16)
y = x 3 − x y = 3 x {\displaystyle {\begin{aligned}&\color {red}\mathbf {y=x^{3}-x} &\color {royalblue}\mathbf {y=3x} \\\end{aligned}}}
x 3 − x = 3 x x 3 = 4 x x 2 ( x ) = 4 ( x ) x 2 = 4 x = 4 x = ± 2 ∫ − 2 2 [ ( x 3 − x ) − ( 3 x ) ] d x = ∫ − 2 2 [ x 3 − 4 x ] d x = [ x 4 4 − 2 x 2 ] | − 2 2 = ( 2 ) 4 4 − 2 ( 2 ) 2 − ( ( − 2 ) 4 4 − 2 ( − 2 ) 2 ) = 16 4 − 2 ( 4 ) − ( 16 4 − 2 ( 4 ) ) = 4 − 8 − ( 4 − 8 ) = − 4 − ( − 4 ) = − 4 + 4 = 0 {\displaystyle {\begin{aligned}x^{3}-x&=3x\\x^{3}&=4x\\x^{2}(x)&=4(x)\\x^{2}&=4\\x&={\sqrt {4}}\\x&=\pm 2\\[2ex]\int _{-2}^{2}[(x^{3}-x)-(3x)]dx=\int _{-2}^{2}[x^{3}-4x]dx\\[2ex]=\left[{\frac {x^{4}}{4}}-2x^{2}\right]{\Bigg |}_{-2}^{2}\\[2ex]={\frac {(2)^{4}}{4}}-2(2)^{2}-\left({\frac {(-2)^{4}}{4}}-2(-2)^{2}\right)\\[2ex]={\frac {16}{4}}-2(4)-({\frac {16}{4}}-2(4))\\[2ex]=4-8-(4-8)\\[2ex]=-4-(-4)\\[2ex]=-4+4\\[2ex]=0\end{aligned}}}
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![{\displaystyle {\begin{aligned}x^{3}-x&=3x\\x^{3}&=4x\\x^{2}(x)&=4(x)\\x^{2}&=4\\x&={\sqrt {4}}\\x&=\pm 2\\[2ex]\int _{-2}^{2}[(x^{3}-x)-(3x)]dx=\int _{-2}^{2}[x^{3}-4x]dx\\[2ex]=\left[{\frac {x^{4}}{4}}-2x^{2}\right]{\Bigg |}_{-2}^{2}\\[2ex]={\frac {(2)^{4}}{4}}-2(2)^{2}-\left({\frac {(-2)^{4}}{4}}-2(-2)^{2}\right)\\[2ex]={\frac {16}{4}}-2(4)-({\frac {16}{4}}-2(4))\\[2ex]=4-8-(4-8)\\[2ex]=-4-(-4)\\[2ex]=-4+4\\[2ex]=0\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/135c08f02f5b5e3c2c67d03427c8b38f7171dbb3)
