Trapezoidal Sums, & Average Values

The correct answer is (C). The interval we are computing the right Riemann sum on is $[0, 4]$. Since there are $4$ sub-intervals, we know that:

$Δx = \dfrac{4 − 0}{4} = 1$

The outputs for $0$, $1$, $2$, $3$, $4$ are shown in the table below:



We can take the sum of the values on the right of each sub-interval, or the sum of the outputs of $1$,$2$,$3$, and $4$. This sum is $6 + 21 + 86 + 261 = 374$.

The correct answer is (B). The interval we are computing the left Riemann sum on is $\left[0, \frac{3π}{2}\right]$. Since there are $3$ sub-intervals, we know that:

$Δx = \dfrac{\frac{3π}{2} − 0}{3} = \dfrac{π}{2}$

The outputs for $0$, $\frac{π}{2}$, $π$, $\frac{3π}{2}$ are shown in the table below:



We can take the sum of the values on the left of each sub-interval, or the sum of the outputs of $0$, $\frac{π}{2}$, and $π$. This sum is $1 + 0 + (−1) = 0$.

The correct answer is (D). Since there are $2$ sub-intervals:

$Δx = \dfrac{4 − 0}{2} = 2$

The table below shows the outputs for the function at $0$, $1$, $2$, $3$, and $4$.



The midpoint of the interval $[0, 2]$ is $1$, and the midpoint of $[2, 4]$ is $3$. The sum of the outputs of $1$ and $3$ is $12$. Since $Δx$ is $2$, the final answer is $2 \cdot 12 = 24$.

The correct answer is (B). The right Riemann sum consists of the outputs at the end of each interval of length $1$. This is the sum of the outputs at $1$, $2$, $3$, and $4$. Therefore the Riemann sum is $7 + 1 + 6 + 8 = 22$.

The correct answer is (A). The left Riemann sum consists of the outputs at the beginning of each interval of length $1$. This is the sum of the outputs of $0$, $1$, $2$, and $3$. Therefore the left Riemann sum is $4 + 8 + 5 + 3 = 20$.

The correct answer is (B). The three intervals are $[0, 2]$, $[2, 4]$, $[4, 6]$. The corresponding outputs for the midpoint of each interval are $8$, $3$, and $4$. The sum of these three outputs, times the length of each interval, is $2(8 + 3 + 4) = 30$.

The correct answer is (C). The table below shows the outputs of the function at $0$, $1$, $2$, $3$, $4$, $5$, and $6$.



The general trapezoidal sum formula for equal sub-intervals is as follows:

$\dfrac{Δx}{2} [f(a) + 2f(a + Δx)~+ $ $2f(a + 2Δx) + ⋯ + f(b)]$

Plugging in the values from the table into the trapezoidal sum formula:

$\dfrac{1}{2} [0 + 2(3) + 2(12) + 2(27)~+ $ $2(48) + 2(75) + 108] = 219$

The correct answer is (A). The table below shows the outputs of the function at $0$, $1$, $2$, $3$ and $4$.



The general trapezoidal sum formula for equal sub-intervals is as follows:

$\dfrac{Δx}{2} [f(a) + 2f(a + Δx)~+$ $2f(a + 2Δx) + ⋯ + f(b)]$

Plugging in the values from the table into the trapezoidal sum formula:

$\dfrac{1}{2} [0 + 2(0) + 2(56)~+$ $2(468) + 1008] = 766$

The correct answer is (A). The value to be evaluated is:

$\dfrac{1}{5−0} \displaystyle\int_0^5(x^3 + 2x^2)dx$

Solving gives the following definite integral:

$\dfrac{1}{5} \left[\dfrac{1}{4} x^4 + \dfrac{2}{3} x^3 \right]$ on the interval $[0, 5]$.

=$\dfrac{1}{5} \left[\dfrac{1}{4} (5)^4 + \dfrac{2}{3} (5)^3 \right]$ $− \dfrac{1}{5} \left[\dfrac{1}{4} (0)^4 + \dfrac{2}{3} (0)^3 \right]$

$= \dfrac{1}{5} \left[\dfrac{625}{4} + \dfrac{250}{3}\right]$ $= \dfrac{125}{4} + \dfrac{50}{3}$

$= \dfrac{375 + 200}{7}$ $= \dfrac{575}{12}$

The correct answer is (C). The integral to be evaluated is:

$\dfrac{1}{3 − 0} \displaystyle\int_0^3(e^{2x} + 5)dx$

Using a simple $u$-substitution and the power rule, the integral is equal to:

$\dfrac{1}{3} \left(\dfrac{1}{2} e^{2x} + 5x \right) + C$

Evaluating the definite integral on the interval $[0, 3]$:

$\dfrac{1}{3} \left[\dfrac{1}{2} e^{2(3)} + 5(3) \right]$ $−\dfrac{1}{3} \left[\dfrac{1}{2} e^{2(0)} + 5(0) \right]$

$= \dfrac{1}{6} e^6 + 5 − \dfrac{1}{6}$

$= \dfrac{1}{6} e^6 + \dfrac{29}{6}$

The correct answer is (A). The integral to be evaluated is:

$\dfrac{1}{\frac{π}{2} −0} \displaystyle\int_0^{\frac{π}{2}} \sin⁡(x) \cos⁡(x)dx$

A simpler way to solve this integral is to rewrite the integrand as

$\frac{1}{2} \sin⁡(2x)$. $\dfrac{2}{π} \cdot \dfrac{1}{2} \displaystyle\int_0^{\frac{π}{2}} \sin⁡(2x)dx$

Performing a $u$-substitution:

$(−1) \cdot \dfrac{2}{π} \cdot {1}{2} \cdot{1}{2} \cdot \cos⁡(2x)$

Applying the first fundamental theorem over the given bounds:

$−\dfrac{1}{2π} [\cos⁡(π) − \cos⁡(0)]$

$= −\dfrac{1}{2π}(−1 − 1)$

$= \dfrac{1}{π}$

The correct answer is (D). The integral to be evaluated is:

$\dfrac{1}{\frac{π}{3} − 0}$ $\displaystyle\int_0^{\frac{π}{3}} \tan⁡(x)dx$

$= \dfrac{3}{π} [−\ln⁡|\cos⁡(x)| ]$

Applying the first fundamental theorem using the given bounds:

$\dfrac{3}{π} \left[−\ln ⁡\left| \cos \left(\dfrac{π}{3} \right) \right| − (−\ln |\cos(0)|) \right]$

$= \dfrac{3}{π} \left[−\ln⁡ \left(\dfrac{1}{2} \right) + \ln⁡(1) \right]$

$= \dfrac{3}{π} \left[\ln(1) − \ln\left(\dfrac{1}{2}\right)\right]$

Using the log rule $\ln(a) − \ln(b) = \ln\left(\dfrac{a}{b}\right)$ we have:

$= \dfrac{3}{π} \left[\ln\left(\dfrac{1}{\frac{1}{2}}\right)\right]$

$= \dfrac{3}{π} \ln(2)$