- #1

yusukered07

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*) =*

**t**

*t***i**-

*t*^{2}**j**+ (

**- 1)**

*t***k**and B(

**) = 2**

*t*

*t*^{2}**i**+ 6

*t***k**, evaluate (

*a*) [tex]\int^{2}_{0}A \cdot B dt,[/tex] (

*b)*[tex]\int^{2}_{0}A \times B dt.[/tex]

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- Thread starter yusukered07
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- #1

yusukered07

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- #2

CompuChip

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[tex]\int a t^n \, dt = \frac{a}{n + 1} t^{n + 1}[/tex]

- #3

yusukered07

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[tex]\int a t^n \, dt = \frac{a}{n + 1} t^{n + 1}[/tex]

The solution I've made is not complicated.

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

- #4

HallsofIvy

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

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

- #5

CompuChip

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I think he asks us to evaluate the integrals.

I just did it.

I just did it.

- #6

yusukered07

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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 [tex]A\cdot B[/tex] and [tex]A\times B[/tex] first. Then integrate the both with respect to their limits.

- #7

CompuChip

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Halls meant: what exactly is your

- #8

yusukered07

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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). [tex]A\cdot B = 2t^{3} + 6t^{2} - 6t[/tex]

Taking its integral with respect to

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

Then, taking its integral with respect to

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