Question

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

❌Nᴏ Sᴘᴀᴍ Aɴsᴡᴇʀs❌​

Answer 1

EXPLANATION.

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

As we know that,

We can write equation as,

⇒ f(x) = (x³ - x).

⇒ f(x) = x(x² - 1).

As we know that,

Formula of :

⇒ (x² - y²) = (x + y)(x - y).

⇒ f(x) = x(x + 1)(x - 1).

Put this point on wavy curve method, we get.

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

As we know that,

Formula of :

⇒ ∫xⁿdx = xⁿ⁺¹/n + 1 + c.

Using this formula in the equation, we get.

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

As we know that,

In definite integration firstly we put the upper limit then we put the lower limit in the equation, we get.

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

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

$\sf \implies \displaystyle - \bigg[ \dfrac{1}{4} - \dfrac{1}{2} \bigg]$

$\sf \implies \displaystyle - \bigg[ \dfrac{1 - 2}{4} \bigg]$

$\sf \implies \displaystyle - \bigg[ \dfrac{-1}{4} \bigg] = \dfrac{1}{4}$

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

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

$\sf \implies \displaystyle \bigg[ \dfrac{1}{2} - \dfrac{1}{4} \bigg]$

$\sf \implies \displaystyle \bigg[ \dfrac{2 - 1}{4} \bigg] = \dfrac{1}{4}$

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

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

$\sf \implies \displaystyle \bigg[ \dfrac{16}{4} - \dfrac{4}{2} \bigg] - \bigg[ \dfrac{1}4} - \dfrac{1}{2} \bigg]$

$\sf \implies \displaystyle \bigg[ 4 - 2 \bigg] - \bigg[ \dfrac{1 - 2}{4} \bigg]$

$\sf \implies \displaystyle \bigg[ 2 \bigg] - \bigg[ \dfrac{-1}{4} \bigg]$

$\sf \implies \displaystyle 2 + \dfrac{1}{4}$

$\sf \implies \displaystyle \dfrac{8 + 1}{4} = \dfrac{9}{4}$

Adding all the values in the equation, we get.

$\sf \implies \displaystyle \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{9}{4} = \dfrac{11}{4}$

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

Answer 2

$\pink{ \boxed{\boxed{\begin{array}{cc} \maltese \bf \: \: given \\ \\ \displaystyle \int_{ - 1}^2 \bf \: | {x}^{3} - 4x | \: dx \\ \\ = \displaystyle \int_{ - 1}^2\bf \: |x( {x}^{2} - 1)| \: dx \\ \\ = \displaystyle \int_{ - 1}^2\bf \: |x(x - 1)(x + 1)| \: dx \\ \\ \bf \: here \: {x}^{3} - x = 0 \\ \bf \: when \: \: \: x =0 \: , \: 1 \: \: and \: \: - 1 \\ \\ \orange{ {\boxed{\begin{array}{c | c} \underline{ \bf \: value \: of \: x }& \underline{\bf \: value \: of \: ( {x}^{3} - x)} \\ \\ \bf - 1 < x < 0& \bf + ve \\ \\ \bf 0 < x < 1& \bf \: - ve \\ \\ \bf \: 1 < x < 2& \bf + ve\end{array}}}}\end{array}}}}$

$\pink{ \boxed{\boxed{\begin{array}{cc} \maltese \bf \: \: again \: \\ \\ \bf | {x}^{3} - x| = {x}^{3} - x \: \: \\ \bf when \: - 1 < x < 0 \: \: and \: 1 < x < 2 \\ \\ \sf \: and \: \\ \\ \bf \: | {x}^{3} - x| = - {x}^{3} + x \: \: \\ \bf \: when \: \: 0 < x < 1\end{array}}}}$

Now,

$\displaystyle \int_{ - 1}^2 \bf \: | {x}^{3} - x| \: dx \\ \\ = \small{ \displaystyle \int_{ - 1}^0 \bf \: ( {x}^{3} - x) \: dx + \displaystyle \int_{0}^1 \bf \: ( - {x}^{3} + x) \: dx + \displaystyle \int_{ 1}^2 \bf \: ( {x}^{3} - x) \: dx} \\ \\ = \bf \: [ \frac{ {x}^{4} }{4} - \frac{ {x}^{2} }{2} ]_ { - 1}^0 \: + [ \frac{ - {x}^{4} }{4} + \frac{ {x}^{2} }{2} ]_{0}^1 \: + [ \frac{ {x}^{4} }{4} - \frac{ {x}^{2} }{2} ]_1^2 \\ \\ = - ( \frac{1}{4} - \frac{1}{2} )\:+\:( \frac{ - 1}{4} + \frac{1}{2} ) + ( \frac{16}{4} - \frac{4}{2} ) - ( \frac{1}{4} - \frac{1}{2} ) \\ \\ = \frac{1}{4} + \frac{1}{4} + 2 + \frac{1}{4} \\ \\ = 2 + \frac{3}{4} \\ \\ = \frac{11}{4}$