3 years ago

“the difference of t and four is negative five” gives you

Mathematics

1 answer:

maria [59]3 years ago

8 0

Yo sup??

t-4=-5. (given)

t=-5+4

t=-1

Hope this helps

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∫c<br> x sin y ds, C is the line segment from (0, 1) to (3, 5)

Mnenie [13.5K]

Parameterize the line segment $C$ by

$\mathbf r(t)=\langle x(t),y(t)\rangle=(1-t)\langle0,1\rangle+t\langle3,5\rangle=\langle3t,1+4t\rangle$

where $t\in[0,1]$ . Then

$\mathrm ds=\|\mathbf r'(t)\|\,\mathrm dt$
$\mathrm ds=\|\langle3,4\rangle\|\,\mathrm dt$
$\mathrm ds=5\,\mathrm dt$

So the integral is

$\displaystyle\int_Cx\sin y\,\mathrm ds=5\int_0^1x(t)\sin y(t)\,\mathrm dt$
$=\displaystyle5\int_0^13t\sin(1+4t)\,\mathrm dt$
$=\dfrac{15}{16}(\sin5-\sin-4\cos5)\approx-2.7516$