I will give and 15 points Question 5 (1 point)
Use the limit process to find the area of the region between the graph of the function y=16-x^2
and x-axis over the interval (-4,4].

The function  will be continuous on the open interval (-4,4) if there is no vertical asymptote or hole on (-4,4).

But the given function has a vertical asymptote at .

Therefore  is not continuous on the interval (-4,4) because x=0 belongs to the given interval.

Step-by-step explanation:

i hope this is right :)

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: