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

Question: Evaluate the definite integral of the vector function \(\mathbf{i} + t\mathbf{j} + 3t^2\mathbf{k}\) with respect to t from -1 to 1. Answer: The definite integral of the vector function is \(2\mathbf{i} + 2\mathbf{k}\).

Step-by-step solution

Identify the components of the vector function

The given vector function is \(\mathbf{i} + t\mathbf{j} + 3t^2\mathbf{k}\). The i-component is a constant function, 1; the j-component is a linear function, \(t\); and the k-component is a quadratic function, \(3t^2\).

Integrate i-component with respect to t

We need to integrate the i-component with respect to t, from -1 to 1. So, $$\int_{-1}^{1} 1 dt$$

Evaluate i-component integral

The integral of a constant (1) with respect to t is linear (t). So, $$\left[t\right]_{-1}^{1} = \left[1 - (-1)\right] = 2$$

Integrate j-component with respect to t

Now, we need to integrate the j-component with respect to t, from -1 to 1. So, $$\int_{-1}^{1} t dt$$

Evaluate j-component integral

The integral of the linear function \(t\) with respect to t is quadratic (\(\frac{1}{2}t^2\)). So, $$\left[\frac{1}{2}t^2\right]_{-1}^{1} = \left[\frac{1}{2}(1^2 - (-1)^2)\right] = \left[\frac{1}{2}(1 - 1)\right] = 0$$

Integrate k-component with respect to t

Finally, we need to integrate the k-component with respect to t, from -1 to 1. So, $$\int_{-1}^1 3t^2 dt$$

Evaluate k-component integral

The integral of the quadratic function \(3t^2\) with respect to t is cubic (\(t^3\)). So, $$\left[t^3\right]_{-1}^1 = \left[1^3 - (-1)^3\right] = \left[1 - (-1)\right] = 2$$

Combine results to form the final vector

Now that we have evaluated the definite integral for each of the components, we can form the resulting vector by combining them: $$\left(2\mathbf{i} + 0\mathbf{j} + 2\mathbf{k}\right) = 2\mathbf{i} + 2\mathbf{k}$$