📝 Integration rules for vector functions (24 MCQs)

📖 From Calculus • 13. Vector Valued Functions • 24 questions available

←Previous Topic:Definite integral of vector valued functions📝 Integration rules for vector functionsAntiderivatives of vector functions:Next Topic→

What is Integration rules for vector functions?

Definition:
Integration rules for vector functions follow linearity, where $\int [c\vec{u}(t) + \vec{v}(t)] \, dt = c\int \vec{u}(t) \, dt + \int \vec{v}(t) \, dt$ .

Example:
Integrating $\langle 2e^t, \cos t \rangle$ gives $\langle 2e^t, \sin t \rangle + \vec{C}$ .

Reason:
Linearity simplifies complex vector integrals into manageable scalar integrations, preserving superposition principles.

Easy

Medium

Hard

📝 All Integration rules for vector functions MCQs

Q1. A particle moves along a space curve with velocity $\mathbf{v}(t) = \langle t^2, e^t, \sin t \rangle$ . If the position at $t=0$ is $\mathbf{r}(0) = \langle 1, 0, -1 \rangle$ , which expression correctly represents the position vector $\mathbf{r}(t)$ after applying integration rules component-wise?

\langle \frac{t^3}{3}, e^t, -\cos t \rangle

\langle \frac{t^3}{3}+1, e^t-1, -\cos t \rangle

\langle \frac{t^3}{3}+1, e^t-1, -\cos t + 1 \rangle

✅

\langle \frac{t^3}{3}+1, e^t, -\cos t + 1 \rangle

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Students must integrate each component separately and apply initial conditions to solve for constants. A common error is forgetting that the constant of integration is a vector, not a scalar, leading to incorrect y or z components when evaluating at t equals zero.

Q2. When integrating a vector-valued function $\int \mathbf{r}(t) \, dt$ , a student claims the result is always parallel to $\mathbf{r}(t)$ . Which scenario best disproves this misconception?

A.Integrating a constant vector yields a linear vector function parallel to the original.

B.Integrating

\langle \cos t, \sin t \rangle

yields

\langle \sin t, -\cos t \rangle

, which is orthogonal, not parallel. ✅

C.Integration always preserves direction because it is a linear operator.

D.The integral of any vector function is a scalar, so parallelism is undefined.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Integration of vector functions does not preserve direction. The antiderivative represents accumulated change, not instantaneous direction. For circular motion, velocity is tangent but position integral relates to displacement, often creating orthogonal relationships between the function and its integral in periodic cases.

Q3. Given $\mathbf{F}'(t) = \langle 2t, 3t^2, 4t^3 \rangle$ and $\mathbf{F}(1) = \langle 2, 3, 4 \rangle$ , a student computes $\mathbf{F}(t) = \langle t^2, t^3, t^4 \rangle$ . What specific error did the student make in applying integration rules?

A.They differentiated instead of integrating.

B.They failed to add the constant vector determined by the initial condition. ✅

C.They integrated only the first component correctly.

D.They used the power rule incorrectly for vector components.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The student found the general antiderivative correctly but neglected to solve for the constant vector using the given point. Without substituting t equals one and solving for C, the specific solution satisfying the initial value problem cannot be determined, making the answer incomplete.

Q4. If $\int_a^b \mathbf{r}'(t) \, dt = \mathbf{0}$ , what can be definitively concluded about the vector-valued function $\mathbf{r}(t)$ over the interval $[a,b]$ ?

A.The function is identically zero on the interval.

B.The net displacement is zero, meaning

\mathbf{r}(b) = \mathbf{r}(a)

. ✅

C.The speed of the particle is constant throughout the interval.

D.The acceleration vector is always perpendicular to velocity.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: By the Fundamental Theorem of Calculus for vector functions, the definite integral of the derivative equals the difference in position vectors. A zero result implies the starting and ending positions are identical, indicating a closed path or return to origin, regardless of the trajectory taken.

Q5. Consider the graph of a vector-valued function's magnitude $\|\mathbf{v}(t)\|$ showing a symmetric bell curve from $t=-2$ to $t=2$ . If $\mathbf{v}(t)$ is an odd vector function, what is $\int_{-2}^{2} \mathbf{v}(t) \, dt$ ?

A.Twice the integral from 0 to 2

B.Zero vector ✅

C.Cannot be determined without explicit components

D.Equal to the area under the magnitude curve

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Even though the magnitude graph is symmetric and positive, the vector function itself being odd means $\mathbf{v}(-t) = -\mathbf{v}(t)$ . Integration of an odd vector function over a symmetric interval cancels out completely, yielding the zero vector despite nonzero speed throughout the interval.

Q6. Which statement correctly distinguishes $\int \|\mathbf{r}'(t)\| \, dt$ from $\| \int \mathbf{r}'(t) \, dt \|$ for a particle traversing a curved path?

A.They are always equal by linearity of norms.

B.The first gives total distance traveled; the second gives magnitude of net displacement. ✅

C.The first gives net displacement; the second gives average speed.

D.Both represent arc length but computed via different methods.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The integral of the norm accumulates scalar speed to find total path length, while the norm of the integral finds the straight-line distance between endpoints. These differ whenever the path is not a straight line segment, highlighting non-commutativity of norms and integration operators.

Q7. A student attempts to integrate $\mathbf{r}(t) = \langle t \mathbf{i} + t^2 \mathbf{j} \rangle$ using substitution $u = t^2$ only on the j-component. Why is this approach problematic when seeking a unified antiderivative?

A.Substitution cannot be used on vector components individually.

B.Each component must be integrated with respect to the same variable t; mixing substitutions creates inconsistent parameterizations. ✅

C.The i-component requires integration by parts instead.

D.There is no problem; component-wise substitution is always valid.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: While component-wise integration is valid, applying different substitutions changes the differential relationship. Each component's antiderivative must remain a function of the original parameter t to maintain the vector function's integrity. Inconsistent parameterization breaks the functional relationship between components in the resulting vector.

Q8. For $\mathbf{r}(t) = \langle e^{-t} \cos t, e^{-t} \sin t \rangle$ , computing $\int_0^\infty \mathbf{r}(t) \, dt$ requires recognizing which convergence property?

A.Each component converges absolutely due to exponential decay dominating oscillation. ✅

B.The integral diverges because trigonometric functions never settle.

C.Convergence depends solely on the cosine component.

D.The vector integral converges only if both components converge conditionally.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: This improper vector integral converges because the exponential envelope forces both oscillating components toward zero rapidly enough. Students must recognize that absolute convergence of each scalar component guarantees convergence of the vector integral, combining knowledge of improper integrals, oscillatory behavior, and vector calculus foundations.

Q9. If $\mathbf{a}(t) = \langle 6t, 2, 0 \rangle$ and initial velocity is $\mathbf{v}(0) = \langle 0, 1, 3 \rangle$ , what is the correct velocity function after integrating acceleration?

\langle 3t^2, 2t, 0 \rangle

\langle 3t^2, 2t+1, 3 \rangle

✅

\langle 6t^2, 2t+1, 3 \rangle

\langle 3t^2+1, 2t, 3 \rangle

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Integrating acceleration component-wise gives $\langle 3t^2+C_1, 2t+C_2, C_3 \rangle$ . Applying initial velocity conditions determines constants as $C_1=0, C_2=1, C_3=3$ . This basic application tests proper use of initial conditions in vector integration without computational complexity.

Q10. When modeling fluid flow with $\mathbf{V}(t) = \langle \sin t, \cos t, t \rangle$ , why can't we simply integrate magnitude to find total fluid displacement vector?

A.Displacement is a vector quantity requiring component-wise integration, not scalar accumulation. ✅

B.Magnitude integration gives volume, not displacement.

C.Fluid flow models never use vector integration.

D.The z-component makes magnitude integration impossible.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Physical displacement requires directional information preserved through vector integration. Integrating magnitude yields total path length or accumulated scalar quantity, losing all directional data essential for displacement vectors. This distinction is critical in physics modeling where vector nature carries physical meaning beyond mere size.

Q11. A peer argues that $\int [\mathbf{f}(t) \times \mathbf{g}(t)] \, dt = [\int \mathbf{f}(t) \, dt] \times [\int \mathbf{g}(t) \, dt]$ . Which counterexample most efficiently refutes this?

A.Use constant vectors where both sides equal zero.

B.Use

\mathbf{f}=\langle t,0,0\rangle, \mathbf{g}=\langle 0,t,0\rangle

; LHS integrates to

\langle 0,0,t^3/3\rangle

, RHS gives

\langle 0,0,t^4/4\rangle

. ✅

C.Cross products are undefined for integrals.

D.The identity actually holds for all smooth vector functions.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This false distributive property fails because integration and cross product don't commute. Testing simple polynomial vectors shows mismatched powers and coefficients, proving the operations aren't interchangeable. Recognizing invalid algebraic manipulations prevents systematic errors in advanced vector calculus computations involving products.

Q12. Given position $\mathbf{r}(t)$ with $\mathbf{r}'(t) = \langle 2\cos 2t, -2\sin 2t, 1 \rangle$ , what geometric insight helps verify your integrated position function without full computation?

A.The xy-projection should trace a circle of radius 1 centered at origin plus constant offset. ✅

B.The z-component must be quadratic in t.

C.Speed should vary sinusoidally with time.

D.Acceleration should be zero in all components.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Recognizing the velocity's xy-components describe uniform circular motion allows prediction that integrated position will have sinusoidal xy-terms forming a circle. This geometric check validates integration results independently, connecting analytical computation with spatial intuition about helical trajectories and their projections.

Q13. In orbital mechanics, if gravitational acceleration is $\mathbf{a}(t) = -\frac{k}{r^3}\mathbf{r}(t)$ , why is direct integration to find $\mathbf{v}(t)$ fundamentally different from integrating constant acceleration?

A.Acceleration depends on position, creating coupled differential equations rather than separable integrals. ✅

B.Gravitational acceleration is always constant near Earth.

C.Vector integration doesn't apply to inverse-square laws.

D.The negative sign prevents integration.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Unlike constant acceleration problems where integration is straightforward, position-dependent acceleration creates nonlinear coupling between components. This requires advanced techniques beyond basic integration rules, illustrating limitations of elementary vector integration and motivating study of differential equations in physics applications.

Q14. Compute $\int_0^{\pi} \langle \sin^2 t, \cos^2 t, \sin t \cos t \rangle \, dt$ efficiently by exploiting symmetry rather than brute-force integration.

A.Use identities: first two components sum to 1, third is odd-symmetric about π/2. ✅

B.All components integrate to zero over full period.

C.Only compute first component; others follow by phase shift.

D.Convert to complex exponentials before integrating.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Recognizing $\sin^2 t + \cos^2 t = 1$ simplifies sum of first two integrands. The third component $\sin t \cos t = \frac{1}{2}\sin 2t$ integrates to zero over [0,π]. Strategic use of trigonometric identities reduces computation significantly compared to term-by-term integration.

Q15. Which condition ensures $\int_a^b \mathbf{r}(t) \, dt$ exists for a vector-valued function?

A.Each component function must be Riemann integrable on [a,b]. ✅

B.The magnitude

\|\mathbf{r}(t)\|

must be continuous.

C.The derivative

\mathbf{r}&#039;(t)

must exist everywhere.

D.The function must be differentiable at endpoints.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Vector integrability reduces to component-wise integrability. Continuity of magnitude or existence of derivatives are sufficient but not necessary conditions. Understanding this foundational criterion prevents overcomplicating existence questions and connects vector calculus back to single-variable integration theory properly.

Q16. A drone's velocity is modeled as $\mathbf{v}(t) = \langle t, t^2, t^3 \rangle$ . To find average velocity over [0,2], a student computes $\mathbf{v}(1)$ . What conceptual error does this reveal?

A.Confusing instantaneous velocity at midpoint with average velocity defined by integral mean. ✅

B.Average velocity requires acceleration, not velocity.

C.Midpoint evaluation is correct for polynomial velocities.

D.Average velocity is undefined for cubic components.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Average velocity equals $\frac{1}{b-a}\int_a^b \mathbf{v}(t)dt$ , not evaluation at interval midpoint. While true for linear functions, this fails for higher-degree polynomials. The error reflects misunderstanding of integral averages versus point evaluations, crucial for accurate kinematic analysis.

Q17. When integrating $\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle$ where f,g,h have different domains of definition, how should the domain of $\int \mathbf{r}(t)dt$ be determined?

A.Take union of all component domains.

B.Take intersection of all component domains. ✅

C.Use domain of the component with largest range.

D.Domain is always all real numbers for integrals.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: A vector function is only defined where all components exist simultaneously. Therefore, the integral's domain must be the intersection of individual component domains. Overlooking this leads to evaluating integrals at points where some components are undefined, producing mathematically meaningless results.

Q18. If $\int \mathbf{r}(t) \, dt = \mathbf{R}(t) + \mathbf{C}$ , and you're given $\mathbf{R}(t_0) = \mathbf{P}_0$ , which step is most frequently missed when solving for $\mathbf{C}$ ?

A.Setting

\mathbf{C} = \mathbf{P}_0 - \mathbf{R}(t_0)

as a vector subtraction. ✅

B.Treating

\mathbf{C}

as three separate scalars solved independently.

C.Assuming

\mathbf{C} = \mathbf{0}

without verification.

D.Differentiating

\mathbf{R}(t)

again to confirm.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Students often forget that the constant of integration is a vector requiring vector arithmetic. Solving $\mathbf{C} = \mathbf{P}_0 - \mathbf{R}(t_0)$ demands component-wise subtraction. Misapplying scalar thinking to vector constants produces systematically wrong particular solutions in initial value problems.

Q19. For a particle with $\mathbf{v}(t) = \langle \cos t, \sin t, 0 \rangle$ , compare $\int_0^{2\pi} \mathbf{v}(t) \, dt$ and $\int_0^{2\pi} \|\mathbf{v}(t)\| \, dt$ . What does their difference signify physically?

A.First is zero (closed loop); second is

2\pi

(circumference). Difference shows path closure vs distance. ✅

B.Both equal

2\pi

; no physical difference.

C.First gives speed; second gives displacement.

D.Difference indicates acceleration magnitude.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The vector integral vanishes due to periodicity, confirming closed circular path. Scalar integral of unit speed gives circumference. Their contrast illustrates fundamental distinction between net displacement and total distance, essential for understanding motion geometry versus kinematics in vector calculus contexts.

Q20. Which integration technique is MOST appropriate for $\int \langle t e^t, t \sin t, \ln t \rangle \, dt$ ?

A.Component-wise integration by parts for first two, standard formula for third. ✅

B.Substitution u=t for all components.

C.Trigonometric identity reduction for all terms.

D.Numerical approximation only.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Each component requires different techniques: integration by parts for products involving exponentials and trig functions, standard logarithmic integral for ln t. Recognizing heterogeneous requirements per component demonstrates flexible application of integration toolbox rather than forcing uniform method across vector function.

Q21. If $\mathbf{r}''(t) = \langle 0, 0, -g \rangle$ with $\mathbf{r}(0)=\langle 0,0,h\rangle$ and $\mathbf{r}'(0)=\langle v_x, v_y, 0\rangle$ , what does double integration reveal about trajectory shape?

A.Parabolic path in vertical plane determined by horizontal velocity and gravity. ✅

B.Linear descent due to constant acceleration.

C.Circular orbit from balanced forces.

D.Exponential decay in altitude.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Double integration of constant gravitational acceleration yields quadratic z(t) and linear x(t),y(t), describing parabolic projectile motion. This classic result connects vector integration directly to physical trajectory prediction, reinforcing how mathematical operations encode geometric and dynamic properties of motion under uniform force fields.

Q22. A student writes $\int \mathbf{r}(t) \cdot \mathbf{s}(t) \, dt = (\int \mathbf{r}(t) \, dt) \cdot (\int \mathbf{s}(t) \, dt)$ . Beyond providing counterexamples, what deeper principle explains why this fails?

A.Dot product is bilinear but integration is a linear functional; their composition doesn't factor. ✅

B.Dot products always yield zero under integration.

C.Integration destroys orthogonality relationships.

D.Only cross products distribute over integration.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The failure stems from integration being a linear operator acting on products, not preserving multiplicative structure. Bilinearity of dot product doesn't commute with integral's averaging effect. Understanding operator algebra prevents false distribution assumptions across various vector operations in advanced calculus and functional analysis contexts.

Q23. When verifying an antiderivative $\mathbf{R}(t)$ of $\mathbf{r}(t)$ , why is differentiation preferred over re-integration as a checking method?

A.Differentiation is algorithmically simpler and avoids reintroducing arbitrary constants. ✅

B.Re-integration is equally valid but slower.

C.Differentiation confirms magnitude only.

D.Checking isn't necessary for vector functions.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Differentiating the proposed antiderivative should recover the original integrand exactly without ambiguity. Re-integrating introduces new constants requiring additional verification steps. Differentiation provides immediate, unambiguous confirmation of correctness, making it the superior validation strategy for vector antiderivatives in practice.

Q24. Given $\mathbf{v}(t) = \langle 3t^2, 2t, 1 \rangle$ and knowing displacement from t=1 to t=3 is $\langle 26, 8, 2 \rangle$ , which reasoning validates this result without recomputing?

A.Antiderivative

\langle t^3, t^2, t \rangle

evaluated at bounds gives exact differences matching stated displacement. ✅

B.Displacement equals average velocity times duration.

C.Velocity magnitude integrated over interval yields displacement.

D.Initial position must be known to validate displacement.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Fundamental theorem guarantees displacement equals antiderivative difference. Computing $\mathbf{R}(3)-\mathbf{R}(1) = \langle 27-1, 9-1, 3-1 \rangle = \langle 26,8,2 \rangle$ confirms result. This validates understanding that definite vector integrals depend only on endpoint values, independent of path details or initial conditions.

🔗 Related Topics (MCQs)

📝 Parametric curves in 3D space MCQs 📝 Introduction to Vector Valued Functions MCQs 📝 Graphing parametric curves with technology MCQs 📝 Parametric equations for surface intersections MCQs 📝 Vector valued functions definition and examples MCQs

←Previous Topic:Definite integral of vector valued functions📝 Integration rules for vector functionsAntiderivatives of vector functions:Next Topic→