Question

Evaluate the following definite integrals. $$\int_{-\pi}^{\pi}(\sin t \mathbf{i}+\cos t \mathbf{j}+2 t \mathbf{k}) d t$$

Short answer

Answer: The definite integral is 2𝑖.

Step-by-step solution

Evaluate the integral of the sin(t) component with respect to i

To find the integral of the sin(t) component, we'll evaluate: $$ \mathbf{i} \int_{-\pi}^{\pi} \sin{t} \ d t $$ Integrating sin(t) with respect to t, we have: $$ \mathbf{i} \left[-\cos{t}\right]_{-\pi}^{\pi} $$ Now, we substitute the bounds of the integral: $$ \mathbf{i} \left[-\cos{(\pi)}- (-\cos{(-\pi)})\right] = \mathbf{i} \left[-(-1)-(-1)\right] = 2\mathbf{i} $$

Evaluate the integral of the cos(t) component with respect to j

Now, we'll find the integral of the cos(t) component: $$ \mathbf{j} \int_{-\pi}^{\pi} \cos{t} \ d t $$ Integrating cos(t) with respect to t, we get: $$ \mathbf{j} \left[\sin{t}\right]_{-\pi}^{\pi} $$ Substitute the bounds of the integral: $$ \mathbf{j} \left[\sin{(\pi)}-\sin{(-\pi)}\right] = \mathbf{j} [0-0] = 0\mathbf{j} $$

Evaluate the integral of the 2t component with respect to k

Finally, we'll evaluate the integral of the 2t component: $$ \mathbf{k} \int_{-\pi}^{\pi} 2t \ d t $$ Integrating 2t with respect to t, we obtain: $$ \mathbf{k} \left[t^2\right]_{-\pi}^{\pi} $$ Substituting the bounds of the integral: $$ \mathbf{k} \left[(\pi)^2-(-\pi)^2\right] = \mathbf{k} [0] = 0\mathbf{k} $$

Combine the results

Now, summing up the results from the previous steps: $$ 2\mathbf{i} + 0\mathbf{j} + 0\mathbf{k} $$ So, the evaluated definite integral is: $$ \int_{-\pi}^{\pi}(\sin t \mathbf{i}+\cos t \mathbf{j}+2 t \mathbf{k}) d t = 2\mathbf{i} $$