Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

ok integral of INT/limits 1 to 0/ (e^x)^2

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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

rewrite?
\[\int_0^1 e^{x^2}dx\]
ah it is error function or something like that I can't do it

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

http://www.wolframalpha.com/input/?i=integration+e^%28x^2%29+from+0+to+1
yeah, i think its like 1/2 sqrt(pi)*error(imaginary) or some such
yep. that's what my calcu says.. 1.462651746 xD
what is this now ?
nonono, (e^x)(e^x) or (e^x)^2, integrate that from 1 to 0
And I have a computer right in front of me as well as a ti-84+ so I dont want your calculator or wolfram alpha answers please :)
Oh, that is much simpler. (e^x)^2 = e^2x so: \[\int\limits_{0}^{1}e^{2x}dx=\frac{1}{2}e^2x\] take it to the limits: [1/2*e^(2)]-[1/2*e^(0)]=1/2e^2-1/2 factor it and you have: \[\frac{1}{2}(e^2-1)\]
oh you sure that (e^x)^2=e^2x?
yes, that is one of the properties of exponents
fsho

Not the answer you are looking for?

Search for more explanations.

Ask your own question