## ksaimouli Group Title f(x) is even and g(x) is odd one year ago one year ago

1. ksaimouli

$\int\limits_{0}^{5}f(x)dx=8$

2. ksaimouli

$\int\limits_{0}^{5}g(x)=4$

3. ksaimouli

find$\int\limits_{-5}^{5}[f(x)+g(x)]dx$

4. ksaimouli

@wio

5. ksaimouli

$\int\limits_{-5}^{0}[f(x)+g(x)] +\int\limits_{0}^{5}[f(x)+g(x)] dx$

6. ksaimouli

@abb0t

7. ksaimouli

after that i have no idea

8. ksaimouli

i know the 2nd part 12

9. ksaimouli

-+12

10. ksaimouli

well even mean f(-x)=f(x) will that mean$\int\limits_{-5}^{0}f(x)=-8$

11. ksaimouli

because$-\int\limits_{5}^{0}f(x)=-8$

12. ksaimouli

even aplies if negative in f(x)? f(-x)=f(x)

13. KingGeorge

Why should $-\int\limits_{5}^{0}f(x)=-8$be true? I think this should be 8 and not -8.

14. ksaimouli

because that is the rule right to flip flop - should be their

15. KingGeorge

$\int_0^5 f(x)=8$$\int_5^0 f(x)=-8$$-\int_5^0 f(x)=8$

16. ksaimouli

$-\int\limits_{-5}^{0}f(x)=$

17. KingGeorge

But you're certainly on the right path to solving this. You know$\int\limits_{0}^{5}[f(x)+g(x)] dx=12$So you just need to find $\int\limits_{-5}^{0}[f(x)+g(x)]dx =\int\limits_{-5}^0f(x)\;dx+\int\limits_{-5}^0g(x)\;dx$First, lets start with the f(x) part.

18. zepp

I drove by to say hello to @KingGeorge!

19. KingGeorge

Since $$f(x)=f(-x)$$, we have that \begin{aligned} \int\limits_{-5}^0f(x)\;dx&=\int\limits_{-5}^0f(x)\;dx \\ &=\int\limits_{-5}^0f(-x)\;dx\\ &=\int\limits_{5}^0f(-u)\;(-du)\qquad\text{this is a u-sub for }u=-x.\\ &=\int\limits_0^5f(-u)\;du\\ &=\int\limits_0^5f(u)\;du\\ &=8 \end{aligned}

20. KingGeorge

This is basically the same thing you do for $$g(x)$$. Instead we have $$-g(x)=g(-x)$$.If we repeat the above argument, we get to the point $\int\limits_0^5g(-u)\;du$Substitute for $$-g(u)$$, and we get $-\int\limits_0^5g(u)\;du=-4$

21. KingGeorge

Did this all make sense? And hello to you too @zepp!

22. ksaimouli

i i got it i did little bit different i took u=-x and then -du=dx

23. KingGeorge

There is a typo in my work two posts above. The second line should not be there. (The line that reads $$=\int_{-5}^0f(-x)\;dx$$).

24. KingGeorge

That works as well.

25. ksaimouli

|dw:1356649511965:dw|

26. ksaimouli

ohk thx a lot

27. KingGeorge

You're welcome.

28. ksaimouli

i will work on this if i have nay questions can i post on this wall

29. KingGeorge

sure.