91Ó°ÊÓ

We are given the vector-valued function: $$ f(t) = t e^{t}(\mathbf{i}+2 \mathbf{j}-\mathbf{k}) $$ The components of the vector-valued function are: $$ f_x(t) = t e^{t} \mathbf{i}, \\ f_y(t) = 2 t e^{t} \mathbf{j}, \\ f_z(t) = -t e^{t} \mathbf{k}. $$

: We are asked to evaluate the definite integral for each component over the interval [0, 2]. So, we find the integral of each component in terms of 't' and then evaluate its value at the given limits: 1. For \(f_x(t) = t e^{t} \mathbf{i}\): $$ \int_{0}^{2} f_x(t) dt = \int_{0}^{2} t e^{t} dt \mathbf{i} $$ 2. For \(f_y(t) = 2 t e^{t} \mathbf{j}\): $$ \int_{0}^{2} f_y(t) dt = \int_{0}^{2} 2 t e^{t} dt \mathbf{j} $$ 3. For \(f_z(t) = -t e^{t} \mathbf{k}\): $$ \int_{0}^{2} f_z(t) dt = \int_{0}^{2} -t e^{t} dt \mathbf{k} $$

1. For the \(f_x(t) = t e^{t} \mathbf{i}\) component, we can use integration by parts: Let $$ u = t, \ dv = e^{t} dt \\ du = dt, \ v = e^t $$ Now we substitute and evaluate the integral: $$ \int_{0}^{2} t e^{t} dt \mathbf{i}= \left[uv\right]_0^2 - \int_{0}^{2} v du \mathbf{i} = \left[te^t\right]_0^2 - \int_{0}^{2} e^t dt \mathbf{i} $$ $$ = \left[2e^2 - 0 \right] - \left[e^t\right]_0^2 \mathbf{i} = 2e^2 - \left[e^2 - 1\right] \mathbf{i} = (2e^2 - e^2 + 1) \mathbf{i} = (e^2 + 1) \mathbf{i} $$ 2. For the \(f_y(t) = 2 t e^{t} \mathbf{j}\) component, we can apply the similar integration by parts as above: $$ \int_{0}^{2} 2 t e^{t} dt \mathbf{j}= 2 \left[\left( te^t \right)_0^2 - \int_{0}^{2} e^t dt\right] \mathbf{j} = 2 (2e^2 - (e^2 -1))\mathbf{j} = 2(e^2 + 1) \mathbf{j}\\ $$ 3. For the \(f_z(t) = -t e^{t} \mathbf{k}\) component, we apply the similar integration by parts as above: $$ \int_{0}^{2} -t e^{t} dt \mathbf{k}= -\left[\left( te^t \right)_0^2 - \int_{0}^{2} e^t dt\right] \mathbf{k} = - (2e^2 - (e^2 -1))\mathbf{k} = -(e^2 + 1) \mathbf{k} $$

Now we combine the results for each component into a single vector: $$ \int_{0}^{2} t e^{t}(\mathbf{i}+2 \mathbf{j}-\mathbf{k}) d t = (e^2 + 1) \mathbf{i} + 2(e^2 + 1) \mathbf{j} - (e^2 + 1) \mathbf{k} $$