AP Calculus BC Quiz: Riemann Sums And Notation

What this quiz covers

This quiz focuses on Riemann Sums And Notation, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\displaystyle \int_{0}^{5} (2-5x)^3\,dx$
2. $\displaystyle \int_{-3}^{2} x^3\,dx$
3. $\displaystyle \int_{0}^{5} (2-x)^3\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{5} (2-x)^3\,5\,dx$
5. $\displaystyle \int_{0}^{5} \left(2-\frac{5x}{n}\right)^3\frac{5}{n}\,dx$

Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The (\frac{5}{n}) acts as (\Delta x), indicating an interval from 0 to 5. The cubed term (2 - \frac{5i}{n}) corresponds to (f(x_i) = (2 - x_i)^3) with (x_i = \frac{5i}{n}). Consequently, the Riemann sum approximates (\int_{0}^{5} (2 - x)^3 , dx). A distractor like choice D, (\int_{0}^{5} (2 - x)^3 \cdot 5 , dx), fails by erroneously including the 5 inside the integrand rather than as part of the differential. To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).

Question 2

1. $\displaystyle \int_{0}^{1} x\cos(2x)\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{2} x\cos(2x)\,dx$
3. $\displaystyle \int_{0}^{1} x\cos(x)\,dx$
4. $\displaystyle \int_{0}^{1} nx\cos(2x)\,dx$
5. $\displaystyle \int_{0}^{1} \left(\frac{x}{n}\right)\cos\!\left(2\frac{x}{n}\right)\,dx$

Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The (\frac{1}{n}) is (\Delta x), pointing to an interval from 0 to 1. The product (\frac{i}{n} \cos(2 \frac{i}{n})) maps to (f(x_i) = x_i \cos(2 x_i)) where (x_i = \frac{i}{n}). The sum thus represents (\int_{0}^{1} x \cos(2x) , dx). Choice B, (\int_{0}^{2} x \cos(2x) , dx), might tempt by extending the upper limit to 2 but fails as the interval is determined by (\Delta x = \frac{1}{n}), not (\frac{2}{n}). To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).

Question 3

1. $\displaystyle \int_{4}^{6} \ln x\,dx$
2. $\displaystyle \int_{0}^{2} \ln(4+x)\,dx$ (correct answer)
3. $\displaystyle \int_{0}^{2} \ln(4+2x)\,dx$
4. $\displaystyle \int_{0}^{2} \ln(4+x)\,2\,dx$
5. $\displaystyle \int_{1}^{n} \ln\!\left(4+\frac{2x}{n}\right)\frac{2}{n}\,dx$

Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The (\frac{2}{n}) is (\Delta x), so the interval is from 0 to 2. The logarithm (\ln(4 + \frac{2i}{n})) matches (f(x_i) = \ln(4 + x_i)) where (x_i = \frac{2i}{n}). This sum therefore converges to (\int_{0}^{2} \ln(4 + x) , dx). Choice D, (\int_{0}^{2} \ln(4 + x) \cdot 2 , dx), is tempting but incorrect as it multiplies the integrand by 2 instead of using it in the dx term. To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).

Question 4

1. $\displaystyle \int_{0}^{1} (5-x)\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{5} (5-x)\,dx$
3. $\displaystyle \int_{4}^{5} x\,dx$
4. $\displaystyle \int_{0}^{1} (5-nx)\,dx$
5. $\displaystyle \int_{0}^{1} (5-x)\,\frac{1}{n}\,dx$

Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The (\frac{1}{n}) serves as (\Delta x), indicating limits from 0 to 1. The expression (5 - \frac{i}{n}) corresponds to (f(x_i) = 5 - x_i) with (x_i = \frac{i}{n}). Therefore, the Riemann sum approximates (\int_{0}^{1} (5 - x) , dx). Choice B, (\int_{0}^{5} (5 - x) , dx), is a distractor that incorrectly expands the interval to 5, missing that the interval length is 1 based on (\Delta x). To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).

Question 5

1. $\displaystyle \int_{0}^{2} \left(5+\sqrt{1+x}\right)\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{2} \left(5+\sqrt{1+0.02x}\right)\,dx$
3. $\displaystyle \int_{0}^{100} \left(5+\sqrt{1+0.02x}\right)\,dx$
4. $\displaystyle \int_{0.02}^{2} \left(5+\sqrt{1+x}\right)\,dx$
5. $\displaystyle \int_{0}^{2} \left(5+\sqrt{1+x}\right)(0.02)\,dx$

Explanation: This problem tests the skill of translating a Riemann sum into its corresponding definite integral notation. The sum ∑_{i=1}^{100} (5 + sqrt(1 + 0.02 i)) (0.02) has Δx = 0.02, x_i = i Δx = 0.02 i, so the term inside sqrt is 1 + x_i, function 5 + sqrt(1 + x). The limits are from 0 to 2, as i=1 to 100, x from 0.02 to 2. Thus, the sum is a right Riemann sum for ∫_0^2 (5 + sqrt(1 + x)) dx. A tempting distractor is ∫_0^2 (5 + sqrt(1 + x)) (0.02) dx, but this incorrectly includes the Δx inside the integral instead of recognizing it as the width multiplier in the sum. To translate any Riemann sum to an integral, identify Δx, express the argument in terms of x_k = a + k Δx, determine the limits from the range of x_k, and form the integral of the resulting function over those limits.

Question 6

1. $\displaystyle \int_{0}^{2} x^3\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{80} \left(\frac{2x}{80}\right)^3\,dx$
3. $\displaystyle \int_{0}^{2} \left(\frac{2x}{80}\right)^3\,dx$
4. $\displaystyle \int_{0}^{2} x^3\,\frac{2}{80}\,dx$
5. $\displaystyle \int_{0}^{80} x^3\,dx$

Explanation: This problem tests the skill of translating a Riemann sum into its corresponding definite integral notation. The sum $\sum_{k=1}^{80} \left( \frac{2k}{80} \right)^3 \frac{2}{80}$ has $\Delta x = \frac{2}{80} = 0.025$, $x_k = k \Delta x = \frac{2k}{80}$. The term is $(x_k)^3$, so function $x^3$. Limits from 0 to 2. A tempting distractor is $\int_0^{80} \left( \frac{2x}{80} \right)^3 dx$, which incorrectly sets the upper limit to 80 without adjusting $\Delta x$. To translate any Riemann sum to an integral, identify $\Delta x$, express the argument in terms of $x_k = a + k \Delta x$, determine the limits from the range of $x_k$, and form the integral of the resulting function over those limits.

Question 7

1. $\displaystyle \int_{2}^{5} \ln(x)\,dx$ (correct answer)
2. $\displaystyle \int_{2}^{3} \ln(x)\,dx$
3. $\displaystyle \int_{0}^{3} \ln(x)\,dx$
4. $\displaystyle \int_{2}^{5} 3\ln(x)\,dx$
5. $\displaystyle \int_{2}^{5} \ln(2+3x)\,dx$

Explanation: This problem asks us to translate a midpoint Riemann sum to integral notation. The sum $\sum_{k=1}^{n} \ln\!\left(2+\frac{3}{n}\left(k-\frac12\right)\right)\frac{3}{n}$ samples at midpoints. When $k=1$, we get $x=2+\frac{3}{n}\cdot\frac{1}{2}$, and when $k=n$, we get approximately $x=2+3\left(1-\frac{1}{2n}\right)$, which approaches $5$ as $n$ increases. The width $\frac{3}{n}$ confirms interval length $5-2=3$. Choice E might seem appealing with $\ln(2+3x)$, but this misinterprets the sampling points as a function transformation. The key is recognizing that midpoint sums still approximate $\int_a^b f(x)\,dx$, not a transformed function.

Question 8

1. $\displaystyle \int_{0}^{\pi} \sin x\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{1} \sin(\pi x)\,dx$
3. $\displaystyle \int_{0}^{\pi} \sin(\pi x)\,dx$
4. $\displaystyle \int_{1}^{n} \sin\!\left(\frac{\pi x}{n}\right)\frac{\pi}{n}\,dx$
5. $\displaystyle \int_{0}^{\pi} \cos x\,dx$

Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The term (\frac{\pi}{n}) is (\Delta x), indicating an interval length of (\pi), typically from 0 to (\pi). The argument of the sine, (\frac{\pi i}{n}), is the sample point (x_i = \frac{\pi i}{n}), so the function is (f(x) = \sin x). The sum thus represents the integral (\int_{0}^{\pi} \sin x , dx) as (n) grows large. Choice C, (\int_{0}^{\pi} \sin(\pi x) , dx), is a common distractor but fails because it mistakenly places (\pi) inside the sine instead of recognizing the direct mapping to (\sin x). To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).

Question 9

1. $\displaystyle \int_{0}^{3} \sqrt{2+3x}\,dx$
2. $\displaystyle \int_{2}^{5} \sqrt{x}\,dx$
3. $\displaystyle \int_{0}^{3} \sqrt{2+x}\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{3} \sqrt{2+x}\,3\,dx$
5. $\displaystyle \int_{0}^{3} \sqrt{2+\frac{3x}{n}}\,\frac{3}{n}\,dx$

Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. Here, (\frac{3}{n}) serves as (\Delta x), implying an interval length of 3, from 0 to 3. The expression (\sqrt{2 + \frac{3i}{n}}) corresponds to (f(x_i) = \sqrt{2 + x_i}) with (x_i = \frac{3i}{n}). As such, the sum converges to (\int_{0}^{3} \sqrt{2 + x} , dx). Choice D, (\int_{0}^{3} \sqrt{2 + x} \cdot 3 , dx), is a distractor that wrongly multiplies the integrand by 3, confusing the role of (\Delta x)'s numerator. To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).

Question 10

1. $\displaystyle \int_{3}^{5} x^2\,dx$
2. $\displaystyle \int_{0}^{2} (3+x)^2\,dx$ (correct answer)
3. $\displaystyle \int_{1}^{n} \left(3+\frac{2x}{n}\right)^2\frac{2}{n}\,dx$
4. $\displaystyle \int_{0}^{2} (3+2x)^2\,dx$
5. $\displaystyle \int_{0}^{2} (3+x)^2\,2\,dx$

Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The term (\frac{2}{n}) represents the width of each subinterval, (\Delta x), so the total interval length is 2, suggesting limits from 0 to 2. The expression inside the sum, (3 + \frac{2i}{n}), corresponds to the sample point (x_i = \frac{2i}{n}) evaluated in the function (f(x) = (3 + x)^2). As (n) approaches infinity, this right Riemann sum approximates the integral (\int_{0}^{2} (3 + x)^2 , dx). A tempting distractor like choice D, (\int_{0}^{2} (3 + 2x)^2 , dx), fails because it incorrectly doubles the coefficient of (x) instead of matching the sum's structure. To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).

Question 11

1. $\displaystyle \int_{0}^{\pi} \cos(x)\,dx$
2. $\displaystyle \int_{\pi}^{2\pi} \cos(x)\,dx$ (correct answer)
3. $\displaystyle \int_{\pi}^{2\pi} \pi\cos(x)\,dx$
4. $\displaystyle \int_{\pi}^{2\pi} \cos(\pi x)\,dx$
5. $\displaystyle \int_{2\pi}^{\pi} \cos(x)\,dx$

Explanation: This problem requires translating a Riemann sum into integral notation. The sum $\sum_{k=1}^{n} \cos\!\left(\pi+\frac{\pi k}{n}\right)\frac{\pi}{n}$ samples cosine at points $\pi+\frac{\pi k}{n}$. When $k=0$, we would get $x=\pi$, and when $k=n$, we get $x=\pi+\pi=2\pi$, establishing interval $[\pi,2\pi]$. The width $\frac{\pi}{n}$ confirms the interval length is $2\pi-\pi=\pi$. Choice D with $\cos(\pi x)$ misinterprets the argument as a function transformation. Remember that the pattern inside the function tells us where to evaluate, not how to transform the function itself.

Question 12

1. $\displaystyle \int_{0}^{1} e^{1+x}\,dx$ (correct answer)
2. $\displaystyle \int_{1}^{2} e^{x}\,dx$
3. $\displaystyle \int_{0}^{1} e^{1+nx}\,dx$
4. $\displaystyle \int_{0}^{1} e^{1+x}\,n\,dx$
5. $\displaystyle \int_{0}^{1} e^{x+\frac{1}{n}}\,dx$

Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The (\frac{1}{n}) is (\Delta x), suggesting limits from 0 to 1. The term (e^{1 + \frac{i}{n}}) maps to (f(x_i) = e^{1 + x_i}) where (x_i = \frac{i}{n}). Thus, the sum represents (\int_{0}^{1} e^{1 + x} , dx). Choice D, (\int_{0}^{1} e^{1 + x} \cdot n , dx), might tempt but fails by incorrectly multiplying by (n), which inverts (\Delta x)'s role. To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).

Question 13

1. $\displaystyle \int_{0}^{4} (1+4x)\,dx$
2. $\displaystyle \int_{1}^{5} x\,dx$
3. $\displaystyle \int_{0}^{4} (1+x)\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{4} (1+x)\,4\,dx$
5. $\displaystyle \int_{0}^{4} \left(1+\frac{4x}{n}\right)\frac{4}{n}\,dx$

Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The (\frac{4}{n}) term is (\Delta x), pointing to an interval from 0 to 4. The summand (1 + \frac{4i}{n}) matches (f(x_i) = 1 + x_i) where (x_i = \frac{4i}{n}). Therefore, the Riemann sum approximates (\int_{0}^{4} (1 + x) , dx). Choice D, (\int_{0}^{4} (1 + x) \cdot 4 , dx), tempts by multiplying by 4 but fails as it incorrectly incorporates (\Delta x)'s numerator inside the integrand instead of as the differential. To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).

Question 14

1. $\displaystyle \int_{0}^{3} \frac{1}{1+x^2}\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{1} \frac{1}{1+9x^2}\,dx$
3. $\displaystyle \int_{0}^{3} \frac{1}{1+9x^2}\,dx$
4. $\displaystyle \int_{0}^{3} \frac{1}{1+\left(\frac{3x}{n}\right)^2}\,dx$
5. $\displaystyle \int_{0}^{3} \frac{3}{1+x^2}\,dx$

Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The term (\frac{3}{n}) represents (\Delta x), implying limits from 0 to 3. The fraction (\frac{1}{1 + (\frac{3i}{n})^2}) corresponds to (f(x_i) = \frac{1}{1 + x_i^2}) with (x_i = \frac{3i}{n}). Hence, the sum approximates (\int_{0}^{3} \frac{1}{1 + x^2} , dx). Choice E, (\int_{0}^{3} \frac{3}{1 + x^2} , dx), distracts by placing the 3 in the numerator but fails to recognize it belongs in the differential. To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).

Question 15

1. $\displaystyle \int_{0}^{3} \sqrt{9-x^2}\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{30} \sqrt{9-\left(\frac{x}{10}\right)^2}\,dx$
3. $\displaystyle \int_{0}^{3} \sqrt{9-\left(\frac{x}{10}\right)^2}\,dx$
4. $\displaystyle \int_{\frac{1}{10}}^{3} \sqrt{9-x^2}\,dx$
5. $\displaystyle \int_{0}^{3} \sqrt{9-x^2}\,\frac{1}{10}\,dx$

Explanation: This problem tests the skill of translating a Riemann sum into its corresponding definite integral notation. The sum ∑_{i=1}^{30} sqrt{9 - left( rac{i}{10} ight)^2 } rac{1}{10} has Δx=0.1, x_i = i Δx = i/10, function sqrt{9 - x^2}. Limits from 0 to 3. Thus, ∫_0^3 sqrt{9 - x^2} dx. A tempting distractor is ∫_0^3 sqrt{9 - x^2} rac{1}{10} dx, mistakenly placing Δx inside the integral. To translate any Riemann sum to an integral, identify Δx, express the argument in terms of x_k = a + k Δx, determine the limits from the range of x_k, and form the integral of the resulting function over those limits.

Question 16

1. $\displaystyle \int_{4}^{8} \frac{1}{\sqrt{1+x}}\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{4} \frac{1}{\sqrt{1+(4+x)}}\,dx$
3. $\displaystyle \int_{4}^{200} \frac{1}{\sqrt{1+\left(4+\frac{x}{50}\right)}}\,dx$
4. $\displaystyle \int_{4}^{8} \frac{1}{\sqrt{1+x}}\,\frac{1}{50}\,dx$
5. $\displaystyle \int_{4.02}^{8} \frac{1}{\sqrt{1+x}}\,dx$

Explanation: This problem tests the skill of translating a Riemann sum into its corresponding definite integral notation. The sum ∑_{k=1}^{200} rac{1}{sqrt{1 + (4 + rac{k}{50}) }} rac{1}{50} has Δx=1/50=0.02, the argument 4 + k/50 = 4 + k Δx. Letting y_k = 4 + k Δx, f(y) = 1/sqrt{1 + y}, limits from 4 to 8. Thus, the sum approximates ∫_4^8 rac{1}{sqrt{1 + x}} dx. A tempting distractor is ∫_4^8 rac{1}{sqrt{1 + x}} rac{1}{50} dx, incorrectly including Δx within the integrand. To translate any Riemann sum to an integral, identify Δx, express the argument in terms of x_k = a + k Δx, determine the limits from the range of x_k, and form the integral of the resulting function over those limits.

Question 17

1. $\displaystyle \int_{3}^{5} x^2\,dx$ (correct answer)
2. $\displaystyle \int_{3}^{5} (3+2x)^2\,dx$
3. $\displaystyle \int_{0}^{2} (3+x)^2\,dx$
4. $\displaystyle \int_{3}^{5} 2x^2\,dx$
5. $\displaystyle \int_{0}^{2} x^2\,dx$

Explanation: This problem requires translating a Riemann sum into integral notation. The sum $\sum_{j=1}^{n} \left(\left(3+\frac{2j}{n}\right)^2\right)\frac{2}{n}$ evaluates the function at points $3+\frac{2j}{n}$. When $j=0$, we would get $x=3$, and when $j=n$, we get $x=3+2=5$, establishing interval $[3,5]$. The width $\frac{2}{n}$ confirms the interval length is $5-3=2$. Choice B might tempt you with $(3+2x)^2$, but this misinterprets the sampling points as part of the function itself. The key insight is that the expression inside the function gives us sample points, not the function's formula.

Question 18

1. $\displaystyle \int_{7}^{9} \sqrt{x}\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{2} \sqrt{7+x}\,dx$
3. $\displaystyle \int_{7}^{9} \sqrt{7+x}\,dx$
4. $\displaystyle \int_{7}^{9} 2\sqrt{x}\,dx$
5. $\displaystyle \int_{0}^{2} \sqrt{x}\,dx$

Explanation: This problem requires identifying the integral matching a Riemann sum. The sum $\sum_{k=0}^{n-1} \sqrt{7+\frac{2k}{n}}\,\frac{2}{n}$ evaluates at points $7+\frac{2k}{n}$. When $k=0$, we get $x=7$, and when $k=n-1$, we approach $x=7+2=9$, establishing interval $[7,9]$. The width $\frac{2}{n}$ matches the interval length $9-7=2$. Choice B shows $\sqrt{7+x}$, which incorrectly interprets the argument as a function composition rather than sampling points. The transferable principle is that $f(a+\frac{(b-a)k}{n})$ samples $f(x)$ on $[a,b]$, not $f(a+x)$.

Question 19

1. $\displaystyle \int_{-1}^{5} g(x)\,dx$ (correct answer)
2. $\displaystyle \int_{-1}^{6} g(x)\,dx$
3. $\displaystyle \int_{0}^{6} g(x)\,dx$
4. $\displaystyle \int_{-1}^{5} 6g(x)\,dx$
5. $\displaystyle \int_{-1}^{5} g(-1+6x)\,dx$

Explanation: This problem asks us to match a Riemann sum to its integral representation. The sum $\sum_{i=1}^{n} g\!\left(-1+\frac{6i}{n}\right)\frac{6}{n}$ samples at points $-1+\frac{6i}{n}$. When $i=0$, we would get $x=-1$, and when $i=n$, we get $x=-1+6=5$, defining interval $[-1,5]$. The width $\frac{6}{n}$ confirms the interval length is $5-(-1)=6$. Choice B extends to $x=6$, which overshoots our endpoint. The key insight is that the coefficient of $i/n$ gives the interval length, and adding it to the constant term gives the right endpoint.

Question 20

1. $\displaystyle \int_{0.5}^{2} v(t)\,dt$ (correct answer)
2. $\displaystyle \int_{0}^{1.5} v(t)\,dt$
3. $\displaystyle \int_{0.5}^{2} 1.5\,v(t)\,dt$
4. $\displaystyle \int_{0.5}^{2} v\!\left(0.5+\frac{1.5t}{n}\right)\,dt$
5. $\displaystyle \int_{0.5}^{1.5} v(t)\,dt$

Explanation: This problem asks us to match a Riemann sum to its integral representation. The sum $\sum_{k=1}^{n} v\!\left(0.5+\frac{1.5k}{n}\right)\frac{1.5}{n}$ evaluates velocity at points $0.5+\frac{1.5k}{n}$. When $k=1$, we get approximately $x=0.5$, and when $k=n$, we get $x=0.5+1.5=2$, defining the interval $[0.5,2]$. The width $\frac{1.5}{n}$ matches the interval length $2-0.5=1.5$. Choice C incorrectly places the factor $1.5$ inside the integral as a coefficient of $v(t)$. Remember that in Riemann sums, the pattern $a+\frac{(b-a)k}{n}$ with width $\frac{b-a}{n}$ corresponds to $\int_a^b f(x)\,dx$.