3 years ago

Or lavendar people I need answers

SAT

2 answers:

riadik2000 [5.3K]3 years ago

3 0

That’s a pretty color. I really like it

slava [35]3 years ago

3 0

Answer:

ooo thats pretty

Explanation:

i like it

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It looks like the integral is

$\displaystyle \int_C xy \, dx + x^2y^3 \, dy$

Let's close the loop by adding a line integral over the line segment joining (1, 4) to (0, 0). Then the closed loop is the triangular region

T = {(x, y) : 0 ≤ x ≤ 1 and 0 ≤ y ≤ 4x}

Since the integrand has no singularities over or on the boundary of T, we have by Green's theorem

$\displaystyle \int_C xy \, dx + x^2y^3 \, dy = \iint_T \left(\frac{\partial(x^2y^3)}{\partial x} - \frac{\partial(xy)}{\partial y}\right) \, dx \, dy = \int_0^1 \int_0^{4x} (2xy^3 - x) \, dy \, dx$

Compute the double integral:

$\displaystyle \int_0^1 \int_0^{4x} (2xy^3 - x) \, dy \, dx = \int_0^1 \int_0^{4x} \left(128x^5 - 4x^2\right) \, dx = \frac{60}3$

From this result, we subtract the line integral over the extra line segment we added. Parameterize this path by

C' : {(1 - t, 4 - 4t) : 0 ≤ t ≤ 1}

The line integral over C' is

$\displaystyle \int_{C'} xy \, dx + x^2y^3 \, dy = \int_0^1 (1-t)(4-4t)(-dt) + (1-t)^2(4-4t)^3(-4\,dt)$

$\displaystyle \int_{C'} xy \, dx + x^2y^3 \, dy = \int_0^1 \left(256 t^5-1280 t^4+2560 t^3-2564 t^2+1288 t-260 \right) \, dt = -44$

so that the line integral over C alone is 60/3 - (-44) = 64.

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