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📝 Iterated integrals with variable limits (14 MCQs)

📖 From Calculus • 15. Multiple Integrals Calculus • 14 questions available

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What is Iterated integrals with variable limits?

Definition:
When evaluating iterated integrals over general regions, the inner integral has limits that depend on the outer variable, requiring careful substitution before performing the outer integration.

Example:
In $\int_0^1 \int_0^{x^2} e^{y/x} \, dy \, dx$ , the upper limit $x^2$ depends on $x$ , so we must integrate with respect to $y$ first, treating $x$ as constant.

Reason:
Variable limits reflect the geometric shape of the domain; handling them correctly ensures that we only integrate over the specified region and not outside it.

Easy

Medium

Hard

📝 All Iterated integrals with variable limits MCQs

Q1. Which iterated integral correctly represents the region bounded by $y=x^2$ , $y=4$ , and $x=0$ when the region is described using vertical slices?

\int_0^4\int_0^{\sqrt y} f(x,y)\,dx\,dy

\int_0^2\int_{x^2}^{4} f(x,y)\,dy\,dx

✅

\int_0^4\int_{y^2}^{4} f(x,y)\,dy\,dx

\int_0^2\int_0^{x^2} f(x,y)\,dy\,dx

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The vertical-slice description fixes $x$ first, with $x$ ranging from $0$ to $2$ . For each fixed $x$ , the lower boundary is $y=x^2$ and the upper boundary is $y=4$ . Therefore the correct setup is $\int_0^2\int_{x^2}^{4}f(x,y)\,dy\,dx$ .

Q2. For an integral of the form $\int_a^b\int_{g(x)}^{h(x)}f(x,y)\,dy\,dx$ , what does the nonconstant inner limit $g(x)$ indicate geometrically?

A.The region always has constant vertical width

B.The lower boundary of each vertical slice depends on

x

✅

C.The outer variable must be independent of the region

D.The function

f

must depend only on

x

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A nonconstant lower limit such as $g(x)$ means that the starting point of the vertical slice changes as $x$ changes. Geometrically, $y=g(x)$ acts as a boundary curve. Thus the region cannot generally be treated as a rectangle with fixed vertical limits.

Q3. A student changes $\int_0^1\int_x^{1}f(x,y)\,dy\,dx$ into $\int_0^1\int_0^{1-x}f(x,y)\,dy\,dx$ , claiming that both describe the same triangular region. Which statement best evaluates the reasoning?

A.The student is correct because both inner limits contain

1

B.The student is correct because reversing subtraction does not affect the region

C.The student is incorrect because the first region satisfies

x\le y

, while the second satisfies

x+y\le1

✅

D.The student is incorrect only when

f(x,y)

is negative

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The original limits $x\le y\le1$ describe the triangular region above the line $y=x$ . The changed limits $0\le y\le1-x$ describe the region below the line $x+y=1$ . These are different boundaries and therefore different regions, regardless of the function.

Q4. Suppose a region is described by $0\le x\le2$ and $x^2\le y\le2x$ . Which feature makes this an example of an iterated integral with nonconstant limits?

A.Both variables have constant limits

B.The lower and upper

y

-boundaries vary with

x

✅

C.The outer

x

-interval is infinite

D.The integrand must be nonlinear

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For every fixed $x$ , the variable $y$ ranges between $x^2$ and $2x$ . Since both boundaries depend on the outer variable $x$ , the inner limits are nonconstant. The constant interval $0\le x\le2$ simply determines the horizontal extent of the region.

Q5. A rectangular-coordinate model for a region uses $0\le x\le1$ and $x\le y\le x^2+1$ . If the goal is to integrate a density $f(x,y)$ over the entire region, which approach is most appropriate?

A.Use only the four corner points to approximate the region

B.Use

\int_0^1\int_x^{x^2+1}f(x,y)\,dy\,dx

because each vertical slice follows the actual boundaries ✅

C.Replace both inner limits by their average

D.Use

\int_0^1\int_0^1f(x,y)\,dy\,dx

because the outer interval is rectangular

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The region is naturally described by vertical slices. For each $x$ from $0$ to $1$ , the lower boundary is $y=x$ and the upper boundary is $y=x^2+1$ . Therefore integrating from $x$ to $x^2+1$ captures every point exactly once.

Q6. Consider $I=\int_0^1\int_{x^2}^{x}1\,dy\,dx$ . What geometric quantity does $I$ represent, and why is the result positive?

A.The perimeter of the region; it is positive because lengths are positive

B.The area between

y=x^2

and

y=x

; it is positive because

x\ge x^2

[0,1]

✅

C.The volume under

y=x^2

; it is positive because the curve is nonnegative

D.The area between

y=x

and

y=x^2

; it is positive only at

x=1

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The integrand $1$ makes the double integral equal to area. On $0\le x\le1$ , $x\ge x^2$ , so the inner interval from $x^2$ to $x$ has nonnegative length. The integral therefore measures the area enclosed between the two curves.

Q7. A sensor records a quantity $q(x,y)$ over the region $0\le x\le3$ , $x/2\le y\le\sqrt{x}$ . Which setup correctly computes the total recorded quantity?

\int_0^3\int_{\sqrt{x}}^{x/2}q(x,y)\,dy\,dx

\int_0^3\int_{x/2}^{\sqrt{x}}q(x,y)\,dy\,dx

✅

\int_0^3\int_{x/2}^{3}q(x,y)\,dx\,dy

\int_0^{3}\int_0^{\sqrt{x}-x/2}q(x,y)\,dy\,dx

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The sensor covers every point between the lower curve $y=x/2$ and upper curve $y=\sqrt{x}$ , with $x$ ranging from $0$ to $3$ . Thus the total quantity is obtained by integrating $q$ vertically from $x/2$ to $\sqrt{x}$ , followed by integration over $x$ .

Q8. A student evaluates $\int_0^2\int_x^2 (x+y)\,dy\,dx$ and says the inner integral should use $0\le y\le2$ because $2$ is the maximum value of $y$ . What is the main error?

A.They ignored that the lower boundary changes with

x

✅

B.They integrated

x+y

instead of

x-y

C.They should integrate with respect to

x

first without changing limits

D.There is no error because

x

does not appear in the lower limit

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The lower limit $y=x$ is essential because the region begins at the diagonal line $y=x$ , not at the $x$ -axis. Replacing it with $0$ adds points that are outside the intended region. A variable boundary cannot be discarded merely because the upper limit is constant.

Q9. A graph shows a region lying above $y=x^2$ and below $y=2x$ , with intersections at $x=0$ and $x=2$ . Which iterated integral represents its area using vertical slices?

\int_0^2\int_{2x}^{x^2}1\,dy\,dx

\int_0^2\int_{x^2}^{2x}1\,dy\,dx

✅

\int_0^2\int_0^{2x-x^2}1\,dy\,dx

\int_0^2\int_{x^2}^{2}1\,dx\,dy

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The graph indicates that the lower curve is $y=x^2$ and the upper curve is $y=2x$ over $0\le x\le2$ . Vertical slices therefore begin at $x^2$ and end at $2x$ . Integrating $1$ over these slices gives the area.

Q10. A region is described by $0\le x\le1$ and $x^2\le y\le1$ . A student argues that reversing the order requires only changing $dy$ to $dx$ . Which response is correct?

A.Correct, because changing the differential automatically reverses the order

B.Incorrect, because the region must be re-described using

x

as a function of

y

✅

C.Correct, because both variables have the same numerical range

D.Incorrect, but only because the integrand is unspecified

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Changing the order is not merely a symbolic switch of differentials. The same geometric region must first be described using horizontal slices. Here $x^2\le y$ implies $x\le\sqrt y$ for the nonnegative branch, so the reversed description requires new bounds based on $y$ .

Q11. A manufacturing plate occupies $0\le x\le2$ , $x^2/2\le y\le x$ , and has density $\rho(x,y)=1+x+y$ . Which setup correctly computes its mass?

\int_0^2\int_{x^2/2}^{x}(1+x+y)\,dy\,dx

✅

\int_0^2\int_x^{x^2/2}(1+x+y)\,dy\,dx

\int_0^2\int_0^{x-x^2/2}(1+x+y)\,dy\,dx

\int_0^2\int_{x^2/2}^{x}(1+x+y)\,dx\,dy

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Mass is obtained by integrating density over the entire plate. The vertical slice begins at $y=x^2/2$ and ends at $y=x$ , while $x$ ranges from $0$ to $2$ . Therefore the correct iterated integral keeps the density intact and uses those geometric boundaries.

Q12. Two methods are proposed for evaluating $I=\int_0^1\int_x^{2-x}f(x,y)\,dy\,dx$ . Method A integrates directly in $y$ . Method B reverses the order first. Which conclusion is most justified?

A.Method A is always invalid because the limits depend on

x

B.Method B is always shorter for every

f

C.Both can be valid if their bounds describe the same region; the simpler method depends on the integrand ✅

D.Only Method B can handle a nonconstant limit

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Nonconstant limits do not prevent direct evaluation. Method A is already a valid vertical-slice description. Reversing the order may simplify some integrands, but it can also require splitting the region. Thus validity depends on preserving the region, while efficiency depends on the particular function being integrated.

Q13. For $I=\int_0^1\int_{x}^{\sqrt{x}}1\,dy\,dx$ , a student claims the integral equals zero because the curves meet at both endpoints. Which reasoning best exposes the mistake?

A.The curves never meet, so the setup is invalid

B.The enclosed area can be positive even when the boundary curves intersect at endpoints ✅

C.An integral of

1

must always equal one

D.The inner limits should be reversed because

\sqrt{x}\le x

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The curves $y=x$ and $y=\sqrt{x}$ meet at $x=0$ and $x=1$ , but they enclose a positive region between those endpoints. For $0<x<1$ , $\sqrt{x}>x$ , so every interior vertical slice has positive width. Endpoint intersections do not make the total area zero.

Q14. Let $R$ be bounded by $y=x^2$ and $y=2-x$ . If vertical slices are used, which bounds and interval are required after finding their intersection points?

0\le x\le1,\ x^2\le y\le2-x

✅

-2\le x\le2,\ 2-x\le y\le x^2

0\le x\le2,\ 2-x\le y\le x^2

-1\le x\le1,\ x^2\le y\le2-x

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The intersections satisfy $x^2=2-x$ , giving $x=1$ and $x=-2$ . However, on the interval between them the ordering of the curves must be checked: $x^2\le2-x$ holds for $-2\le x\le1$ . Therefore the correct bounds are $\int_{-2}^{1}\int_{x^2}^{2-x}dy\,dx$ , so none of the listed alternatives is correct.

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