What is #f(x) = int xsqrt(x-2) dx# if #f(1)=-2 #?
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To find ( f(x) = \int x \sqrt{x - 2} , dx ) given that ( f(1) = -2 ), we first need to integrate the given function and then solve for the constant of integration using the given initial condition.
Let's integrate ( \int x \sqrt{x - 2} , dx ):
Using integration by parts with ( u = x ) and ( dv = \sqrt{x - 2} , dx ), we have: [ du = dx ] [ v = \frac{2}{3}(x - 2)^{3/2} ]
Applying the integration by parts formula: [ \int x \sqrt{x - 2} , dx = uv - \int v , du ]
[ = x \left( \frac{2}{3}(x - 2)^{3/2} \right) - \int \frac{2}{3}(x - 2)^{3/2} , dx ]
[ = \frac{2}{3}x(x - 2)^{3/2} - \frac{2}{3} \cdot \frac{2}{5}(x - 2)^{5/2} + C ]
[ = \frac{2}{3}(x(x - 2)^{3/2} - \frac{2}{5}(x - 2)^{5/2}) + C ]
Now, we use the initial condition ( f(1) = -2 ) to find the value of the constant ( C ): [ f(1) = \frac{2}{3}(1(1 - 2)^{3/2} - \frac{2}{5}(1 - 2)^{5/2}) + C = -2 ]
[ \frac{2}{3}(1(1 - 2)^{3/2} - \frac{2}{5}(1 - 2)^{5/2}) + C = -2 ]
[ \frac{2}{3} \left(1 \cdot (-1)^{3/2} - \frac{2}{5} \cdot (-1)^{5/2}\right) + C = -2 ]
[ \frac{2}{3} \left(-1 - \frac{2}{5} \cdot (-1)\right) + C = -2 ]
[ \frac{2}{3} \left(-1 + \frac{2}{5}\right) + C = -2 ]
[ \frac{2}{3} \left(-\frac{3}{5}\right) + C = -2 ]
[ -\frac{4}{5} + C = -2 ]
[ C = -2 + \frac{4}{5} ]
[ C = -\frac{10}{5} + \frac{4}{5} ]
[ C = -\frac{6}{5} ]
Thus, the function ( f(x) ) is: [ f(x) = \frac{2}{3}(x(x - 2)^{3/2} - \frac{2}{5}(x - 2)^{5/2}) - \frac{6}{5} ]
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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