Question

Help!
Find the area of the region bounded by y=x^3-4x^2+3x and the x-axis

Answer 1

x(x^2 -4x +3)

x(x-3)(x-1) x = 0,1,3

Answer 2

your gonna have a negative area and a positive area....

they either want it as + and - or as a total absolute value...

Answer 3

x^4/4 -4x^3/3 +3x^2/2 ; [0,3]

at x = 0 we get 0 soo thats redundant

at 3 we get

3^4/4 -4(3)^3/3 +3(3)^2/2

Answer 4

-9 if they want a +,- area value....

Answer 5

-9/4.... that is

Answer 6

the othe way is: 5/12 + 32/12 = 37/12...i think

Answer 7

a + b = -9/4

a = 5/12

-b = -9/4 - 5/12

b = 27+5/12

b = 32/12

a+b = 5+32/12 = 37/12.....

Answer 8

a + b = -9/4

a = 5/12

b = -9/4 - 5/12

b = -27-5/12

b = |-32/12|

a+b = 5+32/12 = 37/12 ....same results :)

Answer 9

ok that makes a lot more sense to me! thanks amistre