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📝 Reversing order of integration (15 MCQs)

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

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What is Reversing order of integration?

Definition:
Reversing the order of integration involves switching from $dy \, dx$ to $dx \, dy$ (or vice versa) by re-describing the region boundaries. This is often done to simplify an integral that is difficult to evaluate in its original order.

Example:
For $\int_0^1 \int_y^1 e^{x^2} \, dx \, dy$ , reversing gives $\int_0^1 \int_0^x e^{x^2} \, dy \, dx$ , which is easily solvable since $\int e^{x^2} dx$ has no elementary antiderivative but $\int dy$ is trivial.

Reason:
Some integrands cannot be integrated in one order due to lack of elementary antiderivatives; reversing the order can transform an unsolvable problem into a straightforward calculation.

Easy

Medium

Hard

📝 All Reversing order of integration MCQs

Q1. A region is described by $0\le x\le 2$ and $x^2\le y\le 4$ . Which reversed-order description represents exactly the same region?

0\le y\le 4,\ 0\le x\le \sqrt y

✅

0\le y\le 4,\ \sqrt y\le x\le 2

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

0\le y\le 4,\ y\le x\le 2

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: When $x$ ranges from $0$ to $2$ , the lower boundary is $y=x^2$ and the upper boundary is $y=4$ . After reversing, $y$ ranges from $0$ to $4$ . For a fixed $y$ , the condition $x^2\le y$ gives $0\le x\le\sqrt y$ , which produces the complete original region without adding or removing points.

Q2. Why can reversing an iterated integral sometimes require splitting the integral into two or more integrals instead of simply exchanging the two differentials?

A.Because changing order always changes the value of the integral

B.Because a horizontal or vertical slice may cross different boundary curves in different portions of the region ✅

C.Because the integrand must always be rewritten before changing order

D.Because only rectangular regions permit reversed integration

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The order of integration describes how the region is sliced. A region that is simple with vertical slices may not be simple with horizontal slices. When a horizontal slice encounters different boundary curves at different values of the variable, the limits change, so the region must be partitioned before writing the reversed integral.

Q3. Consider the integral $\int_0^1\int_x^1 f(x,y)\,dy\,dx$ . A student says that reversing the order gives $\int_0^1\int_0^y f(x,y)\,dx\,dy$ . Which reasoning best justifies the student's result?

A.The original inner limit

x

becomes the new outer limit

B.The inequalities

0\le x\le1

and

x\le y\le1

imply

0\le y\le1

and

0\le x\le y

✅

C.The variables can always be interchanged without examining the region

D.The upper limit

1

must become

y

after reversal

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The original bounds describe the region satisfying $0\le x\le1$ and $x\le y\le1$ . Rewriting these inequalities in terms of horizontal slices gives $0\le y\le1$ , while $x$ runs from $0$ to $y$ . Thus the student's reversed integral follows from a correct geometric reinterpretation rather than merely swapping symbols.

Q4. A double integral has bounds $0\le y\le x$ and $0\le x\le1$ . Which reversed description is most appropriate if the goal is to make the inner integral have constant limits?

0\le y\le1,\ y\le x\le1

✅

0\le y\le1,\ 0\le x\le y

0\le x\le1,\ 0\le y\le x

0\le y\le x\le1

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The inequalities are $0\le x\le1$ and $0\le y\le x$ . Since $y$ reaches a maximum of $1$ , the reversed outer range is $0\le y\le1$ . For each fixed $y$ , the condition $y\le x\le1$ determines the horizontal slice. This reversal also makes the inner limits directly reflect the region's boundaries.

Q5. An engineer models the heat over a triangular region using $\int_0^2\int_0^{2-x} T(x,y)\,dy\,dx$ . To integrate with respect to $x$ first, which equivalent formulation should be used?

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

✅

\int_0^2\int_y^2T(x,y)\,dx\,dy

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

\int_{-2}^0\int_0^{2+y}T(x,y)\,dx\,dy

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The original region satisfies $0\le x\le2$ and $0\le y\le2-x$ , which is equivalent to $x+y\le2$ in the first quadrant. Solving the boundary for $x$ gives $0\le x\le2-y$ , while $y$ ranges from $0$ to $2$ . Therefore option A preserves the exact triangular region.

Q6. A student reverses $\int_0^4\int_{\sqrt y}^{2} f(x,y)\,dx\,dy$ as $\int_0^2\int_{x^2}^{4}f(x,y)\,dy\,dx$ . What is the main issue with this transformation?

A.There is no issue; both integrals always have the same limits

B.The new outer variable should range from

0

4

C.The reversal is actually correct because

x^2\le y\le4

corresponds to

0\le x\le2

✅

D.The square root must remain in the reversed integral

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The original region is described by $0\le y\le4$ and $\sqrt y\le x\le2$ . The inequality $\sqrt y\le x$ is equivalent to $y\le x^2$ , so the region is $0\le y\le x^2$ , not $x^2\le y\le4$ . The student's reversal therefore describes a different region, making option C the correct assessment.

Q7. A student claims that $\int_0^1\int_{x}^{2-x} f(x,y)\,dy\,dx$ can be reversed directly as $\int_0^1\int_x^{2-x} f(x,y)\,dx\,dy$ . Why is this reasoning invalid?

A.The integrand changes whenever the order changes

B.The original region is not necessarily described by the same pair of boundary functions when horizontal slices are used ✅

C.The outer limits must always remain unchanged

D.The original integral is undefined

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The original bounds describe vertical slices with $0\le x\le1$ and $x\le y\le2-x$ . Horizontal slices do not generally have the same lower and upper expressions because the lines $y=x$ and $y=2-x$ must be solved for $x$ . The horizontal range is piecewise, so simply exchanging symbols does not preserve the region.

Q8. The region bounded by $y=x$ , $y=2x$ , and $x=2$ is initially described using vertical slices. When the order is reversed, what feature should a correct setup preserve?

A.Every horizontal slice must have the same two boundary formulas for all

y

B.The new limits must describe exactly the same set of points, even if the description requires splitting the region ✅

C.The outer variable must remain

x

D.The numerical value of each individual limit must remain unchanged

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Reversing order is fundamentally a geometric operation: the set of points must remain unchanged. For this region, horizontal slices encounter the lines $y=x$ and $y=2x$ differently depending on the height. A correct setup may therefore require splitting the region, but every resulting piece must collectively reproduce the original region.

Q9. Suppose a graph shows a region in the first quadrant bounded below by $y=x^2$ , above by $y=4$ , and on the left by $x=0$ . Which description correctly represents horizontal slices of the graph?

0\le y\le4,\ 0\le x\le\sqrt y

✅

0\le y\le4,\ y^2\le x\le2

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

0\le y\le4,\ \sqrt y\le x\le2

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The graph is enclosed by the $y$ -axis, the parabola $y=x^2$ , and the horizontal line $y=4$ . At a fixed height $y$ , points begin at $x=0$ and extend to the parabola where $x=\sqrt y$ . Since the highest point has $y=4$ , the horizontal description is exactly option A.

Q10. A manufacturing region is bounded by $y=x$ , $y=3-x$ , and the $x$ -axis. A cost function is $C(x,y)=x+2y$ . Which setup correctly reverses the order of integration for computing total cost?

\int_0^1\int_0^y(x+2y)\,dx\,dy+\int_1^3\int_0^{3-y}(x+2y)\,dx\,dy

✅

\int_0^1\int_y^{3-y}(x+2y)\,dx\,dy

\int_0^3\int_0^y(x+2y)\,dx\,dy

\int_0^1\int_0^{3-y}(x+2y)\,dx\,dy+\int_1^3\int_0^y(x+2y)\,dx\,dy

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The triangular region has vertices $(0,0)$ , $(1,1)$ , and $(0,3)$ . For $0\le y\le1$ , the right boundary is $x=y$ . For $1\le y\le3$ , the right boundary is $x=3-y$ . Thus reversing the order requires two horizontal pieces, exactly as shown in option A.

Q11. Which strategy is generally most reliable when reversing the order of integration for a complicated region?

A.Swap

dx

and

dy

and keep all limits unchanged

B.Ignore the integrand and calculate only the original integral

C.Translate the original bounds into inequalities, sketch or visualize the region, then determine new slices and split the region if necessary ✅

D.Change every lower limit into an upper limit

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The safest approach is to treat the bounds as a description of a geometric region rather than as symbols to be mechanically exchanged. Converting the bounds into inequalities reveals the boundaries, while a sketch or careful visualization shows how new slices intersect them. If the boundary changes, the region should be split.

Q12. An analyst wants to evaluate $\int_0^1\int_{x^2}^{1} e^{y^3}\,dy\,dx$ . The inner integral has no elementary antiderivative. Which approach is most useful for simplifying the calculation?

A.Reverse the order so that the

x

-integration becomes elementary ✅

B.Replace

e^{y^3}

e^y

before integrating

C.Differentiate the integrand with respect to

x

D.Reverse the order only after numerically evaluating the inner integral

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The original region is $0\le x\le1$ and $x^2\le y\le1$ . Reversing gives $0\le y\le1$ and $0\le x\le\sqrt y$ . The new inner integral is $\int_0^{\sqrt y}dx=\sqrt y$ , leaving a one-variable integral in $y$ . Thus changing the order avoids the difficult non-elementary antiderivative.

Q13. For the region satisfying $0\le x\le1$ and $x^2\le y\le x$ , which reversed-order setup is correct?

\int_0^1\int_y^{\sqrt y} f(x,y)\,dx\,dy

✅

\int_0^1\int_{\sqrt y}^{y} f(x,y)\,dx\,dy

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

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

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The region lies between $y=x^2$ and $y=x$ for $0\le x\le1$ . Both curves meet at $0$ and $1$ , so $y$ ranges from $0$ to $1$ . Solving the boundaries for $x$ gives $x=\sqrt y$ and $x=y$ . Since $0\le y\le\sqrt y$ on this interval, the correct horizontal slice runs from $x=y$ to $x=\sqrt y$ .

Q14. A student argues that because $\int\int_R f(x,y)\,dA$ is unchanged by the order of integration, the bounds can always be reversed by merely switching $dx,dy$ to $dy,dx$ . What is the strongest objection?

A.The integrand must always be symmetric

B.The equality concerns the same region and requires the new limits to describe that region correctly; merely switching differentials may describe a different region ✅

C.Double integrals can never be reversed

D.The order of integration changes the region automatically

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Changing the order of integration does not mean mechanically swapping the differentials and retaining the old limits. The mathematical equality applies when both iterated integrals represent the same two-dimensional region. New limits must therefore be derived from the geometry or inequalities. Otherwise, the resulting integral may represent an entirely different set.

Q15. A region is bounded by $y=x^2$ , $y=2x$ , and $x=0$ . Without directly evaluating any integral, which reversed description is correct and why?

0\le y\le4,\ \frac y2\le x\le\sqrt y

✅

0\le y\le4,\ \sqrt y\le x\le\frac y2

0\le y\le2,\ \frac y2\le x\le\sqrt y

0\le y\le4,\ 0\le x\le\frac y2

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The curves $y=x^2$ and $y=2x$ intersect when $x^2=2x$ , giving $x=0$ and $x=2$ , so the maximum $y$ is $4$ . For a fixed $y$ , $y=x^2$ gives $x=\sqrt y$ , while $y=2x$ gives $x=y/2$ . On $0\le y\le4$ , $y/2\le\sqrt y$ , so the horizontal slice is exactly option A.

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