Question

Evaluate the following definite integrals. $$\int_{-1}^{1}\left(\mathbf{i}+t \mathbf{j}+3 t^{2} \mathbf{k}\right) d t$$

Short answer

Answer: The result of the definite integral is \(2\mathbf{i} + 2\mathbf{k}\).

Step-by-step solution

Component Functions

The integrand is given by \((\mathbf{i}+t \mathbf{j}+3 t^{2} \mathbf{k})\). We will need to evaluate the integral of each component function: 1. The i component which is \(1\). 2. The j component which is \(t\). 3. The k component which is \(3t^2\).

Integrating the i Component

$$\int_{-1}^{1} 1 dt$$ To calculate the integral of a constant, multiply the constant by the difference between the integration limits. In our case, the integral of the i component is simply: \(1 * (1 - (-1)) = 1 * 2 = 2\).

Integrating the j Component

$$\int_{-1}^{1} t dt$$ To calculate the integral of \(t\) with respect to \(t\), apply the power rule: add 1 to the exponent and then divide by the new exponent. Then evaluate the integral and apply the Fundamental Theorem of Calculus: $$\int_{-1}^{1} t dt = \left[\frac{t^2}{2}\right]_{-1}^{1} = \frac{(1)^{2}}{2}-\frac{(-1)^{2}}{2} = 0$$.

Integrating the k Component

$$\int_{-1}^{1} 3t^2 dt$$ To calculate the integral of \(3t^2\) with respect to \(t\), apply the power rule again: $$\int_{-1}^{1} 3t^2 dt = \left[t^3\right]_{-1}^{1} = (1)^{3}-(-1)^{3} = 1 - (-1) = 2$$.

Combining the Components

Now that we have evaluated the individual integrals, combine them to get the overall result for the definite integral: $$\int_{-1}^{1}\left(\mathbf{i}+t \mathbf{j}+3 t^{2} \mathbf{k}\right) d t = 2\mathbf{i} + 0\mathbf{j} + 2\mathbf{k} = 2\mathbf{i} + 2\mathbf{k}$$