Evaluate\(\int\limits_{-1}^{1}\) x | x | dx

1.	Evaluate\(\int\limits_{-1}^{1}\) x \| x \| dx
Answer» Let I = \(\int\limits_{-1}^{1} \) x \| x \| dx \|x\|= -x, if x <0 And \|x\|=x, if x ≥ 0 Therefore f(x)=x\|x\|=-x², if x<0 And f(x)=x\|x\|=x², if x ≥ 0 Consider x≥0 ⇒ f(x)=x² Then -x < 0 ⇒ f(-x) = -(-x)² = -f(-x) Now Consider x<0 ⇒ f(x)=-x² Then -x ≥ 0 ⇒ f(-x) =-(-x)²=x²= -f(x) Hence f(x) is an odd function. An odd function is a function which satisfies the propertyf(-x) =-f(-x), ∀ x∈ Domain of f(x) There is a property of integration of odd functions which states that \(\int\limits_{-a}^af(x)dx=0 \) if f(x) is an odd function. Therefore I = \(\int\limits_{-1}^{1} \) x \| x \| dx = 0

Answer»

Let

I = \(\int\limits_{-1}^{1} \) x | x | dx

|x|= -x, if x <0

And |x|=x, if x ≥ 0

Therefore f(x)=x|x|=-x², if x<0

And f(x)=x|x|=x², if x ≥ 0

Consider x≥0 ⇒ f(x)=x²

Then -x < 0 ⇒ f(-x) = -(-x)² = -f(-x)

Now Consider x<0 ⇒ f(x)=-x²

Then -x ≥ 0 ⇒ f(-x) =-(-x)²=x²= -f(x)

Hence f(x) is an odd function. An odd function is a function which satisfies the propertyf(-x) =-f(-x), ∀ x∈ Domain of f(x)

There is a property of integration of odd functions which states that

\(\int\limits_{-a}^af(x)dx=0 \) if f(x) is an odd function.

Therefore I = \(\int\limits_{-1}^{1} \) x | x | dx = 0

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