Question

Evaluate the following definite integrals. $$\int_{0}^{\ln 2}\left(e^{-t} \mathbf{i}+2 e^{2 t} \mathbf{j}-4 e^{t} \mathbf{k}\right) d t$$

Short answer

Question: Evaluate the definite integral of the vector function $$\vec{F}(t) = e^{-t} \mathbf{i} + 2e^{2t} \mathbf{j} - 4e^t \mathbf{k}$$ within the limits [0, ln(2)]. Answer: $$\frac{1}{2} \mathbf{i} + 3 \mathbf{j} - 4 \mathbf{k}$$

Step-by-step solution

Identify the components of the vector function

The given vector function is: $$ \vec{F}(t) = e^{-t} \mathbf{i} + 2e^{2t} \mathbf{j} - 4e^t \mathbf{k} $$ We have three components in the vector function, corresponding to the i, j, and k directions: $$ f(t)_i = e^{-t} $$ $$ f(t)_j = 2e^{2t} $$ $$ f(t)_k = -4e^t $$

Find the definite integral of each component separately

Now we will find the definite integral of each component separately, within the given limits [0, ln(2)]. First component (i-direction): $$ \int_{0}^{\ln 2} e^{-t} dt $$ Second component (j-direction): $$ \int_{0}^{\ln 2} 2e^{2t} dt $$ Third component (k-direction): $$ \int_{0}^{\ln 2} -4e^t dt $$

Evaluate the definite integrals

Evaluate the integrals we found in the previous step. First component (i-direction): $$ \int_{0}^{\ln 2} e^{-t} dt = [- e^{-t}]_{0}^{\ln 2} = -e^{-\ln 2} + e^0 = -\frac{1}{2} + 1 = \frac{1}{2} $$ Second component (j-direction): $$ \int_{0}^{\ln 2} 2e^{2t} dt = [e^{2t}]_{0}^{\ln 2} = e^{2\ln 2} - e^0 = 2^2 - 1 = 3 $$ Third component (k-direction): $$ \int_{0}^{\ln 2} -4e^t dt = -4[e^t]_{0}^{\ln 2} = -4(e^{\ln 2} - e^0) = -4(2 - 1) = -4 $$

Combine the results and form the final vector after integrating

Now that we have the definite integrals of each component, we can combine them to form the final vector. The final vector after integrating is: $$ \frac{1}{2} \mathbf{i} + 3 \mathbf{j} - 4 \mathbf{k} $$