1, find s in terms of t if ds/dt=3t-8/tsquared and s=1and1/2 when t= 1
1, square root of (2-3x)dx
2, given that dy/dx=3/x-2 and y =10 when x=0, find an expression for y in terms of x
1, find( 2e^3x-e^-x) dx
Thanks!
These questions are written badly.
The first one is, I presume, to solve
$\displaystyle \frac{ds}{dt} = 3t - \frac{8}{t^2}$, $\displaystyle s(1) = \frac{3}{2}$
This equation is separable:
$\displaystyle ds = \left ( 3t - \frac{8}{t^2} \right )~dt$
Upon integrating
$\displaystyle s = \frac{3}{2}t^2 + \frac{8}{t} + C$
We know that s(1) = 3/2, so
$\displaystyle \frac{3}{2} = \frac{3}{2} \cdot 1^2 + \frac{8}{1} + C$
Solve for C.
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For the second one I presume you are trying to integrate?
$\displaystyle \int \sqrt{2 - 3x}~dx$
Let $\displaystyle y = 2 - 3x \implies dy = -3~dx$
Then your integral becomes
$\displaystyle \int \sqrt{y}~\frac{dy}{-3}$
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For the third I presume this is
$\displaystyle \frac{dy}{dx} = \frac{3}{x - 2}$, $\displaystyle y(0) = 10$
(Please use parenthesis!)
This is, again, a separable equation:
$\displaystyle dy = \frac{3}{x - 2}~dx$
Integrate both sides by using the substitution u = x - 2.
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I presume the last one is an integration as well and is
$\displaystyle \int (2e^{3x}-e^{-x})~ dx$
(Again, please use parenthesis!)
Split this into two parts:
$\displaystyle \int 2e^{3x}~dx - \int e^{-x}~dx$
For the first integral let y = 3x and the second let u = -x.
-Dan