## mlddmlnog 2 years ago evaluate the following: ( calculus )

1. mlddmlnog

|dw:1358278438909:dw|

2. mlddmlnog

@zepdrix

3. exraven

u can evaluate this easily by using the properties of definite integral $\int\limits_{3}^{5} \left( 2P(x)+Q(x) \right)dx=\int\limits_{3}^{5} 2P(x) dx+\int\limits_{3}^{5}Q(x)dx$$=2\int\limits_{3}^{5}P(x) dx + \int\limits_{3}^{5}Q(x) dx$

4. mlddmlnog

ohh thank you so much! @exraven !

5. exraven

you are welcome :)

6. mlddmlnog

|dw:1358278984085:dw|

7. mlddmlnog

|dw:1358279110805:dw|

8. exraven

for the last one, you can swap the limits of integration and multiply the whole thing by -1$\int\limits_{5}^{1} P(x) dx = -\int\limits_{1}^{5} P(x) dx$

9. zepdrix

|dw:1358279244554:dw|Let's pretend this is what P(x) looks like.

10. mlddmlnog

oh, so when you switch the two, you mulply the integration by -1?

11. mlddmlnog

ok :) @zepdrix

12. zepdrix

P(-x) will be the same function but reflected over the Y-axis.|dw:1358279336129:dw|

13. zepdrix

So what do you think that area will be? c:

14. zepdrix

Integral represents the area under the curve. This part was 4.|dw:1358279422383:dw|

15. zepdrix

Darn they're suppose to be the same, I didn't draw that left curve very well though lol

16. mlddmlnog

@zepdrix - ohh the area stays the same since it's only changed to negative but the mount of space under curve? stays the same !

17. exraven

for this one$\int\limits_{-3}^{-5} Q(-x) dx$ use the substitution rule, let$u = -x$$du = -dx$$x = -3, u = 3$$x = -5, u = 5$the definite integral becomes$-\int\limits_{3}^{5} Q(u) du$

18. zepdrix

ya c: good times.

19. mlddmlnog

but thank you @exraven . and yup thanks as always @zepdrix

20. zepdrix

Hmm rave came up with -4, maybe I made a silly mistake somewhere.. I better check again.

21. mlddmlnog

no the right answer is 4 :) she gave us the answer on the study guide. but just the answers, and no process. :p

22. zepdrix

Oh i see :o

23. mlddmlnog

|dw:1358279740263:dw|

24. zepdrix

Oh that's a fun one c:

25. exraven

$\int\limits_{1}^{5} P(x) dx = \int\limits_{1}^{3} P(x) dx + \int\limits_{3}^{5} P(x) dx$$\int\limits_{1}^{5} P(x) dx = -\int\limits_{3}^{1} P(x) dx + \int\limits_{3}^{5} P(x) dx$solve for $\int\limits_{3}^{1}P(x) dx$

26. mlddmlnog

could you please explain how you did that..? i am completely lost...

27. exraven

okay, first of all suppose P(x) is like this|dw:1358280041569:dw|notice that I divide the area from x = 1 to x = 5 into two parts, and that's why I can write it like this$\int\limits_{1}^{5} P(x) dx = A_1 + A_2 = \int\limits_{1}^{3} P(x) dx + \int\limits_{3}^{5} P(x) dx$

28. exraven

and then, just swap the limit of integration for$\int\limits_{1}^{3} P(x) dx$it becomes$-\int\limits_{3}^{1} P(x) dx$

29. zepdrix

I'm confused why you're swapping the limits D: hmmm

30. zepdrix

Oh cause that's what we need to solve for :D I read it incorrectly haha

31. mlddmlnog

oh i understand that part now, but i still dont understand how to solve for the value when you do not know what $\int\limits_{1}^{3}p(x)dx$ and $\int\limits_{3}^{5}p(x)dx$

32. mlddmlnog

are

33. exraven

actually we know$\int\limits_{3}^{5} P(x) dx$ and $\int\limits_{1}^{5} P(x) dx$

34. mlddmlnog

oh waiat! could you just subtract $\int\limits_{3}^{5}p(x)dx$ from $\int\limits_{1}^{5}p(x)dx$ ?

35. mlddmlnog

so like.. $\int\limits_{1}^{5}p(x)dx - \int\limits_{3}^{5}p(x)dx = -1-3 = -4$ and then, since you have to switch the limits to 3 to 1, you multiply the -4 by -1 to get 4 as our answer?

36. exraven

yes

37. mlddmlnog

oh yay thank you guys! :)

38. exraven

you are welcome :)