Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

LagrangeSon678 Group TitleBest ResponseYou've already chosen the best response.2
you can multiply the x into the parenthese and obtain:dw:1324593176181:dw, then its just routine
 2 years ago

LagrangeSon678 Group TitleBest ResponseYou've already chosen the best response.2
integrate and evaluate at the limits
 2 years ago

virtus Group TitleBest ResponseYou've already chosen the best response.1
could you please further elaborate thankx
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}(xx^2) dx=\frac{x^{1+1}}{1+1}\frac{x^{2+1}}{2+1}+C\] \[\int\limits_{a}^{b}f(x) dx=F(x)_a^b=F(b)F(a)\]
 2 years ago

LagrangeSon678 Group TitleBest ResponseYou've already chosen the best response.2
man, i caught up with something else, or else i would have explained it to you. But the one above me, did a fairly good job
 2 years ago

virtus Group TitleBest ResponseYou've already chosen the best response.1
thank you @ lagrangeson678 ^.^
 2 years ago

virtus Group TitleBest ResponseYou've already chosen the best response.1
myininaya what did you get as your answer?
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
i found the antiderivative now all you have to do is: (just plug in upper limit) then write minus (then plug in lower limit)
 2 years ago

virtus Group TitleBest ResponseYou've already chosen the best response.1
oh because i got a negative answer
 2 years ago

virtus Group TitleBest ResponseYou've already chosen the best response.1
and according to the solution booklet it is a positive answer i.e 17/3 square units
 2 years ago

LagrangeSon678 Group TitleBest ResponseYou've already chosen the best response.2
dw:1324594394344:dw
 2 years ago

LagrangeSon678 Group TitleBest ResponseYou've already chosen the best response.2
dw:1324594437592:dw
 2 years ago

virtus Group TitleBest ResponseYou've already chosen the best response.1
yes and i got 16/3 square units
 2 years ago

LagrangeSon678 Group TitleBest ResponseYou've already chosen the best response.2
Chek out the graph of this fucntion:http://www.wolframalpha.com/input/?i=integral+y+%3D+x%281x%29+%5B2%2C2%5D
 2 years ago

LagrangeSon678 Group TitleBest ResponseYou've already chosen the best response.2
Do you notice anything?
 2 years ago

virtus Group TitleBest ResponseYou've already chosen the best response.1
yes the area is below the x axis, but how come in the solution book to this question, their answer is 17/3?
 2 years ago

virtus Group TitleBest ResponseYou've already chosen the best response.1
OH I SEE!!!!!!!!!! THANKS HEAPS
 2 years ago

LagrangeSon678 Group TitleBest ResponseYou've already chosen the best response.2
but why were you saying 17/3
 2 years ago

virtus Group TitleBest ResponseYou've already chosen the best response.1
oh thats what the solution book says
 2 years ago

LagrangeSon678 Group TitleBest ResponseYou've already chosen the best response.2
The answer is negative
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits\limits_{}^{}(xx^2) dx=\frac{x^{1+1}}{1+1}\frac{x^{2+1}}{2+1}+C \] \[\int\limits_{2}^{2}(xx^2) dx=[\frac{x^2}{2}\frac{x^3}{3}]^2_{2}=[\frac{2^2}{2}\frac{2^3}{3}][\frac{(2)^2}{2}\frac{(2)^3}{3}]\] \[=[2\frac{8}{3}][2+\frac{8}{3}]=2\frac{8}{3}2\frac{8}{3}=22\frac{8}{3}\frac{8}{3}=0\frac{16}{3}\]
 2 years ago

virtus Group TitleBest ResponseYou've already chosen the best response.1
THANK YOU @ MYININAYA
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.