# Try this Vector Integration

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

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arildno
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I have no idea!

Can you help me out, by SHOWING WHAT YOU HAVE DONE SO FAR?

mathman
1. If A (t) = t i - t2j + (t -1) k, evaluate (b) $$\int^{2}_{0} A$$
Integral of t = t2/2 = 2
Integral of t2 = t3/3 = 8/3
Integral of (t-1)=t2/2 - t = 0

Net result 2i -(8/3)j

Integral of t = t2/2 = 2
Integral of t2 = t3/3 = 8/3
Integral of (t-1)=t2/2 - t = 0

Net result 2i -(8/3)j

Sorry for giving a wrong problem....

Redbelly98
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If A(t) = t i - t2 j + (t - 1) k and B(t) = 2t2 i + 6t k, evaluate (a) $$\int^{2}_{0}A \cdot Bdt ,$$ (b) $$\int^{2}_{0}A \times B dt.$$
You need to show an attempt at solving the problem before receiving help!

Start by evaluating A·B and AxB.

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Moderator's note: thread moved from "Calculus & Analysis"

Homework assignments or any textbook style exercises are to be posted in the appropriate forum in our https://www.physicsforums.com/forumdisplay.php?f=152" area. This should be done whether the problem is part of one's assigned coursework or just independent study.

You need to show an attempt at solving the problem before receiving help!

Start by evaluating A·B and AxB.

Yeah.... That's the process I've made... Then I integrate them with respect to t and evaluate from 0 to 2

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For the dot product of A and B, I obtained 2*t3+6*t2-6*t. Then integrate that from 0 to 2.

For the cross product, I obtained -6*t3i+(2*t3-8*t2)j+2*t4k.