# Vector Integration 2

1. Mar 13, 2010

### 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) $$\int^{2}_{0}A \times B dt.$$

2. Mar 14, 2010

### CompuChip

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

3. Mar 14, 2010

### yusukered07

The solution I've made is not complicated.

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

4. Mar 14, 2010

### HallsofIvy

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

5. Mar 14, 2010

### CompuChip

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

6. Mar 14, 2010

### yusukered07

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

7. Mar 14, 2010

### CompuChip

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

8. Mar 14, 2010

### yusukered07

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 $$-24 i- \frac{40}{3}j + \frac{64}{5} k$$