## virtus 3 years ago find the definite integral y = x(1-x) from -2 to 2

1. LagrangeSon678

you can multiply the x into the parenthese and obtain:|dw:1324593176181:dw|, then its just routine

2. LagrangeSon678

integrate and evaluate at the limits

3. virtus

could you please further elaborate thankx

4. myininaya

$\int\limits_{}^{}(x-x^2) dx=\frac{x^{1+1}}{1+1}-\frac{x^{2+1}}{2+1}+C$ $\int\limits_{a}^{b}f(x) dx=F(x)|_a^b=F(b)-F(a)$

5. LagrangeSon678

man, i caught up with something else, or else i would have explained it to you. But the one above me, did a fairly good job

6. virtus

thank you @ lagrangeson678 ^.^

7. virtus

myininaya what did you get as your answer?

8. myininaya

i found the antiderivative now all you have to do is: (just plug in upper limit) then write minus (then plug in lower limit)

9. virtus

oh because i got a negative answer

10. virtus

and according to the solution booklet it is a positive answer i.e 17/3 square units

11. LagrangeSon678

|dw:1324594394344:dw|

12. LagrangeSon678

|dw:1324594437592:dw|

13. virtus

yes and i got -16/3 square units

14. LagrangeSon678

Chek out the graph of this fucntion:http://www.wolframalpha.com/input/?i=integral+y+%3D+x%281-x%29+%5B-2%2C2%5D

15. LagrangeSon678

Do you notice anything?

16. virtus

yes the area is below the x axis, but how come in the solution book to this question, their answer is 17/3?

17. virtus

OH I SEE!!!!!!!!!! THANKS HEAPS

18. LagrangeSon678

but why were you saying 17/3

19. virtus

oh thats what the solution book says

20. LagrangeSon678

The answer is negative

21. myininaya

$\int\limits\limits_{}^{}(x-x^2) dx=\frac{x^{1+1}}{1+1}-\frac{x^{2+1}}{2+1}+C$ $\int\limits_{-2}^{2}(x-x^2) dx=[\frac{x^2}{2}-\frac{x^3}{3}]^2_{-2}=[\frac{2^2}{2}-\frac{2^3}{3}]-[\frac{(-2)^2}{2}-\frac{(-2)^3}{3}]$ $=[2-\frac{8}{3}]-[2+\frac{8}{3}]=2-\frac{8}{3}-2-\frac{8}{3}=2-2-\frac{8}{3}-\frac{8}{3}=0-\frac{16}{3}$

22. virtus

THANK YOU @ MYININAYA