anonymous
  • anonymous
Function f(x)=x^3 has odd symmetry. Fact: if we integrate this function from -k to k for any real number k, we will always get zero, because the negative part (on left) exactly cancels with k=5. Question: Is the fact still true if we use k= infinity? That is, if we integrate f(x)=x^3 from - infinity to infinity, do we get zero? why or why not? You may have to sketch a graph but use the example with k=5 integral from -5 to 5 x^3 dx =0 I guess the only hint is try integrating from - infinity to infinity.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
you would have to do improper integrals to be satisfied with the results
anonymous
  • anonymous
and you have to take the limits separately, so the answer is a resounding no!
anonymous
  • anonymous
it is not \[\int_{-\infty}^{\infty}f(x)dx=\lim_{t\to \infty}\int_{-t}^{t}f(x)dx\] you do not take the limits together

Looking for something else?

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

More answers

anonymous
  • anonymous
you have to take them one at a time, and so since neither converges, you do not get a real number answer
anonymous
  • anonymous
ok but remember although you get zero if you integrate from -t to t, when you take an improper integral you take the limit as t goes to infinity and the limit as t goes to minus infinity separately, not at the same time. in fact you should use two different variables, like \[\lim_{t\to \infty}\int_0^tf(x)dx+\lim_{s\to -\infty}\int_s^0f(x)dx\]

Looking for something else?

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