Evaluate the integral.\(\int_{0}^{2}(4ti - t^{3}j + 3t^{5}k)dt\)

Solution:

Let us evaluate the given integral:\(\int_{0}^{2}(4ti - t^{3}j + 3t^{5}k)dt\)

Using the power rule of integration, ∫ xⁿ dx = (xⁿ⁺¹)/ (n+1)+ C. (Where n ≠ -1)

= \(\int_{0}^{2}4tidt - \int_{0}^{2} t^{3}jdt + \int_{0}^{2}3t^{5}kdt\)

= \(4i[\frac{t^{2}}{2}]_{0}^{2} - j[\frac{t^{4}}{4}]_{0}^{2} + 3k[\frac{t^{6}}{6}]_{0}^{2}\)

Now applying the limits, we get

= 4i[4/2 - 0] - j[16/4 - 0] + 3k[64/6 - 0]

=4i[4/2]-j[16/4] + 3k[64/6]

=8i - 4j + 32k

Evaluate the integral \(\int_{0}^{2}(4ti - t^{3}j + 3t^{5}k)dt\)

Summary:

The integral \(\int_{0}^{2}(4ti - t^{3}j + 3t^{5}k)dt\) results in 8i - 4j + 32k