1



[pic], so [pic] B

1. The x-intercepts are at [pic]; y-intercept is at[pic]. [pic] A

2. The areas are equivalent for all real number triples [pic]. Hence, all the choices are valid. C

3. [pic], which results in the answer of [pic] C

4. [pic] ( [pic], but [pic], so [pic] ( [pic] ( [pic] A

5. Let [pic] be the base of the triangle and [pic] be the height. By similar triangles, [pic] ( [pic], and [pic], [pic] ( [pic]. With vertices at (0,0), (4,0), and (0,2), the centroid is at [pic] D

6. Model the trapezoidal base as pictured. Then, [pic], [pic] ( [pic] ( [pic], [pic], so the optimal base length is [pic] B

7. All are true by way of the Mean Value Theorem. E

8. [pic] B

9. The region is an ellipse whose center we can shift horizontally to the origin without changing the resulting volume. Then, [pic] B

10. The second volume expression is correct via the Shell method. B

11. The formula for average value over a continuous interval gives us [pic]. Given that [pic], the average value is [pic]. D

12. Solve for b: [pic] ( [pic] ( [pic] B

13. The corresponding Riemann integral is [pic]. We can compute this definite integral, or simplify [pic], both of which result in an answer of [pic]. B

14. The graph of [pic] looks like consecutive upward-facing parabolas, with a period of [pic]. We can find the area between 0 and [pic], and then multiply this by 2010. The bounded area can be split into three parts. From 0 to [pic] is a right half of the “parabola,” between [pic] and [pic] is a rectangle with length[pic] and width 1 , and from [pic] to [pic] is a left half of the “parabola.” In one period, [pic]. Then, [pic] B

15. The shape is a circle of radius 3 centered at (-2,1). For a torus, [pic] C

16. The first region of interest is bound above by the x-axis, so to preserve the correct sign of the area, we integrate backwards from [pic] to [pic]:

[pic], and [pic], so

[pic] C

17. By solving systems of equations, the vertices are found to be: (-4,2),(2,1),(-1,-1). To find area of a polygon with vertices (xn,yn):

[pic]. (Can also use the formulas for distance between point and a line and then area for a right triangle to arrive at the same answer.) C

18. [pic] A

19. By the Second Fundamental Theorem—and remembering to take the derivative of the upper limit—the expression is equivalent to [pic] B

20. [pic] [pic] B

21. Using [pic] over [pic], the resulting approximation is [pic], which is equivalent to the given expression, so the answer is B

22. [pic] D

23. [pic], using disk method: [pic] C

24. The sum of the series[pic] = [pic], where [pic] turns out to be an infinite geometric series with [pic] and [pic]. Thus, [pic], resulting in [pic] C

25. The area of an equilateral triangle with side [pic] is [pic]. By cross-sections, [pic] D

26. Using the disk method: [pic] B

27. [pic] The surface describes a hyperboloid because both the [pic]-trace [pic], and the [pic]-trace [pic] define hyperbolas. When y=0, no trace exists (with [pic]), so the hyperboloid is in two sheets instead of one. B

28. [pic] A

29. [pic] D

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