- Integrate the two polynomial components separately using the power rule. - Preserve the component order in the antiderivative vector. - Represent the two independent integration constants by one constant vector \(\mathbf{C}\).

1. Integrate each component separately. 2. The result is \(\int\mathbf{v}(t)\,dt=\left\langle t^{2}, t^{3}\right\rangle+\mathbf{C}\), where \(\mathbf{C}\) is a constant vector.

\(\int\mathbf{v}(t)\,dt=\left\langle t^{2}, t^{3}\right\rangle+\mathbf{C}\)

- Integrate cosine and sine componentwise. - Check the sign of the antiderivative of \(\sin(t)\) by differentiating it. - Include independent constants as a constant vector.

1. Integrate each component separately. 2. The result is \(\int\mathbf{v}(t)\,dt=\left\langle \sin(t), -\cos(t)\right\rangle+\mathbf{C}\), where \(\mathbf{C}\) is a constant vector.

\(\int\mathbf{v}(t)\,dt=\left\langle \sin(t), -\cos(t)\right\rangle+\mathbf{C}\)

- Integrate the exponential component directly. - Use the logarithmic antiderivative of \(1/t\) under the assumption \(t>0\). - Add a constant vector to represent the independent component constants.

1. Integrate each component separately. 2. The result is \(\int\mathbf{v}(t)\,dt=\left\langle e^{t}, \ln(t)\right\rangle+\mathbf{C}\), where \(\mathbf{C}\) is a constant vector.

\(\int\mathbf{v}(t)\,dt=\left\langle e^{t}, \ln(t)\right\rangle+\mathbf{C}\)

- Apply the power rule to the polynomial component. - Account for the inner negative sign when integrating \(e^{-t}\). - Verify both components by differentiating and include \(\mathbf{C}\).

1. Integrate each component separately. 2. The result is \(\int\mathbf{v}(t)\,dt=\left\langle \frac{t^{4}}{4}, -e^{-t}\right\rangle+\mathbf{C}\), where \(\mathbf{C}\) is a constant vector.

\(\int\mathbf{v}(t)\,dt=\left\langle \frac{t^{4}}{4}, -e^{-t}\right\rangle+\mathbf{C}\)

- Recognize the standard antiderivative associated with \(1/(1+t^{2})\). - Recognize \(\sec^{2}(t)\) as the derivative of a basic trigonometric function. - Work on one interval that does not contain a zero of \(\cos(t)\), and include a constant vector.

1. Integrate each component separately. 2. On any interval where \(\cos(t)\ne0\), the result is \(\int\mathbf{v}(t)\,dt=\left\langle \arctan(t),\tan(t)\right\rangle+\mathbf{C}\), where \(\mathbf{C}\) is a constant vector.

\(\int\mathbf{v}(t)\,dt=\left\langle \arctan(t),\tan(t)\right\rangle+\mathbf{C}\) on any interval where \(\cos(t)\ne0\)

- A vector integral is computed one component at a time. - Each constant rate is multiplied by the interval length.

1. Integrate each constant component over three parameter units. 2. The result is \(\langle3,6\rangle\).

\(\langle3,6\rangle\).

- Treat the two component integrals independently. - A zero derivative does not force the original component to be zero.

1. Integrate the first component to get \(t^2+C_1\). 2. Integrate the second component to get a separate constant \(C_2\).

\(\langle t^2+C_1,C_2\rangle\).

- Area between a component-rate graph and the time axis gives signed change in that component. - A rectangle's area is its height times its width.

1. The horizontal component is constantly \(2\) during a two-unit interval. 2. Its integral is the rectangular area \(2\cdot2=4\).

\(4\) coordinate units to the right.

- How does a definite integral connect velocity with change in position? - Include the initial vector when you write the position equation.

1. The change in position is \(\int_{0}^{1}\langle1,-1\rangle\,dt=\langle1,-1\rangle\). 2. Add this integral to the initial position: \(\mathbf{r}(1)=\langle1,2\rangle+\int_{0}^{1}\langle1,-1\rangle\,dt=\langle2,1\rangle\).

\(\mathbf{r}(1)=\langle1,2\rangle+\int_{0}^{1}\langle1,-1\rangle\,dt=\langle2,1\rangle\).

- Integrate the two coordinate functions independently. - For the sine component, account for the derivative of the inner expression \(2t\). - Rewrite \(\sqrt{t}\) as a power, preserve its domain \(t\ge0\), and include a constant vector.

1. On the domain \(t\ge0\), integrate each component separately. 2. The component antiderivatives are \(-\frac{1}{2}\cos(2t)\) and \(\frac{2}{3}t^{3/2}\). 3. Therefore, \(\int\mathbf{v}(t)\,dt=\left\langle -\frac{1}{2}\cos(2t),\frac{2}{3}t^{3/2}\right\rangle+\mathbf{C}\), where \(\mathbf{C}\) is a constant vector.

\(\int\mathbf{v}(t)\,dt=\left\langle -\frac{1}{2}\cos(2t),\frac{2}{3}t^{3/2}\right\rangle+\mathbf{C}\) for \(t\ge0\)

- First integrate each component of \(\mathbf{r}'(t)\) separately. - Keep a separate constant of integration in each component. - Substitute the given parameter value into the antiderivative before solving for the constants.

1. Integrating componentwise gives \(\mathbf{r}(t)=\left\langle t^{2}, t^{3}\right\rangle+\left\langle C_{1}, C_{2}\right\rangle\). 2. Applying \(\mathbf{r}(0)=\left\langle 1, -2\right\rangle\) gives \(\left\langle C_{1}, C_{2}\right\rangle=\left\langle 1, -2\right\rangle\). 3. Therefore, \(\mathbf{r}(t)=\left\langle t^{2}+1, t^{3}-2\right\rangle\).

\(\mathbf{r}(t)=\left\langle t^{2}+1, t^{3}-2\right\rangle\)

- Find an antiderivative for each trigonometric component. - Use separate constants because the two coordinates can shift independently. - Evaluate the antiderivative at \(t=0\) and match both coordinates to the initial vector.

1. Integrating componentwise gives \(\mathbf{r}(t)=\left\langle \sin(t), -\cos(t)\right\rangle+\left\langle C_{1}, C_{2}\right\rangle\). 2. Applying \(\mathbf{r}(0)=\left\langle 2, 1\right\rangle\) gives \(\left\langle C_{1}, C_{2}\right\rangle=\left\langle 2, 2\right\rangle\). 3. Therefore, \(\mathbf{r}(t)=\left\langle \sin(t)+2, 2-\cos(t)\right\rangle\).

\(\mathbf{r}(t)=\left\langle \sin(t)+2, 2-\cos(t)\right\rangle\)

- Integrate the exponential and constant components independently. - Write one integration constant for each coordinate. - Apply the initial vector after forming the general antiderivative.

1. Integrating componentwise gives \(\mathbf{r}(t)=\left\langle e^{t}, t\right\rangle+\left\langle C_{1}, C_{2}\right\rangle\). 2. Applying \(\mathbf{r}(0)=\left\langle 0, 3\right\rangle\) gives \(\left\langle C_{1}, C_{2}\right\rangle=\left\langle -1, 3\right\rangle\). 3. Therefore, \(\mathbf{r}(t)=\left\langle e^{t}-1, t+3\right\rangle\).

\(\mathbf{r}(t)=\left\langle e^{t}-1, t+3\right\rangle\)

- Integrate the polynomial and exponential components separately. - Check the sign when integrating \(e^{-t}\). - Substitute \(t=1\) into both coordinates to determine the constant vector.

1. Integrating componentwise gives \(\mathbf{r}(t)=\left\langle \frac{t^{3}}{3}, -e^{-t}\right\rangle+\left\langle C_{1}, C_{2}\right\rangle\). 2. Applying \(\mathbf{r}(1)=\left\langle 2, 0\right\rangle\) gives \(\left\langle C_{1}, C_{2}\right\rangle=\left\langle \frac{5}{3}, e^{-1}\right\rangle\). 3. Therefore, \(\mathbf{r}(t)=\left\langle \frac{t^{3}}{3}+\frac{5}{3}, e^{-1}-e^{-t}\right\rangle\).

\(\mathbf{r}(t)=\left\langle \frac{t^{3}}{3}+\frac{5}{3}, e^{-1}-e^{-t}\right\rangle\)

- Integrate the two components independently, using the domain to choose the logarithm form. - Keep separate constants of integration for the two coordinates. - Substitute \(t=1\) into both coordinates and match the initial vector.

1. Integrating componentwise gives \(\mathbf{r}(t)=\left\langle \ln(t), t^{2}\right\rangle+\left\langle C_{1}, C_{2}\right\rangle\). 2. Applying \(\mathbf{r}(1)=\left\langle 4, -1\right\rangle\) gives \(\left\langle C_{1}, C_{2}\right\rangle=\left\langle 4, -2\right\rangle\). 3. Therefore, \(\mathbf{r}(t)=\left\langle \ln(t)+4, t^{2}-2\right\rangle\).

\(\mathbf{r}(t)=\left\langle \ln(t)+4, t^{2}-2\right\rangle\)

- Recall the antiderivatives associated with \(\sec^{2}(t)\) and \(\cos(t)\). - Attach an independent integration constant to each component and use the initial vector at \(t=0\). - Identify the nearest parameter values where the tangent or secant expression is undefined.

1. Integrating componentwise gives \(\mathbf{r}(t)=\left\langle \tan(t),\sin(t)\right\rangle+\left\langle C_{1},C_{2}\right\rangle\). 2. Applying \(\mathbf{r}(0)=\left\langle 1,2\right\rangle\) gives \(\left\langle C_{1},C_{2}\right\rangle=\left\langle 1,2\right\rangle\). 3. Therefore, \(\mathbf{r}(t)=\left\langle \tan(t)+1,\sin(t)+2\right\rangle\). 4. The nearest points where \(\sec^{2}(t)\) and \(\tan(t)\) are undefined are \(t=\pm\frac{\pi}{2}\), so the maximal interval containing \(0\) is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).

\(\mathbf{r}(t)=\left\langle \tan(t)+1,\sin(t)+2\right\rangle\) for \(-\frac{\pi}{2}<t<\frac{\pi}{2}\)

- Integrate each polynomial component term by term. - Use a different integration constant in each coordinate. - Substitute the negative parameter value carefully when applying the initial condition.

1. Integrating componentwise gives \(\mathbf{r}(t)=\left\langle t^{3}-t, t^{2}+4t\right\rangle+\left\langle C_{1}, C_{2}\right\rangle\). 2. Applying \(\mathbf{r}(-1)=\left\langle 0, 5\right\rangle\) gives \(\left\langle C_{1}, C_{2}\right\rangle=\left\langle 0, 8\right\rangle\). 3. Therefore, \(\mathbf{r}(t)=\left\langle t^{3}-t, t^{2}+4t+8\right\rangle\).

\(\mathbf{r}(t)=\left\langle t^{3}-t, t^{2}+4t+8\right\rangle\)

- Integrate each component while accounting for the inner derivative \(2\). - Keep the trigonometric and exponential integration constants separate. - Evaluate the general antiderivative at \(t=0\) before solving for both constants.

1. Integrating componentwise gives \(\mathbf{r}(t)=\left\langle -\frac{1}{2}\cos(2t), \frac{1}{2}e^{2t}\right\rangle+\left\langle C_{1}, C_{2}\right\rangle\). 2. Applying \(\mathbf{r}(0)=\left\langle 3, -2\right\rangle\) gives \(\left\langle C_{1}, C_{2}\right\rangle=\left\langle \frac{7}{2}, -\frac{5}{2}\right\rangle\). 3. Therefore, \(\mathbf{r}(t)=\left\langle \frac{7}{2}-\frac{1}{2}\cos(2t), \frac{1}{2}e^{2t}-\frac{5}{2}\right\rangle\).

\(\mathbf{r}(t)=\left\langle \frac{7}{2}-\frac{1}{2}\cos(2t), \frac{1}{2}e^{2t}-\frac{5}{2}\right\rangle\)

- Split the vector integral into one definite integral for each coordinate. - Use the same lower and upper bounds for both components. - Reassemble the two scalar results in their original component order.

1. Apply the bounds to each component: \(\int_{0}^{2}\left\langle t, t^{2}\right\rangle\,dt=\left\langle \int_{0}^{2}t\,dt, \int_{0}^{2}t^{2}\,dt\right\rangle\). 2. Evaluating the scalar integrals gives \(\left\langle 2, \frac{8}{3}\right\rangle\).

\(\left\langle 2, \frac{8}{3}\right\rangle\)

- Separate the vector integral into cosine and sine integrals. - Apply the same angular bounds to both antiderivatives. - Preserve the original component order when forming the result vector.

1. Integrate componentwise: \(\int_{0}^{\pi/2}\left\langle \cos(t), \sin(t)\right\rangle\,dt=\left\langle \int_{0}^{\pi/2}\cos(t)\,dt, \int_{0}^{\pi/2}\sin(t)\,dt\right\rangle\). 2. Evaluating both scalar integrals gives \(\left\langle 1, 1\right\rangle\).

\(\left\langle 1, 1\right\rangle\)

- Integrate each exponential component over the stated interval. - Account for the negative inner derivative when integrating \(e^{-t}\). - Evaluate the endpoints before combining the two scalar results into a vector.

1. Integrate componentwise: \(\int_{0}^{1}\left\langle e^{t}, e^{-t}\right\rangle\,dt=\left\langle \int_{0}^{1}e^{t}\,dt, \int_{0}^{1}e^{-t}\,dt\right\rangle\). 2. Evaluating both scalar integrals gives \(\left\langle e-1, 1-e^{-1}\right\rangle\).

\(\left\langle e-1, 1-e^{-1}\right\rangle\)

- Recognize the inverse-tangent antiderivative in the first component. - Use the power rule on the second component. - Apply the bounds to each scalar antiderivative before rebuilding the vector.

1. Integrate componentwise: \(\int_{0}^{1}\left\langle \frac{1}{t^{2}+1}, 2t\right\rangle\,dt=\left\langle \int_{0}^{1}\frac{1}{t^{2}+1}\,dt, \int_{0}^{1}2t\,dt\right\rangle\). 2. Evaluating both scalar integrals gives \(\left\langle \frac{\pi}{4}, 1\right\rangle\).

\(\left\langle \frac{\pi}{4}, 1\right\rangle\)

- Split the vector integral into two scalar definite integrals. - Rewrite the radical as a fractional power before applying the power rule. - Evaluate the upper-minus-lower difference separately in each component.

1. Integrate componentwise: \(\int_{1}^{4}\left\langle t^{3}-1, \sqrt{t}\right\rangle\,dt=\left\langle \int_{1}^{4}(t^{3}-1)\,dt, \int_{1}^{4}t^{1/2}\,dt\right\rangle\). 2. Evaluating both scalar integrals gives \(\left\langle \frac{243}{4}, \frac{14}{3}\right\rangle\).

\(\left\langle \frac{243}{4}, \frac{14}{3}\right\rangle\)

- Combine the two given intervals before reversing direction. - Reversing integration limits changes the sign of the vector. - Use linearity for the scalar multiple.

1. Add adjacent intervals: \(\int_0^5\mathbf{v}(t)\,dt=\langle3,-1\rangle+\langle-2,4\rangle=\langle1,3\rangle\). 2. Reversing the limits gives \(\int_5^0\mathbf{v}(t)\,dt=\langle-1,-3\rangle\). 3. Multiplying the integrand by \(2\) multiplies the integral by \(2\), giving \(\langle-2,-6\rangle\).

\(\langle-2,-6\rangle\)

- Divide the definite vector integral by the interval length. - Integrate the components independently. - Check whether the interval length changes the resulting vector.

1. The interval length is \(1\), so the average vector is \(\int_0^1\mathbf{v}(t)\,dt\). 2. The first component is \(\int_0^1t^2\,dt=\frac{1}{3}\). 3. The second component is \(\int_0^1\frac{1}{1+t^2}\,dt=\frac{\pi}{4}\). 4. The average value is \(\left\langle\frac{1}{3},\frac{\pi}{4}\right\rangle\).

\(\left\langle\frac{1}{3},\frac{\pi}{4}\right\rangle\)

- Classify the parity of each component separately. - Symmetry can eliminate one integral completely. - Use a familiar trigonometric identity or average value for the remaining component.

1. The first component \(t\cos t\) is odd, so its integral over \([-\pi,\pi]\) is \(0\). 2. The second component \(\sin^2 t\) is even, and \(\int_{-\pi}^{\pi}\sin^2 t\,dt=2\int_0^{\pi}\sin^2 t\,dt=\pi\). 3. Therefore, the vector integral is \(\langle0,\pi\rangle\).

\(\langle0,\pi\rangle\)

- Integrate the components independently. - Recognize a double-angle expression in the second component. - Use the interval endpoints carefully when evaluating each antiderivative.

1. For the first component, \(\int_0^{\pi/2}\sin t\cos t\,dt=\frac{1}{2}\). 2. The second component is \(\int_0^{\pi/2}\cos(2t)\,dt=0\). 3. Therefore, the vector integral is \(\left\langle\frac{1}{2},0\right\rangle\).

\(\left\langle\frac{1}{2},0\right\rangle\)

- Connect the two position vectors with an integral of the derivative. - Evaluate the net change over the full parameter interval. - Add the change vector to the known starting vector.

1. The net change is \(\int_{-1}^{2}\mathbf{r}'(t)\,dt\). 2. The first component is \([t^3-t]_{-1}^{2}=6\), and the second is \([t^2]_{-1}^{2}=3\). 3. Therefore, \(\mathbf{r}(2)=\langle4,2\rangle+\langle6,3\rangle=\langle10,5\rangle\).

\(\mathbf{r}(2)=\langle10,5\rangle\)

- Determine the common subinterval width first. - Add the midpoint vectors componentwise. - Multiply the resulting vector by the common width.

1. Each subinterval has width \(\Delta t=1\). 2. The midpoint-rule sum is \(1[\langle1,2\rangle+\langle3,-1\rangle+\langle2,4\rangle]\). 3. Therefore, \(\int_0^3\mathbf{v}(t)\,dt\approx\langle6,5\rangle\).

\(\int_0^3\mathbf{v}(t)\,dt\approx\langle6,5\rangle\)

- Keep the unknown upper limit in both component integrals. - Equality with the target vector imposes two conditions on the same positive endpoint. - Check the intersection of those conditions rather than treating either coordinate in isolation.

1. The vector integral is \(\int_0^b\langle2t,1\rangle\,dt=\langle b^2,b\rangle\). 2. Matching \(\langle9,3\rangle\) gives the two equations \(b^2=9\) and \(b=3\). 3. Because \(b>0\), the equations are simultaneously satisfied only by \(b=3\).

Integral: \(\langle b^2,b\rangle\) Component equations: \(b^2=9\) and \(b=3\) \(b=3\)

- Use linearity to separate the integral of the vector combination. - Treat the unknown definite integral as a vector variable. - Isolate it with componentwise vector operations.

1. By linearity, \(2\int_0^2\mathbf{r}(t)\,dt+\int_0^2\mathbf{q}(t)\,dt=\langle5,1\rangle\). 2. Subtracting the known integral gives \(2\int_0^2\mathbf{r}(t)\,dt=\langle4,4\rangle\). 3. Dividing by \(2\) gives \(\int_0^2\mathbf{r}(t)\,dt=\langle2,2\rangle\).

\(\langle2,2\rangle\)

- Keep the unknown endpoint in both component antiderivatives. - Turn the target-vector equality into two simultaneous scalar conditions. - Use the positivity restriction only after finding values compatible with both components.

1. The vector integral is \(\int_0^b\langle t,t^2\rangle\,dt=\left\langle\frac{b^2}{2},\frac{b^3}{3}\right\rangle\). 2. Matching \(\left\langle2,\frac{8}{3}\right\rangle\) gives \(\frac{b^2}{2}=2\) and \(\frac{b^3}{3}=\frac{8}{3}\). 3. With \(b>0\), both equations give \(b=2\).

Integral: \(\left\langle\frac{b^2}{2},\frac{b^3}{3}\right\rangle\) Component equations: \(\frac{b^2}{2}=2\) and \(\frac{b^3}{3}=\frac{8}{3}\) \(b=2\)

- Areas below the axis count as negative horizontal change. - Compare the areas on the two sides of the zero crossing.

1. The triangle below the axis between \(0\) and \(1\) contributes \(-\frac{1}{2}\). 2. The equal triangle above the axis between \(1\) and \(2\) contributes \(\frac{1}{2}\). 3. Their signed sum is \(0\).

\(0\).

- Think about which way accumulation runs over the interval. - Does reversing limits affect just one component or both?

1. Reversing the limits changes the sign of every component. 2. The forward integral is \(\langle\int_{0}^{2}t\,dt,\int_{0}^{2}1\,dt\rangle=\langle2,2\rangle\). 3. The reversed integral is \(\langle-2,-2\rangle\).

The integrals are negatives of each other; \(\int_{2}^{0}\langle t,1\rangle\,dt=\langle-2,-2\rangle\).

- Integrate the velocity vector to find the position change. - The integral alone is not the final position unless the starting position is zero.

1. Integrate the components from zero to two: \(\int_{0}^{2}\langle t,2\rangle\,dt=\langle2,4\rangle\). 2. Add the starting position to get \(\mathbf{r}(2)=\langle3,3\rangle\).

\(\langle3,3\rangle\).

- Integrate both velocity components over the stated interval. - How are the start, net displacement, and endpoint related? - Check that adding your displacement to the start gives the given final position.

1. Integrate velocity to get displacement: \(\int_{0}^{2}\langle2t,-1\rangle\,dt=\langle4,-2\rangle\). 2. The final position equals the initial position plus displacement. 3. Therefore \(\mathbf{r}(0)=\langle5,0\rangle-\langle4,-2\rangle=\langle1,2\rangle\).

\(\mathbf{r}(0)=\langle1,2\rangle\).

- Use the Fundamental Theorem of Calculus to relate the two vector values. - Compute the accumulated change by integrating each derivative component from \(0\) to \(2\). - Reverse the change by subtracting it from the known later vector.

1. The accumulated vector change from \(t=0\) to \(t=2\) is \(\int_{0}^{2}\mathbf{r}'(t)\,dt=\left\langle 4, 2\right\rangle\). 2. Since \(\mathbf{r}(2)=\mathbf{r}(0)+\int_{0}^{2}\mathbf{r}'(t)\,dt\), subtracting the change gives \(\mathbf{r}(0)=\left\langle 7, 4\right\rangle-\left\langle 4, 2\right\rangle=\left\langle 3, 2\right\rangle\).

\(\mathbf{r}(0)=\left\langle 3, 2\right\rangle\)

- Express the later vector as the earlier vector plus the integral of \(\mathbf{r}'(t)\). - Integrate the sine and cosine components over \(\left[0,\frac{\pi}{2}\right]\). - Subtract the accumulated vector change from the known later value.

1. The accumulated change from \(t=0\) to \(t=\frac{\pi}{2}\) is \(\int_{0}^{\pi/2}\mathbf{r}'(t)\,dt=\left\langle 1, 1\right\rangle\). 2. Since \(\mathbf{r}\left(\frac{\pi}{2}\right)=\mathbf{r}(0)+\left\langle 1, 1\right\rangle\), subtracting gives \(\mathbf{r}(0)=\left\langle 2, 4\right\rangle\).

\(\mathbf{r}(0)=\left\langle 2, 4\right\rangle\)

- Use the integral of the derivative to represent the change from \(t=0\) to \(t=1\). - Evaluate the exponential and polynomial component integrals separately. - Recover the earlier vector by subtracting the accumulated change from \(\mathbf{r}(1)\).

1. The accumulated change from \(t=0\) to \(t=1\) is \(\int_{0}^{1}\mathbf{r}'(t)\,dt=\left\langle e-1, \frac{1}{2}\right\rangle\). 2. Subtracting this change from \(\mathbf{r}(1)\) gives \(\mathbf{r}(0)=\left\langle e, \frac{3}{2}\right\rangle-\left\langle e-1, \frac{1}{2}\right\rangle=\left\langle 1, 1\right\rangle\).

\(\mathbf{r}(0)=\left\langle 1, 1\right\rangle\)

- Relate \(\mathbf{r}(3)\) and \(\mathbf{r}(1)\) through the integral of the derivative from \(1\) to \(3\). - Compute the two polynomial integrals independently. - Subtract the resulting change vector component by component.

1. The accumulated change from \(t=1\) to \(t=3\) is \(\int_{1}^{3}\mathbf{r}'(t)\,dt=\left\langle \frac{26}{3}, 8\right\rangle\). 2. Subtracting this change from \(\mathbf{r}(3)\) gives \(\mathbf{r}(1)=\left\langle 10, 12\right\rangle-\left\langle \frac{26}{3}, 8\right\rangle=\left\langle \frac{4}{3}, 4\right\rangle\).

\(\mathbf{r}(1)=\left\langle \frac{4}{3}, 4\right\rangle\)

- Antidifferentiate each coordinate function as a separate scalar problem. - Ask whether changing one coordinate's constant must force the other coordinate's constant to change. - Use both coordinates of the initial vector to determine the two constants.

1. Integrating componentwise gives \(\mathbf{r}(t)=\left\langle t^2+C_1,\sin t+C_2\right\rangle\). 2. The two components are independent scalar antiderivatives, so each requires its own integration constant; together, these constants form a constant vector. 3. Applying \(\mathbf{r}(0)=\left\langle1,1\right\rangle\) gives \(C_1=1\) and \(C_2=1\). 4. Therefore, \(\mathbf{r}(t)=\left\langle t^2+1,\sin t+1\right\rangle\).

A constant vector is needed because each component has its own independent integration constant. \(\mathbf{r}(t)=\left\langle t^2+1,\sin t+1\right\rangle\)

- Determine which operation reverses a given derivative. - Integrate the two components separately and include a constant vector. - Use the initial vector only after finding the general antiderivative.

1. Because \(\mathbf{r}'(t)\) is given and \(\mathbf{r}(t)\) is required, integrate each component rather than differentiating. 2. A general antiderivative is \(\mathbf{r}(t)=\left\langle e^t+C_1,\frac{t^3}{3}+C_2\right\rangle\). 3. Applying \(\mathbf{r}(0)=\left\langle0,0\right\rangle\) gives \(C_1=-1\) and \(C_2=0\). 4. Therefore, \(\mathbf{r}(t)=\left\langle e^t-1,\frac{t^3}{3}\right\rangle\).

Chiamaka must antidifferentiate, not differentiate. \(\mathbf{r}(t)=\left\langle e^t-1,\frac{t^3}{3}\right\rangle\)

- Identify the logarithmic and inverse-tangent antiderivatives component by component. - Keep separate constants because the two coordinates are independent. - Substitute \(t=1\) into the general antiderivative and match both coordinates of the initial vector.

1. Integrating componentwise gives \(\mathbf{r}(t)=\left\langle\ln t+C_1,\arctan t+C_2\right\rangle\). 2. Applying \(\mathbf{r}(1)=\left\langle2,3\right\rangle\) gives \(C_1=2\) and \(C_2=3-\frac{\pi}{4}\). 3. Therefore, \(\mathbf{r}(t)=\left\langle\ln t+2,\arctan t+3-\frac{\pi}{4}\right\rangle\).

\(\mathbf{r}(t)=\left\langle\ln t+2,\arctan t+3-\frac{\pi}{4}\right\rangle\)

- Start with the full family of componentwise antiderivatives, including an independent constant in each component. - Use the given vector value at \(t=\frac{\pi}{2}\) to determine those constants. - Verify the final function by differentiating it and checking the initial condition.

1. Integrating componentwise gives \(\mathbf{r}(t)=\left\langle -\cos(t)+C_{1}, \sin(t)+C_{2}\right\rangle\). 2. Applying \(\mathbf{r}\left(\frac{\pi}{2}\right)=\left\langle 1, 0\right\rangle\) gives \(C_{1}=1\) and \(C_{2}=-1\). 3. Therefore, \(\mathbf{r}(t)=\left\langle 1-\cos(t), \sin(t)-1\right\rangle\).

\(\mathbf{r}(t)=\left\langle 1-\cos(t), \sin(t)-1\right\rangle\)

- Treat each component column as a separate table of scalar function values. - Use the spacing between consecutive \(t\)-values as the trapezoid width. - Apply the trapezoidal rule to both columns, then place the approximations in vector order.

1. The subinterval width is \(1\). Apply the trapezoidal rule separately to the two component columns. 2. For the first component, the approximation is \(\frac{1}{2}[(1+3)+(3+2)+(2+4)]=\frac{15}{2}\). 3. For the second component, the approximation is \(\frac{1}{2}[(2+0)+(0-1)+(-1+1)]=\frac{1}{2}\). 4. Therefore, \(\int_{0}^{3}\mathbf{v}(t)\,dt\approx\left\langle \frac{15}{2}, \frac{1}{2}\right\rangle\).

\(\left\langle \frac{15}{2}, \frac{1}{2}\right\rangle\)

- Determine the common width from the listed \(t\)-values before using the rule. - Compute one trapezoidal approximation for each component column. - Keep the first- and second-component results in the same order when forming the vector.

1. The subinterval width is \(0.5\). Apply the trapezoidal rule separately to the two component columns. 2. For the first component, the approximation is \(\frac{0.5}{2}[(0+2)+(2+4)+(4+6)]=\frac{9}{2}\). 3. For the second component, the approximation is \(\frac{0.5}{2}[(2+4)+(4+2)+(2+0)]=\frac{7}{2}\). 4. Therefore, \(\int_{0}^{1.5}\mathbf{v}(t)\,dt\approx\left\langle \frac{9}{2}, \frac{7}{2}\right\rangle\).

\(\left\langle \frac{9}{2}, \frac{7}{2}\right\rangle\)

- Apply the numerical integration rule independently to the two component columns. - Preserve negative table values when averaging adjacent endpoints. - Combine the two scalar approximations into one vector after both sums are complete.

1. The subinterval width is \(1\). Apply the trapezoidal rule separately to the two component columns. 2. For the first component, the approximation is \(\frac{1}{2}[(2+1)+(1+3)+(3+5)]=\frac{15}{2}\). 3. For the second component, the approximation is \(\frac{1}{2}[(-2+1)+(1+2)+(2+0)]=2\). 4. Therefore, \(\int_{1}^{4}\mathbf{v}(t)\,dt\approx\left\langle \frac{15}{2}, 2\right\rangle\).

\(\left\langle \frac{15}{2}, 2\right\rangle\)

- Recover \(\mathbf{r}(t)\) by integrating both derivative components. - Use the condition at \(t=0\) to determine the constant vector before using the later value. - Match corresponding components of the two vectors at \(t=1\).

1. Integrating componentwise gives \(\mathbf r(t)=\left\langle\frac{at^2}{2}+C_1,t^2+C_2\right\rangle\). 2. The condition \(\mathbf r(0)=\langle0,0\rangle\) gives \(C_1=C_2=0\), so before solving for the parameter, \(\mathbf r(t)=\left\langle\frac{at^2}{2},t^2\right\rangle\). 3. Substituting \(t=1\) gives \(\left\langle\frac{a}{2},1\right\rangle=\langle3,1\rangle\), so \(a=6\). 4. The final vector function is \(\mathbf r(t)=\langle3t^2,t^2\rangle\).

Before solving: \(\mathbf r(t)=\left\langle\frac{at^2}{2},t^2\right\rangle\) \(a=6\) Final: \(\mathbf r(t)=\langle3t^2,t^2\rangle\)

- Integrate the two trigonometric components while keeping the parameter \(a\) symbolic. - Apply the initial vector to simplify the constant terms before evaluating at the later time. - Compare corresponding coordinates at \(t=\frac{\pi}{2}\).

1. Integrating componentwise gives \(\mathbf r(t)=\langle\sin t+C_1,-a\cos t+C_2\rangle\). 2. Applying \(\mathbf r(0)=\langle0,0\rangle\) gives \(C_1=0\) and \(C_2=a\), so before solving for the parameter, \(\mathbf r(t)=\langle\sin t,a(1-\cos t)\rangle\). 3. At \(t=\frac{\pi}{2}\), this becomes \(\langle1,a\rangle=\langle1,4\rangle\), so \(a=4\). 4. The final vector function is \(\mathbf r(t)=\langle\sin t,4(1-\cos t)\rangle\).

Before solving: \(\mathbf r(t)=\langle\sin t,a(1-\cos t)\rangle\) \(a=4\) Final: \(\mathbf r(t)=\langle\sin t,4(1-\cos t)\rangle\)

- Integrate each component with \(a\) left as an unknown constant. - Use the vector value at \(t=1\) to determine the complete constant vector in terms of \(a\). - At \(t=3\), the same value of \(a\) must make the entire vector agree with the given vector.

1. Integrating componentwise gives \(\mathbf r(t)=\left\langle at+C_1,\frac{t^3}{3}+C_2\right\rangle\). 2. Applying \(\mathbf r(1)=\langle2,1\rangle\) gives \(C_1=2-a\) and \(C_2=\frac{2}{3}\), so before solving for the parameter, \(\mathbf r(t)=\left\langle at+2-a,\frac{t^3}{3}+\frac{2}{3}\right\rangle\). 3. At \(t=3\), this gives \(\left\langle2a+2,\frac{29}{3}\right\rangle=\left\langle8,\frac{29}{3}\right\rangle\), so \(a=3\). 4. The final vector function is \(\mathbf r(t)=\left\langle3t-1,\frac{t^3}{3}+\frac{2}{3}\right\rangle\).

Before solving: \(\mathbf r(t)=\left\langle at+2-a,\frac{t^3}{3}+\frac{2}{3}\right\rangle\) \(a=3\) Final: \(\mathbf r(t)=\left\langle3t-1,\frac{t^3}{3}+\frac{2}{3}\right\rangle\)

- Treat the two components as separate scalar integrals. - Each integrand contains the derivative of an inner expression. - Apply the original bounds after making each substitution.

1. Integrate componentwise. 2. For the first component, substitute \(u=t^2\) to obtain \(\int_0^1 2te^{t^2}\,dt=e-1\). 3. For the second component, substitute \(u=t^3\) to obtain \(\int_0^1 3t^2\cos(t^3)\,dt=\sin(1)\). 4. Therefore, the vector integral is \(\langle e-1,\sin(1)\rangle\).

\(\langle e-1,\sin(1)\rangle\)

- Locate the zero of each expression inside an absolute value. - The two components need different interval splits. - Integrate each resulting piece before assembling the vector.

1. The first component changes form at \(t=0\): \(\int_{-2}^{1}|t|\,dt=\int_{-2}^{0}(-t)\,dt+\int_0^1t\,dt=\frac{5}{2}\). 2. The second component changes form at \(t=-1\): \(\int_{-2}^{1}|t+1|\,dt=\int_{-2}^{-1}-(t+1)\,dt+\int_{-1}^{1}(t+1)\,dt=\frac{5}{2}\). 3. Therefore, the vector integral is \(\left\langle\frac{5}{2},\frac{5}{2}\right\rangle\).

\(\left\langle\frac{5}{2},\frac{5}{2}\right\rangle\)

- Integrate the vector componentwise before using the target vector. - Translate equality of the two vectors into two scalar constraints on the same value of \(a\). - A valid parameter must satisfy both component constraints simultaneously.

1. Integrating componentwise gives \(\int_0^2\langle at,t^2-a\rangle\,dt=\left\langle2a,\frac{8}{3}-2a\right\rangle\). 2. Matching the target vector gives the two scalar equations \(2a=6\) and \(\frac{8}{3}-2a=-\frac{10}{3}\). 3. The first equation gives \(a=3\), and this value also satisfies the second equation.

Integral: \(\left\langle2a,\frac{8}{3}-2a\right\rangle\) Component equations: \(2a=6\) and \(\frac{8}{3}-2a=-\frac{10}{3}\) \(a=3\)

- Antidifferentiate the two components independently. - Look for an inner expression whose derivative is already present. - Use the initial vector to determine both constants.

1. Integrating componentwise gives \(\mathbf{r}(t)=\langle\sin(t^2)+C_1,\ln(1+t)+C_2\rangle\). 2. Applying \(\mathbf{r}(0)=\langle3,-1\rangle\) gives \(C_1=3\) and \(C_2=-1\). 3. Therefore, \(\mathbf{r}(t)=\langle\sin(t^2)+3,\ln(1+t)-1\rangle\).

\(\mathbf{r}(t)=\langle\sin(t^2)+3,\ln(1+t)-1\rangle\)

- Apply the numerical rule separately to the two data columns. - Check that the subintervals have equal width and that their number is even. - Use the same endpoint, odd-index, and even-index weights in each component.

1. The spacing is \(h=1\), so Simpson’s rule uses the factor \(\frac{1}{3}\). 2. The first-component estimate is \(\frac{1}{3}[0+16+4(1+9)+2(4)]=\frac{64}{3}\). 3. The second-component estimate is \(\frac{1}{3}[1+1+4(2+2)+2(0)]=6\). 4. Therefore, \(\int_0^4\mathbf{v}(t)\,dt\approx\left\langle\frac{64}{3},6\right\rangle\).

\(\int_0^4\mathbf{v}(t)\,dt\approx\left\langle\frac{64}{3},6\right\rangle\)

- Integrate the two components over the stated interval. - A zero vector requires both resulting components to be zero. - Solve the two scalar equations independently.

1. Integrating componentwise gives \(\int_0^{\pi}\mathbf{v}(t)\,dt=\langle a\pi,b\pi+2\rangle\). 2. The first component requires \(a\pi=0\), so \(a=0\). 3. The second component requires \(b\pi+2=0\), so \(b=-\frac{2}{\pi}\).

\(a=0\) and \(b=-\frac{2}{\pi}\)

- The definition changes at an interior parameter value. - Integrate each piece over its own interval. - Add the two resulting vectors componentwise.

1. Split the vector integral at \(t=1\). 2. On \([0,1]\), the integral is \(\left\langle\frac{1}{2},1\right\rangle\). 3. On \([1,2]\), the integral is \(\left\langle\frac{1}{2},\frac{7}{3}\right\rangle\). 4. Adding the two vectors gives \(\left\langle1,\frac{10}{3}\right\rangle\).

\(\left\langle1,\frac{10}{3}\right\rangle\)

- Write the average-value formula using the unknown interval length. - Simplify the vector before comparing components. - Check that both components give the same positive endpoint.

1. The average vector is \(\frac{1}{b}\int_0^b\langle t,2t\rangle\,dt=\left\langle\frac{b}{2},b\right\rangle\). 2. Matching \(\langle3,6\rangle\) gives \(\frac{b}{2}=3\) and \(b=6\). 3. Both equations yield \(b=6\), which satisfies \(b>0\).

Average vector: \(\left\langle\frac{b}{2},b\right\rangle\) Component equations: \(\frac{b}{2}=3\) and \(b=6\) \(b=6\)

- Each component suggests a different inner expression. - Change the bounds along with each substitution. - Keep the two scalar results in their original component order.

1. For the first component, substitute \(u=1+t^4\) to obtain \(\int_0^1\frac{4t^3}{1+t^4}\,dt=\ln2\). 2. For the second component, substitute \(u=1+t^2\) to obtain \(\int_0^1\frac{2t}{(1+t^2)^2}\,dt=\int_1^2u^{-2}\,du=\frac{1}{2}\). 3. Therefore, the vector integral is \(\left\langle\ln2,\frac{1}{2}\right\rangle\).

\(\left\langle\ln2,\frac{1}{2}\right\rangle\)

- Evaluate the vector integral before applying the geometric condition. - Use a scalar condition that characterizes perpendicular vectors. - Solve the resulting equation for the parameter.

1. The vector integral is \(\left\langle1,\frac{a}{2}\right\rangle\). 2. Perpendicularity requires \(\left\langle1,\frac{a}{2}\right\rangle\cdot\langle2,-1\rangle=0\). 3. Thus \(2-\frac{a}{2}=0\), so \(a=4\).

\(a=4\)

- Compute the vector integral componentwise first. - Simplify the magnitude using a trigonometric identity. - Think about cancellation in a signed vector compared with accumulation of nonnegative quantities.

1. Integrating the vector components gives \(\int_0^{2\pi}\mathbf{v}(t)\,dt=\langle0,0\rangle\). 2. The magnitude is \(\|\mathbf{v}(t)\|=\sqrt{\cos^2t+\sin^2t}=1\). 3. Therefore, \(\int_0^{2\pi}\|\mathbf{v}(t)\|\,dt=2\pi\). 4. The vector integral allows directional cancellation, while the integral of the magnitude accumulates nonnegative size.

\(\int_0^{2\pi}\mathbf{v}(t)\,dt=\langle0,0\rangle\), while \(\int_0^{2\pi}\|\mathbf{v}(t)\|\,dt=2\pi\). The vector components cancel over a full cycle, but the magnitudes do not.

- Convert the average derivative vector into a definite integral. - Interpret that integral as net change in the vector function. - Work backward from the later position.

1. Because the interval length is \(2\), \(\int_1^3\mathbf{r}'(t)\,dt=2\langle2,-1\rangle=\langle4,-2\rangle\). 2. Net change gives \(\mathbf{r}(3)-\mathbf{r}(1)=\langle4,-2\rangle\). 3. Therefore, \(\mathbf{r}(1)=\langle7,4\rangle-\langle4,-2\rangle=\langle3,6\rangle\).

\(\mathbf{r}(1)=\langle3,6\rangle\)

- Integrate the components separately. - Each component matches a standard inverse-trigonometric derivative. - Treat the endpoint singularity in the first component with a one-sided limit.

1. The first component is improper at \(t=1\), and \(\lim_{b\to1^-}[\arcsin t]_0^b=\frac{\pi}{2}\). 2. The second component is \([\arctan t]_0^1=\frac{\pi}{4}\). 3. Therefore, the vector integral is \(\left\langle\frac{\pi}{2},\frac{\pi}{4}\right\rangle\).

\(\left\langle\frac{\pi}{2},\frac{\pi}{4}\right\rangle\)

- The lower integration limit should match the parameter of the known vector value. - Compute the componentwise net change over the corrected interval. - Add that change to the known vector.

1. The integral must begin at the parameter of the known position, so \(\mathbf{r}(4)=\mathbf{r}(1)+\int_1^4\mathbf{r}'(t)\,\text{d}t\). 2. The change vector is \(\int_1^4\langle t,2\rangle\,\text{d}t=\left\langle\frac{15}{2},6\right\rangle\). 3. Therefore, \(\mathbf{r}(4)=\langle3,-1\rangle+\left\langle\frac{15}{2},6\right\rangle=\left\langle\frac{21}{2},5\right\rangle\).

The correct limits are \(1\) to \(4\), and \(\mathbf{r}(4)=\left\langle\frac{21}{2},5\right\rangle\).

- Integrate with separate constants in the coordinate components. - Translate the y-axis condition into a restriction on one constant. - The distance condition allows two signed values for the other constant.

1. The general antiderivative is \(\mathbf{r}(t)=\langle t^2+C_1,C_2\rangle\). 2. Since \(\mathbf{r}(0)=(C_1,C_2)\) lies on the y-axis, \(C_1=0\). 3. Its distance from the origin is \(|C_2|=5\), so \(C_2=5\) or \(C_2=-5\). 4. The two functions are \(\langle t^2,5\rangle\) and \(\langle t^2,-5\rangle\).

\(\mathbf{r}(t)=\langle t^2,5\rangle\) or \(\mathbf{r}(t)=\langle t^2,-5\rangle\)

- Treat each adjacent pair of data points as its own trapezoid. - Use the actual width of each subinterval rather than one common width. - Add the vector contributions componentwise.

1. On \([0,1]\), the trapezoidal contribution is \(\frac{1}{2}[\langle0,2\rangle+\langle2,0\rangle]=\langle1,1\rangle\). 2. On \([1,3]\), the width is \(2\), so the contribution is \(\frac{2}{2}[\langle2,0\rangle+\langle4,1\rangle]=\langle6,1\rangle\). 3. Therefore, \(\int_0^3\mathbf{v}(t)\,dt\approx\langle7,2\rangle\).

\(\int_0^3\mathbf{v}(t)\,dt\approx\langle7,2\rangle\)

- Test convergence component by component. - Replace the infinite upper bound with a finite variable before evaluating. - Assemble the vector only after taking both limits.

1. Integrate each component as an improper integral. 2. The first component is \(\lim_{b\to\infty}\left[-\frac{1}{t}\right]_1^b=1\). 3. The second component is \(\lim_{b\to\infty}[-e^{-t}]_1^b=e^{-1}\). 4. Therefore, the vector integral converges to \(\langle1,e^{-1}\rangle\).

\(\langle1,e^{-1}\rangle\)

- Express the change in the vector function as a definite integral. - Use symmetry separately for the two components. - Add the resulting change vector to the known endpoint.

1. Net change gives \(\mathbf{r}(1)-\mathbf{r}(-1)=\int_{-1}^{1}\langle|t|,t\rangle\,dt\). 2. The first component is \(\int_{-1}^{1}|t|\,dt=1\). 3. The second component is \(0\) because \(t\) is odd on a symmetric interval. 4. Therefore, \(\mathbf{r}(1)=\langle0,2\rangle+\langle1,0\rangle=\langle1,2\rangle\).

\(\mathbf{r}(1)=\langle1,2\rangle\)

- What does the area under a velocity-component graph measure? - Treat the two components separately before combining them in a vector. - Each graph has two simple shapes whose areas can be found from the grid.

1. Displacement is \(\int_{0}^{2}\langle v_x(t),v_y(t)\rangle\,dt\). 2. The horizontal component has a triangle of area \(\frac{1}{2}\) on \([0,1]\) and a rectangle of area \(1\) on \([1,2]\), totaling \(\frac{3}{2}\). 3. The vertical component has a rectangle of area \(1\) on \([0,1]\) and a triangle of area \(\frac{1}{2}\) on \([1,2]\), totaling \(\frac{3}{2}\). 4. The displacement vector is \(\langle\frac{3}{2},\frac{3}{2}\rangle\).

\(\int_{0}^{2}\langle v_x(t),v_y(t)\rangle\,dt=\left\langle\frac{3}{2},\frac{3}{2}\right\rangle\).

- Integrate the vector function twice. - Apply the derivative condition after the first integration. - Apply the position condition only after the second integration.

1. Integrating once gives \(\mathbf{r}'(t)=\langle3t^2+C_1,2e^t+C_2\rangle\). 2. The condition \(\mathbf{r}'(0)=\langle1,-2\rangle\) gives \(C_1=1\) and \(C_2=-4\). 3. Integrating again gives \(\mathbf{r}(t)=\langle t^3+t+D_1,2e^t-4t+D_2\rangle\). 4. The condition \(\mathbf{r}(0)=\langle3,4\rangle\) gives \(D_1=3\) and \(D_2=2\).

\(\mathbf{r}(t)=\langle t^3+t+3,2e^t-4t+2\rangle\)

- The two components can both be approached by separating a product. - In the logarithmic component, simplify the rational expression after the first integration step. - Evaluate exact endpoint values rather than using decimals.

1. Integrate the first component by parts: \(\int_0^1te^t\,dt=[(t-1)e^t]_0^1=1\). 2. For the second component, integration by parts gives \(\int_0^1t\ln(1+t)\,dt=\frac{1}{2}\ln2-\frac{1}{2}\int_0^1\frac{t^2}{1+t}\,dt\). 3. Since \(\frac{t^2}{1+t}=t-1+\frac{1}{1+t}\), the remaining integral is \(\ln2-\frac{1}{2}\). 4. The second component is therefore \(\frac{1}{4}\), so the vector integral is \(\left\langle1,\frac{1}{4}\right\rangle\).

\(\left\langle1,\frac{1}{4}\right\rangle\)

- Begin with independent constants in the two antiderivative components. - Translate the geometric condition at \(t=0\) into a relation between those constants. - One free constant should remain because the condition describes a line of possible starting points.

1. The general antiderivative is \(\mathbf{r}(t)=\langle t^2+C_1,-t^2+C_2\rangle\). 2. At \(t=0\), the point is \((C_1,C_2)\). 3. The line condition requires \(C_2=2C_1+1\). 4. Writing \(C_1=c\), all solutions are \(\mathbf{r}(t)=\langle t^2+c,-t^2+2c+1\rangle\), where \(c\) is any real number.

\(\mathbf{r}(t)=\langle t^2+c,-t^2+2c+1\rangle\), where \(c\in\mathbb{R}\)

- Use accumulated component changes from the initial position. - First determine when the vertical accumulated change is zero again. - Evaluate the horizontal component only after finding that time.

1. Integrating the vertical component from \(0\) to \(t\) gives \(y(t)=2+\frac{1-\cos(3t)}{3}\). 2. The condition \(y(t)=2\) requires \(\cos(3t)=1\). The first positive solution is \(t=\frac{2\pi}{3}\). 3. Integrating the horizontal component gives \(x(t)=-1+\frac{e^{2t}-1}{2}=\frac{e^{2t}-3}{2}\). 4. At \(t=\frac{2\pi}{3}\), the x-coordinate is \(\frac{e^{4\pi/3}-3}{2}\).

The first positive time is \(t=\frac{2\pi}{3}\), and the x-coordinate is \(\frac{e^{4\pi/3}-3}{2}\).

- Net change is a definite vector integral, so no initial vector is needed. - Treat each component with the product-integration strategy it requires. - Use the signs of the final components to identify the quadrant.

1. The net change is \(\int_0^{\pi}\langle t\cos t,te^t\rangle\,dt\). 2. Integration by parts gives \(\int_0^{\pi}t\cos t\,dt=-2\). 3. Integration by parts gives \(\int_0^{\pi}te^t\,dt=(\pi-1)e^{\pi}+1\). 4. The change vector is \(\left\langle-2,(\pi-1)e^{\pi}+1\right\rangle\). Its first component is negative and its second is positive, so it lies in Quadrant II.

\(\left\langle-2,(\pi-1)e^{\pi}+1\right\rangle\), in Quadrant II

- Integrate the two squared trigonometric components using equivalent double-angle forms. - Compare the resulting coordinate expressions rather than solving them separately. - Choose the first positive solution before evaluating the position.

1. Power-reduction identities give \(x(t)=\frac{t}{2}-\frac{\sin(2t)}{4}\) and \(y(t)=\frac{t}{2}+\frac{\sin(2t)}{4}\). 2. The coordinates are equal when \(\sin(2t)=0\). 3. The first positive solution is \(t=\frac{\pi}{2}\). 4. At that time, \(x=y=\frac{\pi}{4}\), so the position is \(\left\langle\frac{\pi}{4},\frac{\pi}{4}\right\rangle\).

\(t=\frac{\pi}{2}\), at \(\left\langle\frac{\pi}{4},\frac{\pi}{4}\right\rangle\)

- Start with the full family of componentwise antiderivatives. - Apply the average-value condition to each coordinate function. - The interval length is \(1\), which simplifies the averages.

1. Integrating gives \(\mathbf{r}(t)=\langle t+C_1,t^2+C_2\rangle\). 2. The average first component is \(\int_0^1(t+C_1)\,dt=\frac{1}{2}+C_1\), so \(C_1=\frac{5}{2}\). 3. The average second component is \(\int_0^1(t^2+C_2)\,dt=\frac{1}{3}+C_2\), so \(C_2=\frac{11}{3}\). 4. Therefore, \(\mathbf{r}(t)=\left\langle t+\frac{5}{2},t^2+\frac{11}{3}\right\rangle\).

\(\mathbf{r}(t)=\left\langle t+\frac{5}{2},t^2+\frac{11}{3}\right\rangle\)