Question

Integral troubles. :(
Please help!

Answer 1

\[If \int\limits_{1}^{-1}(ax ^{2}-2ax+b)dx=\int\limits_{3}^{0}(ax ^{2}-2ax+b=6\] find the values for a and b

Answer 2

this is exactly how the question is written out.

Answer 3

\[(ax ^{3}/3-2a(x ^{2}/2)+b(x))_{1}^{-1}=(a(x ^{3}/3)-2a(x ^{2}/2)+bx)_{3}^{0}=6\]

Answer 4

the numbers on the bottom and top of the integral sign should be flipped.

Answer 5

thats what i was thinking because in ur question u have used the wrong limits

Answer 6

lower limit should have been -1 and upper limit 1

Answer 7

hope u can solve it further just substitute the values firstly upper limit in whole equation - put lower limit in full equation
u will get two equations in two variables solve them to get the values of a and b

Answer 8

@bmelyk integral troubles over now ?

Answer 9

not really haha i dont know how to start the problem;

Answer 10

Just integrate normally, why are you stuck? You know integration right ?