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**azn989** 1) Think of estimating the area under the curve of radical(4-x^2) on [-2,0] using right-hand end points. As you use larger and larger values for n, the approximated area will approach:

A. Pi

B. 2pi

C. 4pi

D. 8pi

E. infinity

think about the graph of $\displaystyle y = \sqrt{4-x^2}$ ... what geometric figure is it?

2. (the big S thing in calculus) -> S 2 to -2 ( 2 - radical(4-x^2) ) dx =

A. 12.566

B. 14.2832

C. 7.1416

D. 1.717

E. 0.8585

same clue as before ... sketch the graph of $\displaystyle y = 2 - \sqrt{4-x^2}$

A particle's acceleration along a straight line is given by 3 + 2t at any time t. At t=4, the velocity is 4. Which of these gives the correct function for the velocity of this particle at any time t?

A. v(t) = 3t^2 + t^2- 24

B. v(t) = 3

C. v(t) = 3t + t^2 + 4

D. none of these

E. not enuff info. to decide

the antiderivative of acceleration is velocity ... you are also given an initial condition for v(t) so you may determine the value of the constant of integration ... integrate the acceleration and determine the velocity function.

A paperweight is thrown up with initial vel. of 60 m/s and intial height of 6 meters. The accel(due to gravity) of the object is a(t) = 9.8 m/s^2, where t is in secs.

4. The equation that describes the paperweight's velocity as a function of time t is :

A. v(t) = -9.8t + C

B. v(t) = -9.8t - 60

C. v(t) = -4.9t^2 + 60

D. v(t) = -9.8t + 60

E. v(t) = -4.9t^2 - 60

$\displaystyle y(t) = 6 + 60t - 4.9t^2$ ... velocity is the derivative of position.

5. The maximum height of the paperweight is:

A. 6.122 m

B. 189.673 m

C. -4.9t^2 + 60t + 6 m

D. 3.06 m

E. 144.692 m

when the paperweight achieves its max height v = 0