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VladimirAG [237]
3 years ago
14

Calculus: Help ASAP

Mathematics
1 answer:
Serggg [28]3 years ago
5 0
\bf \displaystyle \int\limits_{-1}^{0}~(4x^6+2x)^3(12x^5+1)\cdot  dx\\\\
-------------------------------\\\\
u=4x^6+2x\implies \cfrac{du}{dx}=24x^5+2\implies \cfrac{du}{2(12x^5+1)}=dx\\\\
-------------------------------\\\\
\displaystyle \int\limits_{-1}^{0}~u^3\underline{(12x^5+1)}\cdot\cfrac{du}{2\underline{(12x^5+1)}}\implies \cfrac{1}{2}\int\limits_{-1}^{0}~u^3\cdot du\\\\
-------------------------------\\\\

\bf \textit{now, we'll change the bounds, using u(x)}
\\\\\\
u(-1)=4(-1)^6+2(-1)\implies u(-1)=2
\\\\\\
u(0)=4(0)^6+2()\implies u(0)=0\\\\
-------------------------------\\\\
\displaystyle \cfrac{1}{2}\int\limits_{2}^{0}~u^3\cdot du\implies \left. \cfrac{1}{2}\cdot \cfrac{u^4}{4}  \right]_{2}^{0}\implies \left. \cfrac{u^4}{8}  \right]_{2}^{0}\implies [0]-[2]\implies -2
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\\ \sf\longmapsto \dfrac{1}{x-1}>4

  • Multiply x-1 on both sides

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\\ \sf\longmapsto 1> 4x-4

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\\ \sf\longmapsto 5> 4x

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ATQ to the equation \bf \dfrac{1}{x-1} x should be greater than 1 .Because if it becomes 1 then the denominator will be 0 which is impossible.

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\\ \sf{:}\!\implies 1

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\\ \therefore\boxed{\bf x\epsilon \left(1,\dfrac{5}{4}\right)}

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