Calculus 1 Quiz: Area Between Curves Y Functions

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Area Between Curves Y FunctionsQuestion 1 of 20

Find the area of the region enclosed by the curves $x = y^4$ and $x = 8y$ .

\frac{24}{5}

\frac{48}{5}

\frac{128}{5}

\frac{64}{5}

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Calculus 1 Quiz

Calculus 1 Quiz: Area Between Curves Y Functions

Practice Area Between Curves Y Functions in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Area Between Curves Y Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Question 1

Find the area of the region enclosed by the curves $x = y^4$ and $x = 8y$ .

$\frac{24}{5}$
$\frac{48}{5}$ (correct answer)
$\frac{128}{5}$
$\frac{64}{5}$

Explanation: Find the intersection points by setting

y^4 = 8y

. This gives

y^4 - 8y = 0

, or

y(y^3 - 8) = 0

. The solutions are

y=0

and

y=2

. These are the limits of integration. For

y

in the interval

(0, 2)

, choose a test point like

y=1

. We have

x = 1^4 = 1

and

x = 8(1) = 8

. Since

8 > 1

, the line

x=8y

is the right curve. The area is

A = \int_{0}^{2} (8y - y^4) \,dy

. Evaluate the integral:

A = [4y^2 - \frac{y^5}{5}]_0^2 = (4(2^2) - \frac{2^5}{5}) - 0 = (16 - \frac{32}{5}) = \frac{80 - 32}{5} = \frac{48}{5}

Question 2

What is the area of the region bounded by the curves $x = e^y$ , $x = 1/e$ , and the lines $y = -1$ and $y = 1$ ?

$e - \frac{2}{e}$
$e - \frac{1}{e}$
$e + \frac{1}{e}$
$e - \frac{3}{e}$ (correct answer)

Explanation: The region is defined by the vertical boundaries

y=-1

and

y=1

. We need to determine the right and left curves,

x_{right}

and

x_{left}

. The curves are

x = e^y

and

x=1/e

. Note that at

y=-1

e^y = e^{-1} = 1/e

. For

y > -1

e^y > 1/e

. Therefore,

x=e^y

is the right curve throughout the interval

[-1, 1]

. The area is given by

A = \int_{-1}^{1} (e^y - \frac{1}{e}) \,dy

. Evaluating the integral:

A = [e^y - \frac{y}{e}]_{-1}^{1} = (e^1 - \frac{1}{e}) - (e^{-1} - \frac{-1}{e}) = (e - \frac{1}{e}) - (\frac{1}{e} + \frac{1}{e}) = e - \frac{1}{e} - \frac{2}{e} = e - \frac{3}{e}

Question 3

A region in the xy-plane is bounded by $x = y^2 + 1$ , $x = 5$ , $y = -1$ , and $y = 2$ . What is the area of this region?

$8$
$9$ (correct answer)
$10$
$12$

Explanation: The integration is with respect to

y

from

y=-1

y=2

. We must determine the right and left boundary curves. The boundaries are

x = y^2 + 1

and

x=5

. The curves intersect when

y^2+1=5

, which is at

y=\pm 2

. Since the interval of integration is

[-1, 2]

, and for any

y

in this interval

y^2+1 \le 5

, the line

x=5

is the right boundary and the parabola

x=y^2+1

is the left boundary. The area is

A = \int_{-1}^{2} (5 - (y^2 + 1)) \,dy = \int_{-1}^{2} (4 - y^2) \,dy

. Evaluating the integral:

A = [4y - \frac{y^3}{3}]_{-1}^{2} = (4(2) - \frac{2^3}{3}) - (4(-1) - \frac{(-1)^3}{3}) = (8 - \frac{8}{3}) - (-4 + \frac{1}{3}) = \frac{16}{3} - (-\frac{11}{3}) = \frac{16+11}{3} = \frac{27}{3} = 9

Question 4

The area of the region bounded by $x=y^2-4y$ and $x=2y-y^2$ is equal to which of the following?

$4$
$9$ (correct answer)
$12$
$18$

Explanation: First, find the points of intersection by setting the two expressions for

x

equal:

y^2 - 4y = 2y - y^2

. This simplifies to

2y^2 - 6y = 0

, or

2y(y-3) = 0

, which gives

y=0

and

y=3

as the limits of integration. To determine the right and left curves, pick a test point in

(0, 3)

, such as

y=1

. For

x = y^2 - 4y

x = 1-4 = -3

. For

x = 2y - y^2

x = 2-1 = 1

. Since

1 > -3

x=2y-y^2

is the right curve. The area is

A = \int_{0}^{3} ((2y - y^2) - (y^2 - 4y)) \,dy = \int_{0}^{3} (6y - 2y^2) \,dy

. Evaluating the integral:

A = [3y^2 - \frac{2y^3}{3}]_0^3 = (3(3^2) - \frac{2(3^3)}{3}) - 0 = (27 - 18) = 9

Question 5

A region's area is given by the integral $\int_{-1}^{3} (2y+3 - y^2) \,dy$ . Which pair of curves bounds this region?

$x=y^2$ and $x=2y+3$ (correct answer)
$y=x^2$ and $y=2x+3$
$x=y^2-2y$ and $x=3$
$x=y^2-3$ and $x=2y$

Explanation: The integral is of the form

\int_c^d (x_{right} - x_{left}) \,dy

. The integrand is

(2y+3) - y^2

. This suggests that the right curve is

x_{right} = 2y+3

and the left curve is

x_{left} = y^2

. To verify this, we check if the limits of integration,

y=-1

and

y=3

, are the intersection points of these two curves. Set

y^2 = 2y+3

, which gives

y^2 - 2y - 3 = 0

. This factors as

(y-3)(y+1) = 0

, yielding

y=3

and

y=-1

. These match the limits of integration. Thus, the curves are

x=y^2

and

x=2y+3

Question 6

A region is bounded by the curve $x=4-y^2$ and the line $x=-5$ . Which of the following integrals represents the area of this region?

$\int_{-3}^{3} (9-y^2) dy$ (correct answer)
$\int_{-2}^{2} (9-y^2) dy$
$\int_{-5}^{4} 2\sqrt{4-x} dx$
$\int_{-3}^{3} (-1-y^2) dy$

Explanation: When finding the area between curves, you need to carefully identify the region's boundaries and set up your integral with the correct limits and integrand. First, let's visualize this region. The curve

x = 4 - y^2

is a parabola opening leftward with vertex at (4, 0). The line

x = -5

is vertical. To find where they could intersect, set

4 - y^2 = -5

, giving

y^2 = 9

, so

y = \pm 3

. However, since the parabola has its rightmost point at

x = 4

and the line is at

x = -5

, they don't actually intersect. The region is bounded by the parabola on the right and extends leftward to the line

x = -5

. For the area calculation using horizontal strips (integrating with respect to

y

), the width of each strip runs from

x = -5

x = 4 - y^2

. This gives a width of

(4 - y^2) - (-5) = 9 - y^2

. The parabola extends from

y = -3

y = 3

, so the area is

\int_{-3}^{3} (9 - y^2) dy

. This matches choice A. Choice B uses the wrong limits (

-2

to 2 instead of

-3

to 3). Choice C attempts integration with respect to

x

, but the setup is incorrect for this region. Choice D has the wrong integrand (

-1 - y^2

instead of

9 - y^2

), which would give a negative area. Strategy tip: Always sketch the region first, identify intersection points to determine limits, and ensure your integrand represents the actual distance between boundaries.

Question 7

Find the area of the region bounded by the curve $x = \tan(y)$ , the y-axis, and the line $y = \pi/3$ .

$\ln(2)$ (correct answer)
$\ln(\sqrt{3})$
$\ln(1/2)$
$1$

Explanation: The region is bounded by

x=\tan(y)

(right curve, since

\tan(y) \ge 0

for

y \in [0, \pi/3]

x=0

(the y-axis, left curve), and the line

y=\pi/3

. The lower bound is implicitly

y=0

since that's where

x=\tan(y)

intersects the y-axis. The area is

A = \int_{0}^{\pi/3} (\tan(y) - 0) \,dy

. The antiderivative of

\tan(y)

-\ln|\cos(y)|

. Evaluating the definite integral:

A = [-\ln|\cos(y)|]_0^{\pi/3} = -\ln|\cos(\pi/3)| - (-\ln|\cos(0)|) = -\ln(1/2) + \ln(1)

. Since

\ln(1)=0

and

-\ln(1/2) = -(\ln(1) - \ln(2)) = \ln(2)

, the area is

\ln(2)

Question 8

To find the area of the region bounded by $y = \arcsin(x)$ , $y=\pi/2$ , and the y-axis, the most effective method is to integrate with respect to $y$ . What is the area?

$1$ (correct answer)
$\frac{\pi}{2}$
$2$
$\frac{\pi}{2} - 1$

Explanation: The equation

y = \arcsin(x)

can be rewritten as

x = \sin(y)

. The region is bounded by

x=\sin(y)

, the y-axis (

x=0

), and the line

y=\pi/2

. The lower bound is where

x=\sin(y)

intersects the y-axis, which is at

y=0

. For

y \in [0, \pi/2]

\sin(y) \ge 0

, so

x=\sin(y)

is the right curve and

x=0

is the left curve. The area is

A = \int_{0}^{\pi/2} (\sin(y) - 0) \,dy

. Evaluating the integral:

A = [-\cos(y)]_0^{\pi/2} = -\cos(\pi/2) - (-\cos(0)) = -0 - (-1) = 1

Question 9

Let $R$ be the region in the first quadrant bounded by the curve $x=y^2$ , the y-axis, and the line $y=k$ for some constant $k>0$ . The vertical line $x=c$ divides the region $R$ into two sub-regions of equal area. Which of the following is the value of $c$ ?

$\frac{k^2}{2}$
$\frac{k^2}{4}$ (correct answer)
$k^2(1 - \frac{1}{\sqrt{2}})$
$\frac{k^2}{3}$

Explanation: The area of the entire region

R

can be found by integrating with respect to

y

A_{total} = \int_0^k y^2 \,dy = [\frac{y^3}{3}]_0^k = \frac{k^3}{3}

. The line

x=c

divides this region into two parts of equal area, so each part has area

\frac{1}{2}A_{total} = \frac{k^3}{6}

. It is easier to calculate the area by integrating with respect to

x

. First, rewrite the boundary curve as

y=\sqrt{x}

. The region

R

is bounded by

y=\sqrt{x}

y=k

, and

x=0

. The total area is

A_{total} = \int_0^{k^2} (k - \sqrt{x}) \,dx = [kx - \frac{2}{3}x^{3/2}]_0^{k^2} = k(k^2) - \frac{2}{3}(k^2)^{3/2} = k^3 - \frac{2}{3}k^3 = \frac{k^3}{3}

. The area of the left sub-region, from

x=0

x=c

, is

A_{left} = \int_0^c (k - \sqrt{x}) \,dx = [kx - \frac{2}{3}x^{3/2}]_0^c = kc - \frac{2}{3}c^{3/2}

. We set this equal to half the total area:

kc - \frac{2}{3}c^{3/2} = \frac{k^3}{6}

. By inspection or by substitution, we can test the options. Let's test

c = k^2/4

k(\frac{k^2}{4}) - \frac{2}{3}(\frac{k^2}{4})^{3/2} = \frac{k^3}{4} - \frac{2}{3}\frac{k^3}{8} = \frac{k^3}{4} - \frac{k^3}{12} = \frac{3k^3 - k^3}{12} = \frac{2k^3}{12} = \frac{k^3}{6}

. This is correct.

Question 10

Which of the following integrals represents the area of the region enclosed by the curves defined by the equations $x + y^2 = 3$ and $x + y = 1$ ?

$\int_{-1}^{2} (2 + y - y^2) \,dy$ (correct answer)
$\int_{-1}^{2} (y^2 - y - 2) \,dy$
$\int_{-1}^{3} (\sqrt{3-x} - (1-x)) \,dx$
$\int_{-2}^{1} (2 + y - y^2) \,dy$

Explanation: First, express

x

in terms of

y

for both equations:

x = 3 - y^2

and

x = 1 - y

. To find the limits of integration, set the expressions for

x

equal to each other:

3 - y^2 = 1 - y

, which simplifies to

y^2 - y - 2 = 0

, or

(y-2)(y+1) = 0

. The points of intersection occur at

y = -1

and

y = 2

. These are the limits of integration. To determine which function is on the right, test a value between -1 and 2, such as

y=0

. For

y=0

x = 3 - 0^2 = 3

and

x = 1 - 0 = 1

. Since

3 > 1

, the curve

x = 3 - y^2

is the right boundary (

x_{right}

) and

x = 1 - y

is the left boundary (

x_{left}

). The area is given by the integral

A = \int_{c}^{d} (x_{right} - x_{left}) \,dy

. Therefore, the area is

\int_{-1}^{2} ((3 - y^2) - (1 - y)) \,dy = \int_{-1}^{2} (2 + y - y^2) \,dy

Question 11

The region enclosed by the parabola $x = y^2$ and the line $x = y + 2$ is considered. Which integral correctly computes its area?

$\int_{-1}^{2} (y^2 - y - 2) \,dy$
$\int_{-2}^{1} (2 + y - y^2) \,dy$
$\int_{-1}^{2} (2 + y - y^2) \,dy$ (correct answer)
$\int_{0}^{4} (\sqrt{x} - (x-2)) \,dx$

Explanation: To set up the integral for the area, first find the points of intersection by setting the expressions for

x

equal:

y^2 = y + 2

. This gives the quadratic equation

y^2 - y - 2 = 0

, which factors as

(y-2)(y+1) = 0

. The y-values of the intersection points are

y = -1

and

y = 2

, which will be our limits of integration. Next, determine which function gives the greater

x

-value (the right boundary) in the interval

(-1, 2)

. We can use a test value, for instance

y=0

. For

x=y^2

, we get

x=0

. For

x=y+2

, we get

x=2

. Since

2 > 0

, the line

x=y+2

is the right boundary. The area is given by

\int_{c}^{d} (x_{right} - x_{left}) \,dy

. Therefore, the integral is

\int_{-1}^{2} ((y+2) - y^2) \,dy = \int_{-1}^{2} (2 + y - y^2) \,dy

Question 12

Consider the region in the first quadrant enclosed by the y-axis and the curve $x = 9 - y^2$ . What is the area of the portion of this region for which $x \ge 5$ ?

$\frac{16}{3}$ (correct answer)
$8$
$\frac{32}{3}$
$12$

Explanation: We are looking for the area in the first quadrant, so

y \ge 0

. The region is bounded by

x=9-y^2

and is restricted to where

x \ge 5

. This means the left boundary is the line

x=5

. To find the y-limits, we find the intersection of

x=9-y^2

and

x=5

, which is

5 = 9-y^2 \implies y^2=4 \implies y=\pm 2

. Since we are in the first quadrant, the y-limits are from 0 to 2. The area is

A = \int_{0}^{2} ((9 - y^2) - 5) \,dy = \int_{0}^{2} (4 - y^2) \,dy

. Evaluating the integral:

A = [4y - \frac{y^3}{3}]_0^2 = (4(2) - \frac{2^3}{3}) - 0 = 8 - \frac{8}{3} = \frac{16}{3}

Question 13

The area between the curves $x=f(y)$ and $x=g(y)$ from $y=a$ to $y=b$ is correctly given by the integral $\int_a^b |f(y)-g(y)| \,dy$ . Which of the following conditions is NOT strictly necessary for this formula to be valid?

$f(y)$ and $g(y)$ are continuous functions on the interval $[a, b]$ .
The curves $x=f(y)$ and $x=g(y)$ must intersect at $y=a$ and $y=b$ . (correct answer)
The integral represents a non-negative value corresponding to the total area.
The value $b$ must be greater than or equal to the value $a$ .

Explanation: The formula for the area between two curves

x=f(y)

and

x=g(y)

on an interval

[a, b]

requires the functions to be continuous so the integral is well-defined. The use of absolute value ensures the integrand

|f(y)-g(y)|

is non-negative, correctly representing the distance between the curves, and thus the integral gives a non-negative total area. The limits of integration

a

and

b

define the interval, and it is standard convention that

a \le b

. However, it is not necessary for the curves to intersect at the endpoints of the interval. The region can be bounded by the lines

y=a

and

y=b

without the curves intersecting at these y-values.

Question 14

A region is bounded by the curves $x = y^3$ , $x = y + 6$ , and the $y$ -axis. What is the area of this region?

$10$
$18$
$24$
$28$ (correct answer)

Explanation: The region's boundaries are defined by

x=y^3

x=y+6

, and

x=0

. We must find the points where these curves intersect.

x=y^3

and

x=y+6

intersect when

y^3=y+6

, which gives

y=2

. The line

x=y+6

intersects the

y

-axis (

x=0

) at

y=-6

. The curve

x=y^3

intersects the

y

-axis at

y=0

. The overall region is split into two parts by the

x

-axis (at

y=0

). For

y \in [-6, 0]

, the region is bounded by

x=y+6

on the right and

x=0

on the left. For

y \in [0, 2]

, the region is bounded by

x=y+6

on the right and

x=y^3

on the left. We must set up two integrals. Area 1 (

y \in [-6, 0]

A_1 = \int_{-6}^{0} (y+6 - 0) dy = \left[ \frac{1}{2}y^2 + 6y \right]_{-6}^{0} = 0 - (\frac{36}{2} - 36) = -(-18) = 18

Area 2 (

y \in [0, 2]

A_2 = \int_{0}^{2} (y+6 - y^3) dy = \left[ \frac{1}{2}y^2 + 6y - \frac{1}{4}y^4 \right]_{0}^{2} = (\frac{4}{2} + 12 - \frac{16}{4}) - 0 = 2 + 12 - 4 = 10

Total Area

A = A_1 + A_2 = 18 + 10 = 28

. Distractor A (

10

) is the area of only the upper part of the region. Distractor B (

18

) is the area of only the lower part. Distractor C (

24

) could result from an integration error, such as using an antiderivative of

\frac{1}{2}y^4

instead of

\frac{1}{4}y^4

Question 15

Which of the following expressions represents the area of the region in the first quadrant bounded by the curves $x = y^2$ , $x = 8 - y^2$ , and the $y$ -axis?

$\int_{0}^{4} (\sqrt{x} - \sqrt{8-x}) dx$
$\int_{0}^{2\sqrt{2}} (8 - 2y^2) dy$
$\int_{0}^{2} y^2 dy + \int_{2}^{2\sqrt{2}} (8 - y^2) dy$ (correct answer)
$\int_{0}^{4} y^2 dy + \int_{4}^{8} (8 - y^2) dy$

Explanation: The region is in the first quadrant (

x \ge 0, y \ge 0

), bounded on the left by the

y

-axis (

x=0

). The other boundaries are

x = y^2

(a parabola opening right) and

x = 8 - y^2

(a parabola opening left). We find the intersection of these two parabolas:

y^2 = 8 - y^2 \implies 2y^2 = 8 \implies y^2 = 4

. In the first quadrant, this gives

y=2

(at which point

x=4

). The parabola

x=8-y^2

intersects the

y

-axis (

x=0

) at

y=\sqrt{8}=2\sqrt{2}

. The structure of the right-hand boundary of the region changes. For

y

-values from

0

2

, the right boundary is the parabola

x=y^2

. For

y

-values from

2

2\sqrt{2}

, the right boundary is the parabola

x=8-y^2

. Therefore, the total area must be expressed as the sum of two integrals: Area = (Area from

y=0

y=2

) + (Area from

y=2

y=2\sqrt{2}

). This corresponds to the expression

\int_{0}^{2} (y^2 - 0) dy + \int_{2}^{2\sqrt{2}} ((8 - y^2) - 0) dy

. Choice A is an incorrect attempt to set up the integral with respect to

x

. Choice B represents the area between the two parabolas, ignoring the

y

-axis boundary and the split nature of the region. Choice D incorrectly uses

x

-values as limits for an integral with respect to

y

Question 16

What is the area of the finite region bounded by the curves $x = 2y^2$ and $x = 3y + 2$ ?

$\frac{33}{8}$
$\frac{65}{6}$
$\frac{125}{24}$ (correct answer)
$\frac{145}{24}$

Explanation: First, find the intersection points by setting the expressions for

x

equal:

2y^2 = 3y + 2

. This gives the quadratic equation

2y^2 - 3y - 2 = 0

. Factoring yields

(2y + 1)(y - 2) = 0

, so the intersection points occur at

y = -1/2

and

y = 2

. These are the limits of integration. To determine the right and left functions, test a value between the limits, like

y=0

. For the parabola

x = 2y^2

x=0

. For the line

x = 3y + 2

x=2

. The line is the right-hand boundary. The area is given by the integral:

A = \int_{-1/2}^{2} ((3y + 2) - 2y^2) dy

A = \left[ \frac{3}{2}y^2 + 2y - \frac{2}{3}y^3 \right]_{-1/2}^{2}

Evaluate at the upper limit:

\frac{3}{2}(2)^2 + 2(2) - \frac{2}{3}(2)^3 = 6 + 4 - \frac{16}{3} = 10 - \frac{16}{3} = \frac{14}{3}

. Evaluate at the lower limit:

\frac{3}{2}(-\frac{1}{2})^2 + 2(-\frac{1}{2}) - \frac{2}{3}(-\frac{1}{2})^3 = \frac{3}{8} - 1 - \frac{2}{3}(-\frac{1}{8}) = \frac{3}{8} - 1 + \frac{1}{12} = \frac{9-24+2}{24} = -\frac{13}{24}

. The area is the difference:

A = \frac{14}{3} - (-\frac{13}{24}) = \frac{112}{24} + \frac{13}{24} = \frac{125}{24}

. Distractor D arises from incorrectly factoring the quadratic as

(2y-1)(y+2)=0

, leading to wrong integration limits. Distractor B arises from an error in the antiderivative (e.g., forgetting the

1/2

\frac{3}{2}y^2

). Distractor A results from an arithmetic error in combining the evaluated limits, such as

\frac{112}{24} - \frac{13}{24}

instead of adding.

Question 17

Find the area of the region bounded by the curve $x=y^2-2y$ and the $y$ -axis.

$\frac{4}{3}$ (correct answer)
$\frac{2}{3}$
$2$
$\frac{8}{3}$

Explanation: When finding areas bounded by curves, you need to set up the correct integral with proper limits and integrand. Since the curve is given as

x = y^2 - 2y

(a function of

y

), you should integrate with respect to

y

. First, find where the curve intersects the

y

-axis by setting

x = 0

0 = y^2 - 2y = y(y-2)

. This gives

y = 0

and

y = 2

as your integration limits. To visualize this region, complete the square:

x = y^2 - 2y = (y-1)^2 - 1

. This parabola opens rightward with vertex at

(-1, 1)

. Between

y = 0

and

y = 2

, the curve lies to the left of the

y

-axis (since

x < 0

in this interval), so the area is:

\text{Area} = \int_0^2 |y^2 - 2y| \, dy = \int_0^2 -(y^2 - 2y) \, dy = \int_0^2 (2y - y^2) \, dy

Evaluating:

\int_0^2 (2y - y^2) \, dy = \left[y^2 - \frac{y^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}

So A is correct. Option B (

\frac{2}{3}

) likely comes from forgetting the absolute value or making an arithmetic error. Option C (

2

) might result from incorrectly using

\int_0^2 2y \, dy

without the

-y^2

term. Option D (

\frac{8}{3}

) could come from using

\int_0^2 (y^2 - 2y) \, dy

without taking the absolute value. Remember: when integrating to find area, always use absolute value to ensure positive results, and choose your variable of integration based on how the curve is parameterized.

Question 18

Find the area of the region bounded by the curves $x=y^3-y$ and $x=0$ .

$0$
$\frac{1}{4}$
$1$
$\frac{1}{2}$ (correct answer)

Explanation: When finding the area between curves, you need to determine which variable to integrate with respect to and identify the correct bounds. Since this problem gives you

x = y^3 - y

and

x = 0

, it's natural to integrate with respect to

y

. First, find where the curves intersect by setting

y^3 - y = 0

. Factoring gives

y(y^2 - 1) = y(y-1)(y+1) = 0

, so the intersection points are at

y = -1, 0, 1

. Next, determine which curve is to the right. For

y \in (-1, 0)

, try

y = -0.5

x = (-0.5)^3 - (-0.5) = -0.125 + 0.5 = 0.375 > 0

. For

y \in (0, 1)

, try

y = 0.5

x = (0.5)^3 - 0.5 = 0.125 - 0.5 = -0.375 < 0

. So the curve

x = y^3 - y

is to the right of

x = 0

(-1, 0)

and to the left on

(0, 1)

. The total area is:

\int_{-1}^{0} (y^3 - y - 0) \, dy + \int_{0}^{1} (0 - (y^3 - y)) \, dy = \int_{-1}^{0} (y^3 - y) \, dy + \int_{0}^{1} (-y^3 + y) \, dy

Evaluating:

\left[\frac{y^4}{4} - \frac{y^2}{2}\right]_{-1}^{0} + \left[-\frac{y^4}{4} + \frac{y^2}{2}\right]_{0}^{1} = (0) - \left(\frac{1}{4} - \frac{1}{2}\right) + \left(-\frac{1}{4} + \frac{1}{2}\right) - (0) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}

The answer is (D) $\frac{1}{2}$ . Choice (A) ignores that area is always positive. Choice (B) likely computed only one of the two regions. Choice (C) probably made a computational error or forgot absolute values. Strategy tip: Always sketch the region and check which curve is "ahead" in each interval—the integrand must be positive for area calculations.

Question 19

Which definite integral correctly represents the area of the region enclosed by the graphs of $x = y^2 - 4y$ and $x = 2y - y^2$ ?

$\int_{0}^{3} (2y^2 - 6y) dy$
$\int_{0}^{3} (6y - 2y^2) dy$ (correct answer)
$\int_{0}^{6} (6y - 2y^2) dy$
$\int_{-4}^{1} ((2-\sqrt{4+x}) - (1-\sqrt{1-x})) dx$

Explanation: To set up the integral for the area between two curves expressed as functions of

y

, we first find their intersection points to determine the limits of integration. Set the expressions for

x

equal:

y^2 - 4y = 2y - y^2

. Rearranging gives

2y^2 - 6y = 0

, which factors as

2y(y - 3) = 0

. The points of intersection are at

y=0

and

y=3

. These will be our limits of integration. Next, we determine which function represents the right boundary and which represents the left boundary within the interval

[0, 3]

. We can pick a test point, such as

y=1

. For

x = y^2 - 4y

, we have

x = 1 - 4 = -3

. For

x = 2y - y^2

, we have

x = 2 - 1 = 1

. Since

1 > -3

, the curve

x = 2y - y^2

is the right boundary. The area integral is

\int_{y_{min}}^{y_{max}} (\text{right function} - \text{left function}) dy

. So, the integral is:

A = \int_{0}^{3} ((2y - y^2) - (y^2 - 4y)) dy = \int_{0}^{3} (6y - 2y^2) dy

Choice A reverses the functions, calculating (left - right), which would yield the negative of the area. Choice C uses an incorrect upper limit of integration, possibly from mis-solving

2y-6=0

. Choice D represents an incorrect and overly complicated attempt to set up the integral with respect to

x

Question 20

The area of the region bounded by the parabola $x = ky^2$ and the line $x=k$ is equal to 16. If $k$ is a positive constant, what is the value of $k$ ?

$8$
$12$ (correct answer)
$16$
$24$

Explanation: The region is bounded by

x=ky^2

and

x=k

. First, find the intersection points:

ky^2 = k

. Since

k>0

, we can divide by

k

to get

y^2 = 1

, so

y = \pm 1

. These are the limits of integration. For

y

(-1, 1)

y^2 < 1

, so

ky^2 < k

. Thus,

x=k

is the right boundary and

x=ky^2

is the left boundary. The area is given by the integral:

A = \int_{-1}^{1} (k - ky^2) dy = 16

We can factor out the constant

k

k \int_{-1}^{1} (1 - y^2) dy = 16

Now, evaluate the integral:

k \left[ y - \frac{y^3}{3} \right]_{-1}^{1} = 16

k \left( (1 - \frac{1}{3}) - (-1 - \frac{-1}{3}) \right) = 16

k \left( \frac{2}{3} - (-\frac{2}{3}) \right) = 16

k \left( \frac{4}{3} \right) = 16

Solving for

k

k = 16 \cdot \frac{3}{4} = 12

. Distractor D,

k=24

, results from calculating the area over half the interval,

[0,1]

, and setting that equal to 16 (i.e.,

k(2/3)=16

). Distractor C,

k=16

, might be chosen if one equates the area with the parameter

k

without performing the integration. Distractor A,

k=8

, could arise from an arithmetic or integration error.