Question

Integrate |x^3 - x | with respect to x,from limits -1 to 2​

Answer 1

EXPLANATION.

$\implies \displaystyle \int_{-1}^{2} |x^{3} - x | dx$

As  we know that,

⇒ f(x) = x³ - x.

⇒ f(x) = x(x² - 1).

As we know that,

Formula of ;

⇒ (x² - y²) = (x - y)(x + y).

Using this formula in equation, we get.

⇒ f(x) = x(x - 1)(x + 1).

Put this point on wavy curve method, we get.

$\implies \displaystyle \int_{-1}^{0} (x^{3} - x)dx \ + \int_{0}^{1} -(x^{3} - x)dx \ + \int_{1}^{2} (x^{3} - x)dx$

$\implies \displaystyle \bigg[\dfrac{x^{4} }{4} - \dfrac{x^{2} }{2} \bigg]_{-1}^{0} + \bigg[\dfrac{-x^{4} }{4} + \dfrac{x^{2} }{2} \bigg]_{0}^{1} \ + \bigg[ \dfrac{x^{4} }{4} - \dfrac{x^{2} }{2} \bigg]_{1}^{2}$

As we know that,

First we put the upper limit then we put the lower limit in definite integration, we get.

$\implies \displaystyle \bigg[\dfrac{x^{4} }{4} - \dfrac{x^{2} }{2} \bigg]_{-1}^{0} = \bigg[ \bigg(\dfrac{0^{4} }{4} - \dfrac{0^{2} }{2} \bigg) - \bigg(\dfrac{(-1)^{4} }{4} - \dfrac{(-1)^{2} }{2} \bigg) \bigg]$

$\implies \displaystyle \bigg[\dfrac{x^{4} }{4} - \dfrac{x^{2} }{2} \bigg]_{-1}^{0} \ = - \bigg( \dfrac{1}{4} - \dfrac{1}{2} \bigg)$

$\implies \displaystyle \bigg[\dfrac{x^{4} }{4} - \dfrac{x^{2} }{2} \bigg]_{-1}^{0} \ = - \bigg(\dfrac{1 - 2}{4} \bigg) \ = \dfrac{1}{4}$

$\implies \displaystyle \bigg[ \dfrac{x^{2} }{2} - \dfrac{x^{4} }{4} \bigg]_{0}^{1} \ = \bigg[ \bigg( \dfrac{(1)^{2} }{2} - \dfrac{(1)^{4} }{4} \bigg) - \bigg( \dfrac{(0)^{2} }{2} - \dfrac{(0)^{4} }{4} \bigg) \bigg]$

$\implies \displaystyle \bigg[ \dfrac{x^{2} }{2} - \dfrac{x^{4} }{4} \bigg]_{0}^{1} \ = \bigg[ \dfrac{1}{2} - \dfrac{1}{4} \bigg]$

$\implies \displaystyle \bigg[ \dfrac{x^{2} }{2} - \dfrac{x^{4} }{4} \bigg]_{0}^{1} \ = \bigg[ \dfrac{2 - 1}{4} \bigg] \ = \dfrac{1}{4}$

$\implies \displaystyle \bigg[ \dfrac{x^{4} }{4} - \dfrac{x^{2} }{2} \bigg]_{1}^{2} \ = \bigg[ \bigg( \dfrac{(2)^{4} }{4} - \dfrac{(2)^{2} }{2} \bigg) - \bigg( \dfrac{(1)^{4} }{4} - \dfrac{(1)^{2} }{2} \bigg) \bigg]$

$\implies \displaystyle \bigg[ \dfrac{x^{4} }{4} - \dfrac{x^{2} }{2} \bigg]_{1}^{2} \ = \bigg[ \dfrac{16}{4} - \dfrac{4}{2} \bigg] - \bigg[ \dfrac{1}{4} - \dfrac{1}{2} \bigg]$

$\implies \displaystyle \bigg[ \dfrac{x^{4} }{4} - \dfrac{x^{2} }{2} \bigg]_{1}^{2} \ = (4 - 2) - \bigg( \dfrac{1 - 2}{4} \bigg)$

$\implies \displaystyle \bigg[ \dfrac{x^{4} }{4} - \dfrac{x^{2} }{2} \bigg]_{1}^{2} \ = 2 + \dfrac{1}{4} \ = \dfrac{8 + 1}{4} = \dfrac{9}{4}$

Adding three equations, we get.

⇒ 1/4 + 1/4 + 9/4.

⇒ 1 + 1 + 9/4 = 11/4.

$\implies \displaystyle \int_{-1}^{2} |x^{3} - x | dx = \dfrac{11}{4}$