Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Let f(x) be defined by f(x) = x if x 0. the value of the integral from -2 to 1 of xf(x)dx = a.3/2 b.5/2 c.3 d. 7/2 e. 11/2

See more answers at
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.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

|dw:1355956366325:dw| \[\int\limits_{-2}^{1} xf(x)dx\]
basically the pieces \[x, x \le 0\] \[x+1, x > 0\]
i think you supposed to break it up? \[\int\limits_{1}^{1} x(x+1)dx\] = 0 \[\int\limits_{0}^{-2} x^2dx\] = -8/3 i'm doing something wrong...

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Limits of integration should be [-2,0] and [0,1] respectively.
ohhh that's what i thought but cuz it's not greater than or equal to, i didn't put the zero...
Also, for the first part you seem to have taken the integral from 0 to -2. It should be from -2 to 0.
oops my bad.
ok yeah thanks. i got 7/2 as the answer
That's what I got too.
k thanks.
\[\int\limits\limits_{1}^{1} x(x+1)dx \]Looks like you got this already, but as for this part with the limits of integration that you were confused on - you can still pick the first limit on the integral as 0, even though the function is defined as x>0. The picture might help. Even though x>0, the area under the curve is still effectively from x=0 to x=1. If you only integrate from 1 to 1, the area under that is zero. You could pick a value of x just slightly greater than x=0, like 0.0000000000001, and do the integration, and you'll get essentially the same amount for the area from x=0.0000000000001 to x=1 as you do for x=0 to x=1. Sorry if that sounded confusing..
1 Attachment

Not the answer you are looking for?

Search for more explanations.

Ask your own question