Question

Evaluate the definite integral. $$ \int_{-1}^{1}\left(t \mathbf{i}+t^{3} \mathbf{j}+\sqrt[3]{t} \mathbf{k}\right) d t $$

Short answer

The definite integral evaluates to 0 for \(i\) and \(j\) components, and is undefined for the \(k\) component.

Step-by-step solution

Perform r Integration

First, perform the integral \(\int_{-1}^{1} t dt\) for the term corresponding to \(\mathbf{i}\), then perform the integral \(\int_{-1}^{1} t^3 dt\) for the term corresponding to \(\mathbf{j}\), and finally perform the integral \(\int_{-1}^{1} \sqrt[3]{t} dt\) for the term corresponding to \(\mathbf{k}\).

Evaluate Integral for i Component

The integral \(\int_{-1}^{1} t dt\) is 0. This can be solved by remembering the power rule for integration, which states the integral of \(t^n\) with respect to \(t\) is \(\frac{t^{n+1}}{n+1}\), plus the constant of integration C. Evaluating this integral from -1 to 1 gives 0.

Evaluate Integral for j Component

The integral \(\int_{-1}^{1} t^3 dt\) is also 0. This is because we are integrating an odd function over a symmetric interval, which always gives a result of 0.

Evaluate Integral for k Component

The integral \(\int_{-1}^{1} \sqrt[3]{t} dt\) is undefined, because the function \(\sqrt[3]{t}\) isn't defined for negative \(t\).