anonymous
  • anonymous
find the area enclosed between the curve y=x^3 and the line y=x
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
did you sketch the region?
anonymous
  • anonymous
im not sure on how to
anonymous
  • anonymous
you're not sure how to sketch x^3 ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
no i mean i sketched it, i dont know what to do after
anonymous
  • anonymous
say, for example, you're asking about the area enclosed for positive values of x..: |dw:1355190249043:dw|
anonymous
  • anonymous
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)
anonymous
  • anonymous
aka:\[\int\limits_{0}^{1} x-x ^{3} dx\]
anonymous
  • anonymous
can you evaluate that integral?
anonymous
  • anonymous
and then how do we find the anti derivitve of that hat
anonymous
  • anonymous
the usual way:\[\int\limits_{ }^{ } x^n dx = \frac{ x^{n+1} }{ n+1 }\]
anonymous
  • anonymous
so for \[\int\limits_{ }^{} x^1 dx\] \[\int\limits_{ }^{} x^1 dx = \frac{ x^{1+1} }{ 1+1}\]
anonymous
  • anonymous
try it for x^3...
anonymous
  • anonymous
x^4/4
anonymous
  • anonymous
yes
anonymous
  • anonymous
so \[\frac{ x^2 }{ 2 } -\frac{ x^4 }{4 } \]
anonymous
  • anonymous
ok?
anonymous
  • anonymous
and whats next?
anonymous
  • anonymous
plug in the limts
anonymous
  • anonymous
for the region on the interval x =0...+inf ( 1/2 -1/4 ) - (0/2 -0/4)
anonymous
  • anonymous
kk

Looking for something else?

Not the answer you are looking for? Search for more explanations.