`int_(0)^(10)|x(x-1)(x-2)|dx` is equal to

Options

A. `160.05`
B. `1600.5`
C. `16.005`
D. none of these

Answer

B

Explanation

Let `f(x)= (x-1)(x-2)`. The signs of f(x) for differenta values of x are as shown belown

`{:(-x(x-1)(x-2)=(-x^(3)-3x^(2)+2x)"," ,"if" x lt 0),(x(x-1)(x-2)=x^(3)-3x^(2)+2x"," , "if" 0 le x lt 1),(-x(x-1)(x-2)=-(x^(3)-3x^(2)+2x)"," , " if" 1 le x lt 2),(x(x-1)(x-2)=x^(3)-3x^(2)+2x",", "if" x gt 2):}`
`therefore underset(0)overset(10)int|x(x-1)(x-2)|dx`
`rArr I = overset (1) underset (0 ) ( int) |x(x-1 ) (x - 2 ) | dx + overset (2 ) underset (1) (int) |x(x- 1 ) (x - 2 )| dx `
`rArr I = overset (1) underset (0) int (x ^(3) - 3x ^(2) + 2x ) dx + overset (2)underset (1) int - (x ^(3) - 3x ^(2) + 2 x ) dx + overset (10) underset (2) int (x ^(3) - 3x ^(2) + 2x ) dx `
`rArr I = [ (x ^(4))/( 4) - x ^(3) + x ^(2) ]_(0) ^(1) - [ (x ^(4)) /( 4 ) - x ^(3) + x ^(2) ] _(1) ^(2) + [ (x ^(4))/( 4) - x ^(3)+ x ^(2)] _(2) ^(10)`
` rArr I = (1)/(2) + 1600 = 1600.5`