Integration 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 our expert's

answer on brainly

SEE EXPERT ANSWER

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

A community for students.

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

|dw:1440354731938:dw|
The integral of x is just (x^2)/2, right?
yea

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

well +c but we don't care about +c really with definite integrals
So I have|dw:1440354893717:dw|
or b^2/2 - a^2/2 (1/b-a)?
right, but nobody leaves it like that.. it can be simplified...
|dw:1440355025079:dw| can you help with the bottom?
\[\frac{1}{2} \frac{b^2-a^2}{b-a}\] hint difference of squares
Oh ok. Not too bad then. Can more be done? The book says b+a / 2
(b+a)/2 is correct
and you can't do anything else unless you want to write it as b/2+a/2
can you show me how you get from 1/2 (b^2-a^2 / 2(b-a) to that?
do you know the difference of squares formula...
that was my hint
so maybe not.. b^2-a^2=(b-a)(b+a)
Ok thanks that was the missing piece for me I guess.
hmmm how did you get that extra 2 ?
\[\frac{1}{b-a} \int\limits_a^b x dx \\ \frac{1}{b-a} \cdot \frac{x^2}{2}|_a^b \\ \frac{1}{b-a}(\frac{b^2}{2}-\frac{a^2}{2} ) \\ \frac{1}{b-a} (\frac{b^2-a^2}{2}) \\ \frac{1}{2} \frac{b^2-a^2}{b-a} \\ \frac{1}{2} \frac{(b-a)(b+a)}{b-a}\] you should just have one 2 on bottom not two 2's
"can you show me how you get from 1/2 (b^2-a^2 / 2(b-a) to that?" or was the extra 2 just a type-o?
Oh me! Yes, I accidentally put in an extra 2 on the bottom, forgetting the 1/2!
oh okay

Not the answer you are looking for?

Search for more explanations.

Ask your own question