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Evaluate ∫2−1 ∣x^3−x∣ dx - Mathematics

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प्रश्न

 

Evaluate `int_(-1)^2|x^3-x|dx`

 

उत्तर

 

Let:

`I=int_(-1)^2|x^3-x|dx`

f(x)=x3x

f(x)=x3x=x(x1)(x+1)

The signs of f(x) for the different values are shown in the figure given below:

f(x)>0 for all x(1,0)(1,2)

f(x)<0 for all x(0,1)

Therefore

`|x^3-x|={(x^3-x,","xepsilon"(-1,0)"UU"(1,2)"),(-(x^3-x),","xepsilon(0,1)):}`

`:.I=int_(-1)^2|x^3-x|dx`

`=int_(-1)^0|x^3-x|dx+int_0^1|x^3-x|dx+int_1^2|x^3-x|dx`

`=int_(-1)^0(x^3-x)dx-int_0^1(x^3-x)dx+int_1^2(x^3-x)dx`

`=[x^4/4-x^2/2]_(-1)^0+[x^4/4-x^2/2]_0^1+[x^4/4-x^2/2]_1^2`

`=-(1/4-1/2)-(1/4-1/2)+(16/4-4/4)-(1/4-1/2)`

 `=3/4+(4-2)`

 `=11/4`

 
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