# Vector Integration 2

yusukered07
If A(t) = t i - t2 j + (t - 1) k and B(t) = 2t2 i + 6t k, evaluate (a) $$\int^{2}_{0}A \cdot B dt,$$ (b) $$\int^{2}_{0}A \times B dt.$$

## Answers and Replies

Homework Helper
Doesn't sound too complicated, just plug in A and B, work out the vector products and do the integration using
$$\int a t^n \, dt = \frac{a}{n + 1} t^{n + 1}$$

yusukered07
Doesn't sound too complicated, just plug in A and B, work out the vector products and do the integration using
$$\int a t^n \, dt = \frac{a}{n + 1} t^{n + 1}$$

The solution I've made is not complicated.

You try first to evaluate the vectors and then take the integral of them.

Homework Helper
The solution I've made is not complicated.

You try first to evaluate the vectors and then take the integral of them.
What do you mean "the solution I've made"? You did not show any work or solution at all. What, exactly, are you asking?

Homework Helper
I think he asks us to evaluate the integrals.
I just did it.

yusukered07
What do you mean "the solution I've made"? You did not show any work or solution at all. What, exactly, are you asking?

I evaluate the values of $$A\cdot B$$ and $$A\times B$$ first. Then integrate the both with respect to their limits.

Homework Helper
Yes, that is how you solve it.
Halls meant: what exactly is your question (as in, problem, what you need our help with)?

yusukered07
I think he asks us to evaluate the integrals.
I just did it.

I didn't pretend that I know the answer.

Here is my solution to the problem I posted.

To letter (a). $$A\cdot B = 2t^{3} + 6t^{2} - 6t$$

Taking its integral with respect to t from 0 to 2 will give an answer of 12.

To letter (b). $$A\times B = -6t^{3}i + [2t^{2} (t -1) - 6t^{2}]j - 2t^{4}k$$

Then, taking its integral with respect to t from 0 to 2 will result to $$-24 i- \frac{40}{3}j + \frac{64}{5} k$$