## d92292 2 years ago find the area enclosed between the curve y=x^3 and the line y=x

1. Algebraic!

did you sketch the region?

2. d92292

im not sure on how to

3. Algebraic!

you're not sure how to sketch x^3 ?

4. d92292

no i mean i sketched it, i dont know what to do after

5. Algebraic!

say, for example, you're asking about the area enclosed for positive values of x..: |dw:1355190249043:dw|

6. Algebraic!

you can see that it's the area under the line x (from 0 to 1) minus the area under x^3 (from 0 to 1)

7. Algebraic!

aka:$\int\limits_{0}^{1} x-x ^{3} dx$

8. Algebraic!

can you evaluate that integral?

9. d92292

and then how do we find the anti derivitve of that hat

10. Algebraic!

the usual way:$\int\limits_{ }^{ } x^n dx = \frac{ x^{n+1} }{ n+1 }$

11. Algebraic!

so for $\int\limits_{ }^{} x^1 dx$ $\int\limits_{ }^{} x^1 dx = \frac{ x^{1+1} }{ 1+1}$

12. Algebraic!

try it for x^3...

13. d92292

x^4/4

14. Algebraic!

yes

15. Algebraic!

so $\frac{ x^2 }{ 2 } -\frac{ x^4 }{4 }$

16. Algebraic!

ok?

17. d92292

and whats next?

18. Algebraic!

plug in the limts

19. Algebraic!

for the region on the interval x =0...+inf ( 1/2 -1/4 ) - (0/2 -0/4)

20. d92292

kk