AP Calculus AB Quiz

AP Calculus AB Quiz: Area Between Curves Functions Of Y

Practice Area Between Curves Functions Of Y in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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An expression for area between $x=y-4$ (left) and $x=y^2$ (right) on $1\le y\le 4$ is

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

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Question 1

An expression for area between $x=y-4$ (left) and $x=y^2$ (right) on $1\le y\le 4$ is

$\displaystyle \int_{1}^{4}\big[y^2-(y-4)\big]dy$ (correct answer)
$\displaystyle \int_{1}^{4}\big[(y-4)-y^2\big]dy$
$\displaystyle \int_{4}^{1}\big[y^2-(y-4)\big]dy$
$\displaystyle \int_{1}^{4}\big[(y+4)-y^2\big]dy$
$\displaystyle \int_{1}^{4}\big[y^2-(y+4)\big]dy$

Explanation: This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = y² and the left curve is x = y - 4, so the integrand is y² - (y - 4). The limits of integration are from y = 1 to y = 4, as specified. A tempting distractor is choice B, which reverses the order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.

Question 2

Which integral gives the area between $x=y^2-1$ (left) and $x=3y$ (right) for $0\le y\le 2$ ?

$\displaystyle \int_{0}^{2}\big[3y-(y^2-1)\big]dy$ (correct answer)
$\displaystyle \int_{0}^{2}\big[(y^2-1)-3y\big]dy$
$\displaystyle \int_{2}^{0}\big[3y-(y^2-1)\big]dy$
$\displaystyle \int_{0}^{2}\big[3y-(y^2+1)\big]dy$
$\displaystyle \int_{0}^{2}\big[(3y-1)-(y^2-1)\big]dy$

Explanation: This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = 3y and the left curve is x = y² - 1, so the integrand is 3y - (y² - 1). The limits of integration are from y = 0 to y = 2, as specified. A tempting distractor is choice B, which reverses the order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.

Question 3

Which integral represents the area between $x=\sqrt{y}$ (left) and $x=\frac{y}{2}+1$ (right) for $0\le y\le 4$ ?

$\displaystyle \int_{0}^{4}\left[\sqrt{y}-\left(\frac{y}{2}+1\right)\right]dy$
$\displaystyle \int_{4}^{0}\left[\left(\frac{y}{2}+1\right)-\sqrt{y}\right]dy$
$\displaystyle \int_{0}^{4}\left[\left(\frac{y}{2}+1\right)-\sqrt{y}\right]dy$ (correct answer)
$\displaystyle \int_{0}^{4}\left[\left(\frac{y}{2}-1\right)-\sqrt{y}\right]dy$
$\displaystyle \int_{0}^{4}\left[\left(\frac{y}{2}+1\right)-\sqrt{y+1}\right]dy$

Explanation: This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = y/2 + 1 and the left curve is x = √y, so the integrand is (y/2 + 1) - √y. The limits of integration are from y = 0 to y = 4, as specified. A tempting distractor is choice A, which reverses the order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.

Question 4

Which integral gives area between $x=2y^2$ (left) and $x=8$ (right) for $-2\le y\le 2$ ?

$\displaystyle \int_{-2}^{2}\big[(2y^2)-8\big]dy$
$\displaystyle \int_{-2}^{2}\big[8-(2y^2)\big]dy$ (correct answer)
$\displaystyle \int_{2}^{-2}\big[8-(2y^2)\big]dy$
$\displaystyle \int_{-2}^{2}\big[8-(y^2)\big]dy$
$\displaystyle \int_{-2}^{2}\big[8-(2y)\big]dy$

Explanation: This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = 8 and the left curve is x = 2y², so the integrand is 8 - 2y². The limits of integration are from y = -2 to y = 2, as specified. A tempting distractor is choice A, which reverses the order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.

Question 5

Set up the area between $x=\ln(y+1)+2$ (right) and $x=1$ (left) for $0\le y\le 3$ .

$\displaystyle \int_{0}^{3}\big[1-(\ln(y+1)+2)\big]dy$
$\displaystyle \int_{3}^{0}\big[(\ln(y+1)+2)-1\big]dy$
$\displaystyle \int_{0}^{3}\big[(\ln(y+1)+2)-1\big]dy$ (correct answer)
$\displaystyle \int_{0}^{3}\big[(\ln(y-1)+2)-1\big]dy$
$\displaystyle \int_{0}^{3}\big[(\ln(y+1)-2)-1\big]dy$

Explanation: This problem tests the skill of finding areas between curves by integrating with respect to y. The area is given by the integral from the lower to upper y-value of (x_right - x_left) dy. Here, x = ln(y+1) + 2 is the right curve and x = 1 is the left curve, so the integrand is (ln(y+1) + 2) - 1. The limits are from y=0 to y=3, matching choice C. A tempting distractor is choice A, which subtracts left minus right, yielding a negative value instead of the positive area. Always evaluate the functions at a point within the interval to confirm which curve is on the right.

Question 6

Which integral gives the area between $x=\frac{y+1}{2}$ (right) and $x=y-2$ (left) for $-1\le y\le 3$ ?

$\displaystyle \int_{-1}^{3}\left[\frac{y+1}{2}-(y-2)\right]dy$ (correct answer)
$\displaystyle \int_{-1}^{3}\left[(y-2)-\frac{y+1}{2}\right]dy$
$\displaystyle \int_{3}^{-1}\left[\frac{y+1}{2}-(y-2)\right]dy$
$\displaystyle \int_{-1}^{3}\left[\frac{y-1}{2}-(y-2)\right]dy$
$\displaystyle \int_{-1}^{3}\left[\frac{y+1}{2}-(y+2)\right]dy$

Explanation: This problem tests the skill of finding areas between curves by integrating with respect to y. The area is given by the integral from the lower to upper y-value of (x_right - x_left) dy. Here, x = (y+1)/2 is the right curve and x = y-2 is the left curve, so the integrand is (y+1)/2 - (y-2). The limits are from y=-1 to y=3, matching choice A. A tempting distractor is choice B, which subtracts left minus right, yielding a negative value instead of the positive area. Always evaluate the functions at a point within the interval to confirm which curve is on the right.

Question 7

Which integral gives the area between $x=2y+1$ (left) and $x=y^2+3$ (right) from $y=-1$ to $y=2$ ?

$\displaystyle \int_{-1}^{2}\big[(2y+1)-(y^2+3)\big]\,dy$
$\displaystyle \int_{-1}^{2}\big[(y^2+3)-(2y+1)\big]\,dy$ (correct answer)
$\displaystyle \int_{-2}^{1}\big[(y^2+3)-(2y+1)\big]\,dy$
$\displaystyle \int_{-1}^{2}\big[(y^2-3)-(2y+1)\big]\,dy$
$\displaystyle \int_{-1}^{2}\big[(y^2+3)-(2y-1)\big]\,dy$

Explanation: This question requires setting up an area integral with respect to y using the fundamental principle of right minus left. When integrating along the y-axis, we subtract the left boundary function from the right boundary function to get positive area. The problem states that x = 2y + 1 is on the left and x = y² + 3 is on the right, meaning we need (y² + 3) - (2y + 1) as our integrand. Choice A incorrectly uses (2y + 1) - (y² + 3), which would yield negative area since it subtracts the rightmost function from the leftmost. To verify which function is truly on the right, pick a y-value like y = 0: the left curve gives x = 1 while the right curve gives x = 3, confirming our setup.

Question 8

Set up the $dy$ integral for area between $x=\ln(y+2)$ (left) and $x=2$ (right) on $-1 \le y \le 2$ .

$\displaystyle \int_{-1}^{2}\big[\ln(y+2)-2\big]\,dy$
$\displaystyle \int_{-1}^{2}\big[2-\ln(y+2)\big]\,dy$ (correct answer)
$\displaystyle \int_{-2}^{1}\big[2-\ln(y+2)\big]\,dy$
$\displaystyle \int_{-1}^{2}\big[2-\ln(y-2)\big]\,dy$
$\displaystyle \int_{-1}^{2}\big[3-\ln(y+2)\big]\,dy$

Explanation: This question tests your ability to set up area integrals with respect to y, applying the fundamental right-minus-left subtraction principle. When computing area using dy integration, we must subtract the left boundary function from the right boundary function to obtain positive area. The problem identifies $x = \ln(y + 2)$ as the left curve and $x = 2$ as the right curve, so our integrand becomes $2 - \ln(y + 2)$ . Choice A incorrectly uses $\ln(y + 2) - 2$ , which reverses the subtraction and would yield negative area—a typical error when students forget to consider relative positions. To confirm the ordering, evaluate at $y = 0$ : the left curve gives $x = \ln(2) \approx 0.693$ while the right curve gives $x = 2$ , verifying that the constant function $x = 2$ is indeed to the right.

Question 9

Set up the area integral (in $y$ ) between $x=-y^2+4$ (right) and $x=y-2$ (left) for $0\le y\le 3$ .

$\displaystyle \int_{0}^{3}\big[(y-2)-(-y^2+4)\big]\,dy$
$\displaystyle \int_{0}^{3}\big[(-y^2+4)-(y-2)\big]\,dy$ (correct answer)
$\displaystyle \int_{0}^{3}\big[(-y^2-4)-(y-2)\big]\,dy$
$\displaystyle \int_{-3}^{0}\big[(-y^2+4)-(y-2)\big]\,dy$
$\displaystyle \int_{0}^{3}\big[(-y^2+4)-(y+2)\big]\,dy$

Explanation: This problem tests your understanding of area integrals with respect to y, specifically the right-minus-left subtraction order. When finding area between curves using dy integration, we always compute (right function) - (left function) where right means larger x-values. The problem identifies x = -y² + 4 as the right curve and x = y - 2 as the left curve, so our integrand must be (-y² + 4) - (y - 2). Choice A reverses this order to (y - 2) - (-y² + 4), which would produce a negative result—a frequent error when students mechanically subtract without considering position. To remember the correct order, visualize or test: at y = 1, the left curve gives x = -1 while the right curve gives x = 3, confirming that -y² + 4 is indeed rightmost.

Question 10

Which integral represents the area between $x=4$ (right) and $x=\sin y$ (left) for $0\le y\le \pi$ ?

$\displaystyle \int_{0}^{\pi}\big[\sin y-4\big]dy$
$\displaystyle \int_{\pi}^{0}\big[4-\sin y\big]dy$
$\displaystyle \int_{0}^{\pi}\big[4-\sin y\big]dy$ (correct answer)
$\displaystyle \int_{0}^{\pi}\big[4-\cos y\big]dy$
$\displaystyle \int_{0}^{\pi}\big[4+\sin y\big]dy$

Explanation: This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = 4 and the left curve is x = sin y, so the integrand is 4 - sin y. The limits of integration are from y = 0 to y = π, as specified. A tempting distractor is choice A, which reverses the order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.

Question 11

Which integral gives the area between $x=\frac{y}{3}$ (left) and $x=\frac{9}{y+3}$ (right) for $0\le y\le 3$ ?

$\displaystyle \int_{0}^{3}\left[\frac{y}{3}-\frac{9}{y+3}\right]dy$
$\displaystyle \int_{0}^{3}\left[\frac{9}{y+3}-\frac{y}{3}\right]dy$ (correct answer)
$\displaystyle \int_{3}^{0}\left[\frac{9}{y+3}-\frac{y}{3}\right]dy$
$\displaystyle \int_{0}^{3}\left[\frac{9}{y-3}-\frac{y}{3}\right]dy$
$\displaystyle \int_{0}^{3}\left[\frac{9}{y+3}-\frac{3}{y}\right]dy$

Explanation: This problem tests the skill of finding areas between curves by integrating with respect to y. The area is given by the integral from the lower to upper y-value of (x_right - x_left) dy. Here, x = 9/(y+3) is the right curve and x = y/3 is the left curve, so the integrand is 9/(y+3) - y/3. The limits are from y=0 to y=3, matching choice B. A tempting distractor is choice A, which subtracts left minus right, yielding a negative value instead of the positive area. Always evaluate the functions at a point within the interval to confirm which curve is on the right.

Question 12

Find an area expression between $x=3-y^2$ (left) and $x=5$ (right) for $-1 \le y \le 1$ .

$\displaystyle \int_{-1}^{1}\big[(3-y^2)-5\big]dy$
$\displaystyle \int_{-1}^{1}\big[5-(3-y^2)\big]dy$ (correct answer)
$\displaystyle \int_{1}^{-1}\big[5-(3-y^2)\big]dy$
$\displaystyle \int_{-1}^{1}\big[5-(3+y^2)\big]dy$
$\displaystyle \int_{-1}^{1}\big[5-(3-y)\big]dy$

Explanation: This problem tests the skill of finding areas between curves by integrating with respect to y. The area is given by the integral from the lower to upper y-value of $(x_{\text{right}} - x_{\text{left}}) \, dy$ . Here, $x = 5$ is the right curve and $x = 3 - y^2$ is the left curve, so the integrand is $5 - (3 - y^2)$ . The limits are from $y=-1$ to $y=1$ , matching choice B. A tempting distractor is choice A, which subtracts left minus right, yielding a negative value instead of the positive area. Always evaluate the functions at a point within the interval to confirm which curve is on the right.

Question 13

Which integral represents area between $x=1-y$ (left) and $x=3+y^2$ (right) for $-2\le y\le 1$ ?

$\displaystyle \int_{-2}^{1}\big[(1-y)-(3+y^2)\big]dy$
$\displaystyle \int_{1}^{-2}\big[(3+y^2)-(1-y)\big]dy$
$\displaystyle \int_{-2}^{1}\big[(3+y^2)-(1-y)\big]dy$ (correct answer)
$\displaystyle \int_{-2}^{1}\big[(3-y^2)-(1-y)\big]dy$
$\displaystyle \int_{-2}^{1}\big[(3+y^2)-(1+y)\big]dy$

Explanation: This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = 3 + y² and the left curve is x = 1 - y, so the integrand is (3 + y²) - (1 - y). The limits of integration are from y = -2 to y = 1, as specified. A tempting distractor is choice A, which reverses the order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.

Question 14

Which integral represents the area between $x=y^2$ (right) and $x=y^2-4$ (left) for $-2\le y\le 2$ ?

$\displaystyle \int_{-2}^{2}\big[(y^2-4)-y^2\big]dy$
$\displaystyle \int_{2}^{-2}\big[y^2-(y^2-4)\big]dy$
$\displaystyle \int_{-2}^{2}\big[y^2-(y^2-4)\big]dy$ (correct answer)
$\displaystyle \int_{-2}^{2}\big[y^2-(y^2+4)\big]dy$
$\displaystyle \int_{-2}^{2}\big[(y^2-4)-(y^2+4)\big]dy$

Explanation: This problem tests the skill of finding areas between curves by integrating with respect to y. The area is given by the integral from the lower to upper y-value of (x_right - x_left) dy. Here, x = y^2 is the right curve and x = y^2 - 4 is the left curve, so the integrand is y^2 - (y^2 - 4). The limits are from y=-2 to y=2, matching choice C. A tempting distractor is choice A, which subtracts left minus right, yielding a negative value instead of the positive area. Always evaluate the functions at a point within the interval to confirm which curve is on the right.

Question 15

Find an area setup between $x=1$ (left) and $x=y^2+1$ (right) for $-3\le y\le 3$ .

$\displaystyle \int_{-3}^{3}\big[(y^2+1)-1\big]dy$ (correct answer)
$\displaystyle \int_{-3}^{3}\big[1-(y^2+1)\big]dy$
$\displaystyle \int_{3}^{-3}\big[(y^2+1)-1\big]dy$
$\displaystyle \int_{-3}^{3}\big[(y^2-1)-1\big]dy$
$\displaystyle \int_{-3}^{3}\big[(y+1)-1\big]dy$

Explanation: This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = y² + 1 and the left curve is x = 1, so the integrand is (y² + 1) - 1. The limits of integration are from y = -3 to y = 3, as specified. A tempting distractor is choice B, which reverses the order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.

Question 16

Set up the area between $x=2\cos y$ (left) and $x=3$ (right) for $-\frac{\pi}{2}\le y\le \frac{\pi}{2}$ .

$\displaystyle \int_{-\pi/2}^{\pi/2}\big[2\cos y-3\big]dy$
$\displaystyle \int_{\pi/2}^{-\pi/2}\big[3-2\cos y\big]dy$
$\displaystyle \int_{-\pi/2}^{\pi/2}\big[3-2\cos y\big]dy$ (correct answer)
$\displaystyle \int_{-\pi/2}^{\pi/2}\big[3-2\sin y\big]dy$
$\displaystyle \int_{-\pi/2}^{\pi/2}\big[3+2\cos y\big]dy$

Explanation: This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is $x = 3$ and the left curve is $x = 2 \cos y$ , so the integrand is $3 - 2 \cos y$ . The limits of integration are from $y = -\frac{\pi}{2}$ to $y = \frac{\pi}{2}$ , as specified. A tempting distractor is choice A, which reverses the order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.

Question 17

Set up the area between $x=y^3+2$ (right) and $x=y$ (left) on $-1 \le y \le 2$ .

$\displaystyle \int_{-1}^{2}\big[(y^3+2)-y\big]dy$ (correct answer)
$\displaystyle \int_{-1}^{2}\big[y-(y^3+2)\big]dy$
$\displaystyle \int_{2}^{-1}\big[(y^3+2)-y\big]dy$
$\displaystyle \int_{-1}^{2}\big[(y^3-2)-y\big]dy$
$\displaystyle \int_{-1}^{2}\big[(y^2+2)-y\big]dy$

Explanation: This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is $x = y^3 + 2$ and the left curve is $x = y$ , so the integrand is $(y^3 + 2) - y$ . The limits of integration are from $y = -1$ to $y = 2$ , as specified. A tempting distractor is choice B, which reverses the order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.

Question 18

Find the area setup (with respect to $y$ ) between $x=y^2-1$ (left) and $x=3-y$ (right) for $-1 \le y \le 2$ .

$\displaystyle \int_{-1}^{2}\big[(3-y)-(y^2-1)\big]\,dy$ (correct answer)
$\displaystyle \int_{-1}^{2}\big[(y^2-1)-(3-y)\big]\,dy$
$\displaystyle \int_{-2}^{1}\big[(3-y)-(y^2-1)\big]\,dy$
$\displaystyle \int_{-1}^{2}\big[(3+y)-(y^2-1)\big]\,dy$
$\displaystyle \int_{-1}^{2}\big[(3-y)-(y^2+1)\big]\,dy$

Explanation: This problem tests your ability to set up area integrals with respect to y by identifying which function is on the right versus left. When integrating with respect to y, we use the formula (right function) - (left function), where "right" means larger x-values and "left" means smaller x-values. The problem explicitly tells us that $x = y^2 - 1$ is on the left and $x = 3 - y$ is on the right, so we need $(3 - y) - (y^2 - 1)$ in our integrand. Choice B reverses this subtraction, which would give a negative area—a common mistake when students forget the right-minus-left rule. To avoid confusion, always sketch the curves or evaluate at a test point: at $y = 0$ , the left curve gives $x = -1$ and the right curve gives $x = 3$ , confirming that $3 - y$ is indeed to the right.

Question 19

Set up the area (in $y$ ) between $x=2-y$ (left) and $x=\sqrt{4-y}$ (right) for $0\le y\le 3$ .

$\displaystyle \int_{0}^{3}\big[(2-y)-\sqrt{4-y}\big]\,dy$
$\displaystyle \int_{0}^{3}\big[\sqrt{4-y}-(2-y)\big]\,dy$ (correct answer)
$\displaystyle \int_{0}^{4}\big[\sqrt{4-y}-(2-y)\big]\,dy$
$\displaystyle \int_{-3}^{0}\big[\sqrt{4-y}-(2-y)\big]\,dy$
$\displaystyle \int_{0}^{3}\big[\sqrt{4+y}-(2-y)\big]\,dy$

Explanation: This problem tests your ability to construct area integrals with respect to y using the fundamental right-minus-left principle. For dy integration, we always compute (right function) - (left function) where right indicates the function with larger x-values at each y-level. Given that x = $2 - y$ is on the left and x = $\sqrt{4 - y}$ is on the right, the correct integrand is $\sqrt{4 - y} - (2 - y)$ . Choice A reverses this to $(2 - y) - \sqrt{4 - y}$ , which would produce negative area—a frequent error when students mechanically subtract without considering relative positions. To confirm the ordering, evaluate at $y = 0$ : the left curve gives $x = 2$ while the right curve gives $x = 2$ as well, but at $y = 3$ : left gives $x = -1$ while right gives $x = 1$ , verifying that $\sqrt{4 - y}$ is indeed rightmost.

Question 20

An integral for the area between $x=\frac{1}{y+1}$ (right) and $x=0$ (left) on $1\le y\le 3$ is

$\displaystyle \int_{1}^{3}\left[0-\frac{1}{y+1}\right]dy$
$\displaystyle \int_{3}^{1}\left[\frac{1}{y+1}-0\right]dy$
$\displaystyle \int_{1}^{3}\left[\frac{1}{y+1}-0\right]dy$ (correct answer)
$\displaystyle \int_{1}^{3}\left[\frac{1}{y-1}-0\right]dy$
$\displaystyle \int_{1}^{3}\left[\frac{1}{1-y}-0\right]dy$

Explanation: This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = 1/(y + 1) and the left curve is x = 0, so the integrand is 1/(y + 1) - 0. The limits of integration are from y = 1 to y = 3, as specified. A tempting distractor is choice A, which subtracts in the wrong order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.

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