Question

finding values for a and b using integrals

Answer 1

\[\int\limits_{-1}^{1} ax^{2}-2ax+b) dx=\int\limits_{0}^{3} (ax^{2}-2ax+b)dx = 6\]

Answer 2

How do I go about solving for a and b?

Answer 3

take the anti derivative, evaluate both integrals and set the results equal to 6

Answer 4

I did that.. but I must have done something wrong
when I try to solve for a and b in each integral it doesn't work out

Answer 5

for example the anti derivative of the first one is
\[\frac{ax^3}{3}-ax+bx\] when you evaluate at 1 you get
\[\frac{a}{3}-a+b\] and when you evaluate at\(-1\) you get \[-\frac{a}{3}+a-b\]

Answer 6

subtract the second from the first gives
\[\frac{2a}{3}-2a+2b=6\] as one equation

Answer 7

or if you prefer
\[-\frac{4}{3}a+2b=6\]

Answer 8

second one is the same, replace \(x\) by 3 and get
\[6a+3b=6\]

Answer 9

if you replace \(x\) by \(0\) you get 0 so you can ignore that part

Answer 10

How did you get the derivative ax^3/3-ax+bx
I got ax^3/3-ax^2+bx

Answer 11

because i made a mistake

Answer 12

anti

Answer 13

\[\frac{ax^3}{3}-ax^2+bx\]

Answer 14

damn now i have to start over

Answer 15

I'm getting the idea though!

Answer 16

but now it is much much easier

Answer 17

in the second integral, replacing \(x\) by \(3\) gives \(3b=6\implies b=2\) because the \(a\) term drops out

Answer 18

via \(9a-9a=0\)

Answer 19

now you can solve for \(a\) from by setting \(b=2\) and the result equal to 6 and solve for \(a\)

Answer 20

and a= -3/2 correct?