This problem requires integrating a vector-valued acceleration to find velocity, a key skill in vector calculus. To solve, integrate each component separately, starting with the x-component where ∫6t dt is 3t² plus a constant. For the y-component, ∫-4 dt is -4t plus a constant. Apply the initial condition at t=0 to get constants 1 and 2, resulting in ⟨3t² + 1, -4t + 2⟩. A tempting distractor like choice B omits the constants, ignoring the initial velocity. Always integrate vector functions component-wise and incorporate initial conditions to find the specific solution.

This problem requires finding the indefinite integral of a vector-valued function, a key skill in vector calculus. To solve, integrate each component separately, with the x-component where $∫e^t$ dt is $e^t$ plus a constant C1. For the y-component, ∫1/(1 + t²) dt is arctan t plus a constant C2. This gives the antiderivative $⟨e^t$ + C1, arctan t + C2⟩. A tempting distractor like choice D substitutes ln(1 + t²) for arctan t, but its derivative is 2t/(1 + t²), not matching. Always integrate vector functions component-wise, recalling standard antiderivatives like arctan for 1/(1 + t²) and including separate constants.

This problem requires evaluating a definite integral of a vector function. For the first component: ∫₀² (t-1)² dt = ∫₀² (t² - 2t + 1) dt = [t³/3 - t² + t]₀² = 8/3 - 4 + 2 = 2/3. For the second component: ∫₀² 1/(1+t) dt = [ln|1+t|]₀² = ln 3 - ln 1 = ln 3. For the third component: ∫₀² sin(πt) dt = [-(1/π)cos(πt)]₀² = -(1/π)(cos 2π - cos 0) = 0. Choice C incorrectly evaluates the first component as 8/3, perhaps computing only the t³/3 term. When integrating polynomials, expand first and integrate term by term to avoid errors.

This problem involves integrating a vector-valued velocity function to find the position function. For the x-component, integrate 2t with respect to t to obtain t² plus a constant. For the y-component, integrate cos t to get sin t plus a constant. For the z-component, integrate 3 to yield 3t plus a constant, then use the initial position at t=0 to solve for the constants, resulting in +1, +0, and -2 respectively. A tempting distractor like choice B includes 2t² +1 for the x-component, which fails because it incorrectly doubles the integral of 2t instead of properly computing t². When integrating vector functions, always perform component-wise integration and apply initial conditions to determine the constant vector.

This problem requires evaluating a definite integral of a vector-valued function from 0 to 2. For the x-component, integrate 3t² to get t³ evaluated from 0 to 2, yielding 8. For the y-component, integrate 4 to obtain 4t evaluated from 0 to 2, resulting in 8. For the z-component, integrate $e^t$ to $e^t$ evaluated from 0 to 2, giving e² - 1. A tempting distractor like choice B shows 12 for the x-component, which fails because it mistakenly uses 4t³/4 instead of t³ for the antiderivative of 3t². Remember, for vector integrals, compute each component's definite integral separately and combine them into the resulting vector.

This problem involves evaluating a definite integral of a vector-valued function. We integrate each component separately over the interval [0, π]: ∫₀^π sin t dt = [-cos t]₀^π = -cos π - (-cos 0) = -(-1) - (-1) = 2. For the second component: ∫₀^π 2t dt = [t²]₀^π = π² - 0 = π². For the third component: ∫₀^π 3 dt = [3t]₀^π = 3π - 0 = 3π. Choice C incorrectly evaluates the second component as 2π instead of π², confusing the antiderivative of 2t. When integrating vector functions, apply the fundamental theorem of calculus to each component separately.

This problem requires finding position from velocity with an initial condition. Integrating the velocity components: $∫e^t$ dt = $e^t$ + C₁, ∫sin t dt = -cos t + C₂, and ∫2t dt = t² + C₃. Applying r⃗(0) = ⟨0, 1, -3⟩: $e^0$ + C₁ = 0 gives C₁ = -1; -cos 0 + C₂ = 1 gives C₂ = 2; and 0² + C₃ = -3 gives C₃ = -3. Choice A incorrectly calculates C₂ = 1 instead of C₂ = 2, missing that -cos 0 = -1. The crucial step in initial value problems is carefully evaluating each component at t = 0 to find the correct constants.

This problem requires computing a definite integral of a vector-valued function from 0 to π. For the x-component, integrate cos t to sin t evaluated from 0 to π, yielding 0. For the y-component, integrate sin t to -cos t evaluated from 0 to π, resulting in 2. For the z-component, integrate 2t to t² evaluated from 0 to π, giving π². A tempting distractor like choice A shows 0 for y, which fails because it neglects the sign change in evaluating -cos t at the limits. When performing definite integrals on vectors, integrate and evaluate each component separately to form the final vector.

This problem involves integrating trigonometric velocity components to find position. Integrating $\vec{v}(t) = \langle 4\cos t, 5\sin t \rangle$ gives $\vec{r}(t) = \langle 4\sin t + C_1, -5\cos t + C_2 \rangle$. Using $\vec{r}(0) = \langle -1, 3 \rangle$: we get $4\sin(0) + C_1 = -1$, so $C_1 = -1$, and $-5\cos(0) + C_2 = 3$, so $C_2 = 8$. Choice D incorrectly has $5\cos t$ instead of $-5\cos t$, forgetting the negative sign when integrating sine. Remember that $\int \sin t,dt = -\cos t + C$, and apply initial conditions after integration.

This problem asks for the indefinite integral of a vector-valued trigonometric function. Integrating component by component: ∫5cos t dt = 5sin t + C₁ and ∫(-4sin t) dt = 4cos t + C₂. The antiderivative is ⟨5sin t + C₁, 4cos t + C₂⟩. Choice E incorrectly shows -4cos t for the y-component, which would result from forgetting that integrating -sin t gives +cos t (the negative sign is already accounted for). When integrating vector functions, carefully track signs and remember that each component needs its own integration constant.