anonymous
  • anonymous
can anyone help me with integration
Mathematics
chestercat
  • chestercat
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

JamesJ
  • JamesJ
Quite a few people on to night are more than competent to help with your likely question. Post it.
anonymous
  • anonymous
|dw:1327030921905:dw|
anonymous
  • anonymous
Separate it into two integrals. Divide the sqrt(2) into the top and then you have 2 trivial integrals.

Looking for something else?

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

More answers

anonymous
  • anonymous
can you show how you would do it
anonymous
  • anonymous
|dw:1327031115564:dw|
anonymous
  • anonymous
|dw:1327031166888:dw|
anonymous
  • anonymous
Make sense?
anonymous
  • anonymous
no
anonymous
  • anonymous
isnt like this
anonymous
  • anonymous
|dw:1327031304975:dw|
anonymous
  • anonymous
I'm not sure what you did there. How did you get a square root 2? That would just make it harder. Remember that since constants don't change with the variables, you can always pull them out of the integral. You can even do the following:
anonymous
  • anonymous
|dw:1327031471323:dw|
anonymous
  • anonymous
\[\frac{1}{\sqrt{2}}\int_1^4 (u-2)du = \frac{1}{\sqrt{2}}\left(\frac12u^2-2u\right|_1^4=\frac{3}{2\sqrt{2}}\]
anonymous
  • anonymous
thats worng because my book says the answer is 2/3
anonymous
  • anonymous
Your book is incorrect. This answer is correct, with wolfram as additional evidence to boot: http://www.wolframalpha.com/input/?i=integrate+%28x-2%29%2Fsqrt%282%29+from+1+to+4
anonymous
  • anonymous
no i dont think so because iam using the fundemental thereom of calculus
anonymous
  • anonymous
Does your question specifically ask to solve \[\int_1^4\frac{u-2}{\sqrt{2}}du\] Because the answer to that is what I stated. If the problem in the text is different, let me know and we'll go from there.
anonymous
  • anonymous
thats exactly how it reads and it asks to evaluate the definite integral
anonymous
  • anonymous
Mhm, the answer is what I said.
anonymous
  • anonymous
Think of it this way, since u-2 is a linear equation, you are just calculating the areas of a couple of triangles with rational side lengths. However, dividing by the square root of 2 (an irrational number) will necessitate that your integral is irrational. 2/3 is not irrational, so that cannot be the answer.

Looking for something else?

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