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**DarkestEvil** Tomorrow's my test and there are a few concepts that I have to work on.

1. Sketch the region bounded by the graphs of the equations, and determine its area.

$\displaystyle y=\frac{4}{\sqrt x}$

$\displaystyle y=0, x=1, x=9$

Answer: A=16

My main problem with this is getting rid of the $\displaystyle \sqrt x$ from under so I can begin to integrate.

rewrite as $\displaystyle \textcolor{red}{4x^{-1/2}}$ , then integrate.

2. Find the particular solution of the differential equation $\displaystyle f''(x)=6(x-1)$ whose graph passes through the point (2,1) and is tangent to the line $\displaystyle 3x-y-5=0$ at that point.

I already found that $\displaystyle f'(x)=3x^2-6x$ and $\displaystyle f(x)=x^3-3x^2$. What now? I don't get what the question is asking me.

f''(x) = 6x-6

f'(x) = 3x^2 - 6x + C ... did you calculate the constant of integration?

what information did they give you that might help you find it?

3. Use the Second Fundamental Theorem of Calculus to find $\displaystyle F'(x)$.

F(x)=(integration sign, x between 0 and x)$\displaystyle t^2\sqrt {1+t^3} dt$

Answer=$\displaystyle x^2+3x+2$

I know that $\displaystyle F(x)$ is the antiderivative of the inside of the integration symbol, but I don't understand how to find $\displaystyle F'(x)$.

your "answer" is incorrect.

$\displaystyle \textcolor{red}{\frac{d}{dx} \int_a^x f(t) \, dt = f(x)}$