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LuckyWell [14K]
3 years ago
6

Evaluate the double integral i=∫∫dxyda where d is the triangular region with vertices (0,0),(4,0),(0,4).

Mathematics
1 answer:
mario62 [17]3 years ago
8 0
I=\displaystyle\iint_D\mathrm dA=\int_{x=0}^{x=4}\int_{y=0}^{y=4-x}\mathrm dy\,\mathrm dx
I=\displaystyle\int_{x=0}^{x=4}(4-x)\,\mathrm dx
I=8

To verify this, we can simply find the area of the triangle using the well-known formula \dfrac12bh, where b and h are the base and height, respectively of the triangular region D. We have b=h=4, so the area is \dfrac{4^2}2=8, as expected.
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X + 2y =6<br> X- y =3<br> Solution?
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Since one equation has a negative y and the other has a positive y, I'm going to use those since they cancel each other out. Before that, the two y's need to be equal to each other.

x+2y=6
x-y=3

Multiply the bottom equation by two so then you have:

x+2y=6
2x-2y=6

The y's now cancel out:

x=6
2x=6

Add them together

3x=12

Divide

x=4.

To find y, plug x into either equation (*don't have to do both, but I will)

(4)+2y=6
(4)-y=3

Subtract four

2y=2
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Divide each

2y/2 = 2/2
y=1

-y/-1 = -1/-1
y=1

The answer is:
x=4
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I hope that helps!
4 0
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Read 2 more answers
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