AP Calculus AB Quiz: Area Between Curves With Multiple Intersections

What this quiz covers

This quiz focuses on Area Between Curves With Multiple Intersections, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

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

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at x=0 because that's where y=e^x-1 crosses y=0. From -1 to 0, y=0 is above y=e^x-1 since e^x-1<0. From 0 to 1, y=e^x-1 is above y=0 since e^x-1>0. One tempting distractor, such as choice A, fails because it computes the net area, allowing positive and negative regions to cancel out. A transferable strategy is to always locate all intersection points, divide the interval accordingly, and in each subinterval integrate the absolute difference by subtracting the bottom function from the top function.

Question 2

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

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at x=π/2 and x=3π/2 because those are where the order changes for y=cos x and y=-cos x. From 0 to π/2, y=cos x is above y=-cos x. From π/2 to 3π/2, y=-cos x is above y=cos x in parts. From 3π/2 to 2π, y=cos x is above y=-cos x. One tempting distractor, such as choice A, fails because it computes the net area, allowing positive and negative regions to cancel out. A transferable strategy is to always locate all intersection points, divide the interval accordingly, and in each subinterval integrate the absolute difference by subtracting the bottom function from the top function.

Question 3

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

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = sin x intersects y = 0 at x = 0, π, 2π, dividing [0, 2π] into subintervals. In [0, π], sin is above; in [π, 2π], below, requiring a split. The negative in the second integral ensures positive area. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel. To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

Question 4

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

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = x³ - 4x intersects y = 0 at x = -2, 0, 2, dividing [-2, 2] into subintervals. In [-2, 0], the cubic is above, as at x = -1, (-1)³ - 4(-1) = 3 > 0. In [0, 2], it is below, as at x = 1, 1 - 4 = -3 < 0, requiring a split to ensure positive integrands. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel. To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

Question 5

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

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = x⁴ - x² intersects y = 0 at x = -1, 0, 1, but is below throughout [-1, 1]. In [-1, 1], y < 0, as x⁴ - x² = x²(x² - 1) < 0 for |x| < 1. The split at 0 is harmless since integrand same, giving positive area via x² - x⁴ > 0. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel (though none here). To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

Question 6

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

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = ln x intersects y = 0 at x = 1 within [1/e, e], dividing into subintervals. In [1/e, 1], ln x < 0; in [1, e], > 0, requiring a split. The negative in the first ensures positive area. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel. To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

Question 7

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

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at x=-2 and x=2 because those are where y=x^2-4 crosses y=0 within [-3,3]. From -3 to -2, y=x^2-4 is above y=0 since x^2-4>0 for |x|>2. From -2 to 2, y=0 is above y=x^2-4 since x^2-4<0. From 2 to 3, y=x^2-4 is above y=0 again. One tempting distractor, such as choice A, fails because it computes the net area, allowing positive and negative regions to cancel out. A transferable strategy is to always locate all intersection points, divide the interval accordingly, and in each subinterval integrate the absolute difference by subtracting the bottom function from the top function.

Question 8

1. $\displaystyle \int_{-2}^{2}\left[(1-x^2)-(x^2-1)\right]dx$
2. $\displaystyle \int_{-2}^{2}\left[(x^2-1)-(1-x^2)\right]dx$
3. $\displaystyle \int_{-2}^{2}\left[(1-x^2)+(x^2-1)\right]dx$
4. $\displaystyle \int_{-2}^{-1}\left[(x^2-1)-(1-x^2)\right]dx+\int_{-1}^{1}\left[(1-x^2)-(x^2-1)\right]dx+\int_{1}^{2}\left[(x^2-1)-(1-x^2)\right]dx$ (correct answer)
5. $\displaystyle \int_{-2}^{2}\left[(1-x^2)(x^2-1)\right]dx$

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at $x=-1$ and $x=1$ because those are the intersection points of $y=1-x^2$ and $y=x^2-1$ within $[-2,2]$. From $-2$ to $-1$, $y=x^2-1$ is above $y=1-x^2$. From $-1$ to $1$, $y=1-x^2$ is above $y=x^2-1$. From $1$ to $2$, $y=x^2-1$ is above $y=1-x^2$ again. One tempting distractor, such as choice A, fails because it computes the net area, allowing positive and negative regions to cancel out. A transferable strategy is to always locate all intersection points, divide the interval accordingly, and in each subinterval integrate the absolute difference by subtracting the bottom function from the top function.

Question 9

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

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve $y = \cos x$ intersects $y = \tfrac{1}{2}$ at $x = \pi/3$, $5\pi/3$ within $[0, 2\pi]$, dividing into three subintervals. In $[0, \pi/3]$ and $[5\pi/3, 2\pi]$, $\cos x > \tfrac{1}{2}$; in $[\pi/3, 5\pi/3]$, $\cos x < \tfrac{1}{2}$, requiring splits. This setup captures positive areas. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel. To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

Question 10

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

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at x=π because that's where y=sin x crosses y=x-π within [0,2π]. From 0 to π, y=sin x is above y=x-π since their difference is non-negative. From π to 2π, y=x-π is above y=sin x since their difference reverses. One tempting distractor, such as choice A, fails because it computes the net area, allowing positive and negative regions to cancel out. A transferable strategy is to always locate all intersection points, divide the interval accordingly, and in each subinterval integrate the absolute difference by subtracting the bottom function from the top function.

Question 11

1. $\displaystyle \int_{-2}^{2}\left[x^2-(2-x^2)\right]dx$
2. $\displaystyle \int_{-2}^{2}\left[(2-x^2)-x^2\right]dx$
3. $\displaystyle \int_{-2}^{2}\left[x^2+(2-x^2)\right]dx$
4. $\displaystyle \int_{-2}^{-1}\left[x^2-(2-x^2)\right]dx+\int_{-1}^{1}\left[(2-x^2)-x^2\right]dx+\int_{1}^{2}\left[x^2-(2-x^2)\right]dx$ (correct answer)
5. $\displaystyle \int_{-2}^{2}\left[x^2(2-x^2)\right]dx$

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at x=-1 and x=1 because those are the intersection points of y=x^2 and y=2-x^2 within [-2,2]. From -2 to -1, y=x^2 is above y=2-x^2. From -1 to 1, y=2-x^2 is above y=x^2. From 1 to 2, y=x^2 is above y=2-x^2 again. One tempting distractor, such as choice A, fails because it computes the net area, allowing positive and negative regions to cancel out. A transferable strategy is to always locate all intersection points, divide the interval accordingly, and in each subinterval integrate the absolute difference by subtracting the bottom function from the top function.

Question 12

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

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at x=-1 and x=1 because those are where y=x^2-1 crosses y=0 within [-2,2]. From -2 to -1, y=x^2-1 is above y=0 since x^2-1>0 for |x|>1. From -1 to 1, y=0 is above y=x^2-1 since x^2-1<0. From 1 to 2, y=x^2-1 is above y=0 again. One tempting distractor, such as choice A, fails because it computes the net area, allowing positive and negative regions to cancel out. A transferable strategy is to always locate all intersection points, divide the interval accordingly, and in each subinterval integrate the absolute difference by subtracting the bottom function from the top function.

Question 13

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

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = x³ - x intersects y = 0 at x = -1, 0, 1, dividing [-1, 1] into subintervals. In [-1, 0], the cubic is above the x-axis, as at x = -0.5, (-0.5)³ - (-0.5) = 0.375 > 0. In [0, 1], it is below, as at x = 0.5, (0.5)³ - 0.5 = -0.375 < 0, necessitating a split for positive area. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel. To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

Question 14

1. $\displaystyle \int_{0}^{2\pi}(\sin x-\tfrac12)\,dx$
2. $\displaystyle \int_{0}^{2\pi}(\tfrac12-\sin x)\,dx$
3. $\displaystyle \int_{0}^{2\pi}(\sin x+\tfrac12)\,dx$
4. $\displaystyle \int_{0}^{\pi/6}(\tfrac12-\sin x)dx+\int_{\pi/6}^{5\pi/6}(\sin x-\tfrac12)dx+\int_{5\pi/6}^{2\pi}(\tfrac12-\sin x)dx$ (correct answer)
5. $\displaystyle \int_{0}^{2\pi}(\sin x\cdot \tfrac12)\,dx$

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = sin x intersects y = 1/2 at x = π/6, 5π/6 within [0, 2π], dividing into three subintervals. In [0, π/6] and [5π/6, 2π], sin < 1/2; in [π/6, 5π/6], sin > 1/2, requiring splits. This ensures positive differences throughout. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel. To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

Question 15

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

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = x⁴ - 5x² + 4 intersects y = 0 at x = ±1, ±2, dividing [-2, 2] into subintervals. In [-2, -1] and [1, 2], y < 0; in [-1, 1], y > 0, requiring multiple splits. The negatives in outer integrals ensure positive area. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel. To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

Question 16

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

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = sin(2x) intersects y = 0 at x = 0, π/2, π, dividing [0, π] into subintervals. In [0, π/2], sin(2x) > 0; in [π/2, π], < 0, requiring a split. The negative in the second ensures positive area. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel. To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

Question 17

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

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at x=0 because that's where y=x^3 crosses y=x within [-1,1], in addition to the endpoints. From -1 to 0, y=x^3 is above y=x since x^3 > x for x in (-1,0). From 0 to 1, y=x is above y=x^3 since x > x^3 for x in (0,1). One tempting distractor, such as choice A, fails because it computes the net area, allowing positive and negative regions to cancel out. A transferable strategy is to always locate all intersection points, divide the interval accordingly, and in each subinterval integrate the absolute difference by subtracting the bottom function from the top function.

Question 18

1. $\displaystyle \int_{-2}^{2}\left[(1-x^2)-(x^2-1)\right]dx$
2. $\displaystyle \int_{-2}^{2}\left[(x^2-1)-(1-x^2)\right]dx$
3. $\displaystyle \int_{-2}^{2}\left[(1-x^2)+(x^2-1)\right]dx$
4. $\displaystyle \int_{-2}^{-1}\left[(x^2-1)-(1-x^2)\right]dx+\int_{-1}^{1}\left[(1-x^2)-(x^2-1)\right]dx+\int_{1}^{2}\left[(x^2-1)-(1-x^2)\right]dx$ (correct answer)
5. $\displaystyle \int_{-2}^{2}\left[(1-x^2)(x^2-1)\right]dx$

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at x=-1 and x=1 because those are the intersection points of y=1-x^2 and y=x^2-1 within [-2,2]. From -2 to -1, y=x^2-1 is above y=1-x^2. From -1 to 1, y=1-x^2 is above y=x^2-1. From 1 to 2, y=x^2-1 is above y=1-x^2 again. One tempting distractor, such as choice A, fails because it computes the net area, allowing positive and negative regions to cancel out. A transferable strategy is to always locate all intersection points, divide the interval accordingly, and in each subinterval integrate the absolute difference by subtracting the bottom function from the top function.

Question 19

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

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = ln x intersects y = 1 at x = e within [1, e²], dividing into subintervals. In [1, e], ln x < 1; in [e, e²], > 1, requiring a split. This ensures positive differences. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel. To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

Question 20

1. $\displaystyle \int_{0}^{2\pi}\left(\sin x-\tfrac12\sin x\right)dx$
2. $\displaystyle \int_{0}^{2\pi}\left(\tfrac12\sin x-\sin x\right)dx$
3. $\displaystyle \int_{0}^{2\pi}\left(\sin x+\tfrac12\sin x\right)dx$
4. $\displaystyle \int_{0}^{\pi}\left(\sin x-\tfrac12\sin x\right)dx+\int_{\pi}^{2\pi}\left(\tfrac12\sin x-\sin x\right)dx$ (correct answer)
5. $\displaystyle \int_{0}^{2\pi}\left(\sin x\cdot \tfrac12\sin x\right)dx$

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at x=π because that's where y=sin x crosses y=1/2 sin x within [0,2π], in addition to endpoints. From 0 to π, y=sin x is above y=1/2 sin x since sin x >0. From π to 2π, y=1/2 sin x is above y=sin x since sin x <0. One tempting distractor, such as choice A, fails because it computes the net area, allowing positive and negative regions to cancel out. A transferable strategy is to always locate all intersection points, divide the interval accordingly, and in each subinterval integrate the absolute difference by subtracting the bottom function from the top function.