find the integral of |x-x^2| limits are a=-1 and b=2 [i posted this qusetion earlier but the limits werewrng

Question

please include some explanation and some steps

Answer 1

find the integral of |x-x^2| limits are a=-1 and b=2

1st thing remove the absolute value.

x -x^2 = -x( x -1)
roots are x =0 and x =1

|x-x^2| = x^2 -x when x is in the interval
] - infinity , 0 ] and [1 , + infinity[

|x-x^2| = -x^2 +x when x is in the interval
[ 0, 1 ]

integration interval is = [-1, 2]
we need to divide this into two 3 intervals
[-1, 0] and [0,1] and [1, 2]
The integral of |x-x^2| from -1 to 2 is =
integral of (x^2-x) from -1 to 0 call this M1
+integral of (-x^2 +x) from 0 to 1 call this M2
+integral of (x^2 -x) from 1 to 2 call this M3

M1= (x^3/3 -x^2/2)(0) - (x^3/3 -x^2/2)(-1)
= 0 -(-1/3 -1/2) = 1/3 +1/2 =5/6

M2 = ( -x^3/3 +x^2/2)(1) = -1/3 +1/2 = 1/6

M3 = (x^3/3 -x^2/2)(2) - (x^3/3 -x^2/2)(1)
= 8/3 -4/2 -(1/3 -1/2)
=8/3 -2 +1/6 = 16/6 -12/6 +1/6 = 5/6

Integral = M1 +M2 +M3 =
5/6 +1/6 +5/6 = 11/6 = 1.8333

Answer 2

You have to do this in stages.
x - x^2 = x*(1-x)

So |x - x^2| =
1) x^2 - x, if -1 < x < 0
2) x - x^2, if 0 < x < 1
3) x^2 - x, if 1 < x < 2

Now do each integral separately to get
(x^3 /3 - x^2 /2) | {-1,0} - {0,1} + {1,2}
= -(-1/3 - 1/2) - (1/3 - 1/2) + (8/3 - 2) - (1/3 - 1/2)
= 11/6

Check over for errors

Answer 3

First, you have to write the absolute value as a piecewise function.

|x-x^2|=(x-x^2) between zero and one.

it equals -(x-x^2) when x>1 and when x<0

you'll spit it up into different integrals then.

0 1 2
- ( (x-x^2) + ( (x-x^2) - ( (x-x^2)
) ) )
-1 0 1

Work them separately and you've got it.

0 1 2
- (1/2 x^2 - 1/3 x^3)| + (1/2 x^2 - 1/3 x^3)| - (1/2 x^2 - 1/3 x^3)|
-1 0 1

Plug in, subtract (watch your signs) and solve!
Hope this helps!

Answer 4

With no abs, graph is negative from -1 to 0, pos from 0 to 1, and neg again from 1 to 2. Do in three pieces!

int(x-x^2) = x^2/2 - x^3/3, from -1 to 2

int1 = - (x^2/2 - x^3/3), from -1 to 0
int1 = - (0^2/2 - 0^3/3) - - ((-1)^2/2 - (-1)^3/3)
int1 = 0 + (1/2 - -1/3)
= 5/6

int2 = (x^2/2 - x^3/3), from 0 to 1
int2 = (1^2/2 - 1^3/3) - ((0)^2/2 - (0)^3/3)
int2 = (1/2 - 1/3 ) - 0
int2 = (3/6 - 2/6 )
int2 = 1/6

int3 = - (x^2/2 - x^3/3), from 1 to 2
int3 = - (2^2/2 - 2^3/3) - - ((1)^2/2 - (1)^3/3)
int3 = - (2 - 8/3 ) + ( 1/2 - 1/3 )
int3 = - (6/3 - 8/3 ) + ( 3/6 - 2/6 )
int3 = - ( -2/3 ) + (1/6 )
int3 = 2/3 + 1/6
int3 = 4/6 + 1/6
int3 = 5/6

Total area: 11/6.

Whew!

Answer 5

The function x - x^2 is negative from -1 to 0 and again negative from 1 to 2. Therefore you should do
integ(-1,0)[x^2 - x]dx + integ(0,1)[x - x^2]dx + integ(1,2)[x^2 - x]dx.