$\displaystyle\int_C\mathbf F(x,y)\cdot\mathrm d\mathbf r=-\iint_D\left(\frac{\partial(xy+x\cos x)}{\partial x}-\frac{\partial(y\cos x-xy)}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

where D is the triangule region with C as its boundary, given by the set

$D=\{(x,y)\mid0\le x\le2\land0\le y\le8-4x\}$

So we have

$\displaystyle\int_C\mathbf F(x,y)\cdot\mathrm d\mathbf r=-\int_0^2\int_0^{8-4x}((y+\cos x-x\sin x)-(\cos x-x\sin x))\,\mathrm dy\,\mathrm dx$

$\displaystyle\int_C\mathbf F(x,y)\cdot\mathrm d\mathbf r=-\int_0^2\int_0^{8-4x}y\,\mathrm dy\,\mathrm dx=\boxed{-\dfrac{64}3}$

5 0

2 years ago

MARKING AS BRAINLIEST!!

devlian [24]

Answer: They are inverses of each other

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Explanation:

We need to check two things:

f(g(x)) = x
g(f(x)) = x

If both of those equations are true, then f and g are inverses of each other.

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Part 1

$f(x) = -3x-12\\\\f(g(x)) = -3( g(x) )-12\\\\f(g(x)) = -3\left( \frac{-12-x}{3} \right)-12\\\\f(g(x)) = 12+x-12\\\\f(g(x)) = x\\\\$

So far, so good. Now we need to check the other equation mentioned.

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Part 2

$g(x) = \frac{-12-x}{3}\\\\g(f(x)) = \frac{-12-( f(x) )}{3}\\\\g(f(x)) = \frac{-12-( -3x-12 )}{3}\\\\g(f(x)) = \frac{-12+3x+12}{3}\\\\g(f(x)) = \frac{3x}{3}\\\\g(f(x)) = x\\\\$

That works out as well. Both equations mentioned are true, which confirms the two original functions are inverses of one another.

3 0

2 years ago