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📝 Fubini's theorem for double integrals (14 MCQs)

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

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What is Fubini's theorem for double integrals?

Definition:
Fubini's Theorem states that if $f(x, y)$ is continuous on a rectangular region $R = [a, b] \times [c, d]$ , then $\iint_R f(x, y) \, dA = \int_a^b \int_c^d f(x, y) \, dy \, dx = \int_c^d \int_a^b f(x, y) \, dx \, dy$ .

Example:
For $f(x,y) = x+y$ on $[0,1]\times[0,1]$ , integrating first with respect to $y$ or first with respect to $x$ yields the same result: $1$ .

Reason:
This theorem guarantees that the order of integration does not affect the final value for continuous functions on rectangles, giving us flexibility to choose the easier order of calculation.

Easy

Medium

Hard

📝 All Fubini's theorem for double integrals MCQs

Q1. A function $f(x,y)$ is continuous on the rectangle $R=[0,2]\times[1,3]$ . Which conclusion is most justified when evaluating $\iint_R f(x,y)\,dA$ ?

A.Either order of integration can be used, and both iterated integrals have the same value. ✅

B.Only integration with respect to

x

first is valid.

C.Only integration with respect to

y

first is valid.

D.The value depends on which variable is integrated first.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Continuity on a rectangular region guarantees the integrability conditions needed for reversing the order of integration. Therefore, integrating first with respect to $x$ or first with respect to $y$ produces the same double-integral value. The theorem is especially useful when one order makes the calculation substantially simpler.

Q2. Suppose $f(x,y)=x^2y$ on $0\le x\le1$ and $0\le y\le2$ . A student claims that integrating with respect to $x$ first must give a different answer from integrating with respect to $y$ first because the antiderivatives look different. What is the best evaluation of this claim?

A.The claim is correct because different antiderivatives always produce different double integrals.

B.The claim is incorrect because both valid iterated integrals evaluate to the same double integral. ✅

C.The claim is correct unless

f(x,y)

is symmetric.

D.The claim is incorrect only because

x^2y

is positive.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The order of integration changes the intermediate algebra but not the value of the double integral when the required integrability conditions hold. Here $f(x,y)=x^2y$ is continuous on the rectangle, so either order is valid. Different-looking antiderivatives can still produce exactly the same final result.

Q3. A rectangular region is described by $1\le x\le4$ and $0\le y\le x-1$ . Which statement best explains why a direct application of fixed rectangular limits in both orders is inappropriate?

A.The region is not rectangular, so at least one variable generally requires limits depending on the other variable. ✅

B.The integrand must be zero on the boundary.

C.Fubini's theorem applies only to circular regions.

D.The region cannot be represented by an iterated integral.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The region is triangular rather than rectangular. Consequently, its iterated-integral description requires a variable-dependent bound, such as $0\le y\le x-1$ , or an equivalent reversed description. Fubini's theorem still permits iterated integration, but the limits must accurately represent the geometry of the region.

Q4. Consider $f(x,y)=e^{x+y}$ over $0\le x\le1$ , $0\le y\le2$ . Which strategy is most efficient for evaluating the double integral?

A.Integrate with respect to

x

first, then

y

, because the exponential separates into factors. ✅

B.Integrate only with respect to

x

, since

y

is independent.

C.Integrate only with respect to

y

, because

x

occurs in the exponent.

D.Convert immediately to polar coordinates because exponentials require polar coordinates.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The expression $e^{x+y}$ can be viewed as $e^xe^y$ , so integrating in either order is straightforward. Integrating first with respect to $x$ produces a factor involving only $y$ , after which the remaining integral is elementary. Polar coordinates would add unnecessary complexity.

Q5. A manufacturing model uses $f(x,y)=xy$ to represent a quantity distributed over $0\le x\le3$ , $0\le y\le2$ . An analyst evaluates $\int_0^3\int_0^2xy\,dy\,dx$ . Which alternative calculation must produce the same result?

\int_0^2\int_0^3xy\,dx\,dy

✅

\int_0^3\int_0^2xy\,dx\,dy

\int_0^2\int_0^2xy\,dx\,dy

\int_0^3\int_0^3xy\,dy\,dx

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The original region is the rectangle $0\le x\le3$ , $0\le y\le2$ . Reversing the order requires the outer variable $y$ to range from $0$ to $2$ , while $x$ ranges from $0$ to $3$ . Thus $\int_0^2\int_0^3xy\,dx\,dy$ represents the same double integral.

Q6. A student evaluates $\int_0^1\int_x^1 (x+y)\,dy\,dx$ and obtains $1/2$ . Which reasoning best identifies the likely issue if the correct value is checked independently?

A.The student may have treated the lower limit

y=x

as a constant when integrating with respect to

y

. ✅

B.The student must have reversed the integration order, which is never allowed.

C.The integrand cannot be integrated because it contains two variables.

D.The value must be zero because the region touches the line

y=x

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: When integrating with respect to $y$ , $x$ is treated as a constant, but the lower limit $y=x$ must remain dependent on $x$ . The region is triangular, so careless substitution of the variable-dependent limit can easily produce an incorrect value. Reversing the order is possible but requires correctly changing the limits.

Q7. A sensor measures $f(x,y)=x+y$ over the triangular region $0\le x\le1$ , $x\le y\le1$ . Which reversed-order description represents the same region?

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

✅

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

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

0\le y\le1,\;1\le x\le y

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The original inequalities are $0\le x\le1$ and $x\le y\le1$ . Geometrically, the region lies above the line $y=x$ and below $y=1$ . For a fixed $y$ , $x$ therefore ranges from $0$ to $y$ , while $y$ ranges from $0$ to $1$ .

Q8. A numerical integration program evaluates a continuous function over a rectangle by integrating first in $x$ and then in $y$ . A second program reverses the order and obtains a slightly different numerical answer because of rounding. Which conclusion is mathematically appropriate?

A.Fubini's theorem is false for numerical calculations.

B.The two exact iterated integrals must agree, while small numerical differences can arise from approximation or rounding. ✅

C.The first integration order is always more accurate.

D.The function cannot be integrated using two different orders.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a continuous function on a rectangular region, the exact iterated integrals have the same value. Numerical algorithms approximate these integrals and may introduce truncation, discretization, or rounding errors. Therefore, a small computational discrepancy does not contradict the mathematical equality guaranteed under the appropriate conditions.

Q9. A graph shows a region bounded by $y=x^2$ , $y=4$ , and $x=0$ . A student chooses $0\le y\le4$ and $0\le x\le\sqrt y$ . Why is this setup advantageous for reversing the integration order?

A.Every horizontal slice intersects the region in one continuous interval. ✅

B.It removes the need to consider the parabola.

C.It changes the region into a rectangle.

D.It makes

y=x^2

irrelevant to the calculation.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The region lies between the $y$ -axis and the parabola $x^2=y$ , up to $y=4$ . For each fixed $y$ between $0$ and $4$ , the horizontal slice runs continuously from $x=0$ to $x=\sqrt y$ . This gives a clean reversed-order description without splitting the region.

Q10. An engineer models total load by $\iint_R (2x+3y)\,dA$ , where $R$ is a rectangle. One integration order produces several algebraic terms, while the reversed order gives a shorter calculation. What should determine the engineer's choice?

A.The order should be selected based on which valid description gives the simpler or more stable calculation. ✅

B.The order must always be

dy\,dx

C.The order must always match the order in which the variables appear in the function.

D.Changing the order changes the physical quantity being modeled.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The order of integration is a computational choice when the hypotheses for changing order are satisfied. Both orders represent the same accumulated quantity over the same region. Choosing the simpler order can reduce algebraic work and potential errors, especially in applied models involving complicated bounds or integrands.

Q11. A student writes $\int_0^1\int_0^x f(x,y)\,dy\,dx=\int_0^1\int_0^y f(x,y)\,dx\,dy$ . Which assessment is most accurate?

A.The equality is valid because both expressions describe the same triangular region. ✅

B.The equality is invalid because the order of integration can never be changed.

C.The equality is valid for every function without considering integrability.

D.The equality is invalid because

x

and

y

cannot both serve as limits.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Both iterated integrals describe the triangular region bounded by $0\le y\le x\le1$ , just from opposite slicing directions. Thus the equality is justified when the integrability requirements are satisfied. The important point is that reversing order requires changing the limits to describe exactly the same geometric region.

Q12. A function is positive and continuous on a rectangular region. Two students obtain $12$ and $18$ for the same double integral after integrating in opposite orders. Which diagnostic is most useful first?

A.Check whether the bounds and treatment of the other variable as a constant were handled correctly. ✅

B.Assume both values are possible because integration order changes area.

C.Conclude that the function violates continuity.

D.Replace the double integral by a triple integral.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For a continuous function on a rectangle, correctly formed iterated integrals must agree. A discrepancy therefore strongly suggests an error in antiderivatives, substitution of limits, or the treatment of the other variable as constant. Checking the bounds and intermediate calculations is more appropriate than questioning the theorem.

Q13. Let $f(x,y)=x-y$ on the square $0\le x,y\le1$ . Without fully computing both iterated integrals, what structural observation can predict the value?

A.The region and function change sign under exchanging

x

and

y

, so contributions cancel and the double integral is zero. ✅

B.The integrand is always positive, so the integral is

1

C.The integrand is independent of both variables, so the integral is zero.

D.Changing the order necessarily doubles the integral.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The square is symmetric under exchanging $x$ and $y$ , while $f(y,x)=y-x=-(x-y)$ . Every contribution at $(x,y)$ is therefore canceled by the corresponding contribution at $(y,x)$ . This symmetry predicts a zero double integral and provides a useful check without performing lengthy calculations.

Q14. Suppose a region is split into two nonoverlapping parts $R_1$ and $R_2$ , and $f$ is continuous on the entire region. An analyst evaluates the integral over each part separately and adds the results. Which principle supports this approach?

A.The double integral is additive over nonoverlapping subregions, and each part can be represented by suitable iterated integrals. ✅

B.Fubini's theorem requires every region to remain rectangular.

C.Splitting a region always changes the value of the integral.

D.Only one of the two subregions may be integrated.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: A continuous function is integrable over the region, and the integral is additive when the region is partitioned into suitable nonoverlapping pieces. This is especially useful when reversing integration order creates complicated bounds that require splitting the region. The separate contributions can then be evaluated and summed without changing the total quantity.

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