📝 Fundamental theorem of line integrals (15 MCQs)

📖 From Calculus • 16. Topics in vector Calculus • 15 questions available

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What is Fundamental theorem of line integrals?

Fundamental theorem of line integrals:
If $\mathbf{F} = \nabla f$ , then $\int_C \mathbf{F} \cdot d\mathbf{r} = f(B) - f(A)$ , independent of the path from $A$ to $B$ .

Example:
For $\mathbf{F} = \langle y, x \rangle$ , $f = xy$ , so $\int_C \mathbf{F} \cdot d\mathbf{r} = x_B y_B - x_A y_A$ .

Reason:
This theorem generalizes the Fundamental Theorem of Calculus to vector fields, providing a powerful evaluation tool for conservative fields.

Easy

Medium

Hard

📝 All Fundamental theorem of line integrals MCQs

Q1. A scalar potential is given by $f(x,y)=x^2y+3y$ , and a vector field is defined by $\mathbf F=\nabla f$ . What is the value of $\int_C \mathbf F\cdot d\mathbf r$ along any smooth curve from $(1,2)$ to $(3,4)$ ?

A.66

B.70

C.76 ✅

D.82

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Because $\mathbf F$ is the gradient of $f$ , the line integral depends only on the endpoints. Thus $\int_C\mathbf F\cdot d\mathbf r=f(3,4)-f(1,2)$ . We obtain $f(3,4)=48+12=60$ and $f(1,2)=2+6=8$ , giving $60-8=52$ . Therefore none of the listed values is correct.

Q2. Which statement best captures the central computational advantage provided by a potential function $f$ when $\mathbf F=\nabla f$ ?

A.The curve must always be a straight line

B.The integral can be evaluated using only the potential values at the endpoints ✅

C.The vector field must have constant magnitude everywhere

D.The parameterization of the curve must be linear

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: When a vector field is the gradient of a scalar potential, its line integral is determined entirely by the change in that potential between the initial and terminal points. This eliminates the need to parameterize the curve or integrate separately along each segment, provided the required conditions are satisfied.

Q3. A robot moves through a force field $\mathbf F=\nabla f$ , where $f$ represents an energy-related scalar quantity. Two different routes connect the same starting and ending positions. Which prediction is most justified?

A.The route with greater length always produces greater work

B.Both routes produce the same line integral ✅

C.The route with more turns produces zero work

D.Only the route with constant speed can be evaluated

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a conservative gradient field, the Fundamental Theorem of Line Integrals states that the line integral equals the change in the potential function between endpoints. Therefore, two different paths joining identical endpoints give exactly the same integral, regardless of their lengths, shapes, or parameterization speeds.

Q4. Suppose $\mathbf F=\langle 2xy+1,x^2+4y\rangle$ . A student wants to use the Fundamental Theorem of Line Integrals. Which potential function is appropriate?

f(x,y)=x^2y+x+2y^2

✅

f(x,y)=2x^2y+x+4y^2

f(x,y)=x^2y+2x+4y^2

f(x,y)=xy^2+x+2y^2

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: To find the potential, integrate the first component $2xy+1$ with respect to $x$ , giving $x^2y+x+g(y)$ . Differentiating with respect to $y$ gives $x^2+g'(y)$ . Matching the second component $x^2+4y$ requires $g'(y)=4y$ , so $g(y)=2y^2$ .

Q5. For $\mathbf F=\langle 3x^2,4y^3\rangle$ , a curve starts at $(1,-1)$ and ends at $(2,2)$ . Without parameterizing the curve, what is the line integral?

A.24

B.28

C.32 ✅

D.36

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A potential function is $f(x,y)=x^3+y^4$ , because its gradient is $\langle3x^2,4y^3\rangle$ . Therefore the integral equals $f(2,2)-f(1,-1)$ . The endpoint values are $8+16=24$ and $1+1=2$ , respectively, so the integral is $24-2=22$ . Hence the numerical choices do not match the correct result.

Q6. A student evaluates a gradient-field integral by parameterizing a complicated three-segment path and obtains a nonzero answer. Another student evaluates the potential at the endpoints and obtains zero. The endpoints are identical for both calculations. Which conclusion is strongest?

A.The complicated path must always produce zero only when it is straight

B.The endpoint method is invalid for piecewise curves

C.The discrepancy indicates an error in the parameterized calculation ✅

D.Both answers are valid because different paths can change the result

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For a gradient field, the line integral depends only on the initial and final points, and this remains true for piecewise smooth paths. If the endpoint potential values are equal, the integral must be zero. A different nonzero result from direct parameterization therefore signals an algebraic, orientation, or substitution error.

Q7. A potential function satisfies $f(0,0)=5$ and $f(4,-2)=17$ . If $\mathbf F=\nabla f$ , what can be concluded about any smooth path connecting these two points?

A.The integral is

12

for every such path ✅

B.The integral is

22

only for straight paths

C.The integral is

85

for every such path

D.The integral depends on the path length

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The Fundamental Theorem of Line Integrals converts the line integral of a gradient field into the difference between potential values at the terminal and initial points. Therefore $\int_C\mathbf F\cdot d\mathbf r=17-5=12$ , independent of the geometry, length, or parameterization of the chosen path.

Q8. Consider a graph of a potential function along a one-dimensional motion. A particle moves from position $x=a$ where $f(a)=10$ to position $x=b$ where $f(b)=3$ . If the force is $F=f'(x)$ , what does the graph imply about the work?

A.The work is

13

B.The work is

7

C.The work is

-7

✅

D.The work must be zero because the force varies

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Since the force is the derivative of the potential, the one-dimensional version of the theorem gives the work as $f(b)-f(a)$ . The graph shows the potential decreases from $10$ to $3$ , so the work is $3-10=-7$ . The negative sign indicates a net decrease in potential.

Q9. A contour plot of a scalar potential shows two points lying on the same contour level. A vector field is the gradient of that potential. A student claims that the line integral between the points must be positive because the path is not straight. How should the claim be evaluated?

A.Correct, because curved paths increase the integral

B.Correct, because gradients always point in the direction of motion

C.Incorrect, because equal potential values make the integral zero ✅

D.Incorrect, because gradients always produce negative integrals

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Points on the same contour have equal potential values. For a gradient field, the line integral equals the terminal potential minus the initial potential. Therefore the integral is zero regardless of whether the connecting path is straight, curved, long, or piecewise smooth. Path shape does not override endpoint dependence.

Q10. A field is $\mathbf F=\langle 2x,2y\rangle$ . A student computes the integral around a closed circular path and argues that it must be positive because the field points outward everywhere. What is the correct analysis?

A.The student is correct because outward fields always give positive circulation

B.The integral is zero because the field is a gradient field and the path is closed ✅

C.The integral depends on the radius but never vanishes

D.The integral is zero only for a circle centered at the origin

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The field is $\nabla f$ for $f(x,y)=x^2+y^2$ . On any closed curve, the starting and ending points coincide, so the potential difference is zero. Although the field points outward and may have a positive dot product on portions of a path, the total line integral around the closed curve is zero.

Q11. A field is $\mathbf F=\langle y,x\rangle$ . A student proposes $f(x,y)=xy$ and concludes that the line integral from $(1,2)$ to $(4,5)$ is $f(4,5)-f(1,2)$ . Is the reasoning valid?

A.Yes, because

f_x=y

and

f_y=x

, so the field is exactly

\nabla f

✅

B.No, because the field must contain only positive components

C.No, because the endpoints are too far apart

D.No, because a potential function cannot contain a product

xy

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The proposed potential is valid because $\nabla(xy)=\langle y,x\rangle$ , exactly matching the vector field. Therefore the Fundamental Theorem applies and the integral equals the endpoint potential difference. The form of the potential is not restricted to sums of single-variable functions; mixed products are completely acceptable.

Q12. A field is known to be a gradient field on a region containing two possible paths between $P$ and $Q$ . Path A is straight and Path B consists of five curved segments. Which method is generally most efficient for finding the line integral?

A.Parameterize Path A because straight paths are always required

B.Parameterize Path B because more information gives greater accuracy

C.Find a potential and evaluate its values at

P

and

Q

✅

D.Average the integrals obtained from both paths

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: When the field is known to be a gradient field, direct parameterization is unnecessary. The most efficient method is to identify a potential function and subtract its value at the initial point from its value at the terminal point. This approach works equally well for straight, curved, and piecewise smooth paths.

Q13. A potential surface has values $f(P)=14$ and $f(Q)=9$ . A path from $P$ to $Q$ first climbs to a region where the potential is 20 and then descends to $Q$ . For $\mathbf F=\nabla f$ , what is the total line integral?

A.20

B.11

C.-5 ✅

D.5

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Intermediate increases and decreases in potential cancel when the complete path is considered. The Fundamental Theorem depends only on the initial and final potential values. Thus the total integral is $f(Q)-f(P)=9-14=-5$ , even though the path temporarily reaches a higher potential value before ending below its starting value.

Q14. A student argues: 'If $\mathbf F=\nabla f$ , then $\mathbf F\cdot d\mathbf r=df$ , so the line integral is always equal to $f$ at the endpoint.' What essential correction is needed?

A.The result should use the endpoint potential minus the initial potential ✅

B.The result should be the average of the endpoint potentials

C.The result should use only the initial potential

D.The result should multiply the endpoint potential by the path length

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The differential relation $\mathbf F\cdot d\mathbf r=df$ means that integrating along a path accumulates the change in $f$ . Therefore the correct result is $f(\text{final})-f(\text{initial})$ , not merely the final value. Forgetting the initial contribution is a common conceptual error when applying the theorem.

Q15. For a smooth scalar function $f$ , suppose $\mathbf F=\nabla f$ . Three paths connect $A$ to $B$ : one has length 2, another length 10, and the third forms several loops before reaching $B$ . Which statement is necessarily true?

A.The longest path has the greatest integral

B.The shortest path has the greatest integral

C.All three integrals are equal to

f(B)-f(A)

✅

D.The path containing loops has integral zero

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The theorem makes the geometry of the path irrelevant when the vector field is a gradient. Every path begins at $A$ and ends at $B$ , so each integral equals the same potential difference $f(B)-f(A)$ . Even additional loops do not alter the total value because each closed portion contributes zero.

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