[tex]\bf \displaystyle \int_{-1}^{0}~(2x^4+8x)^3(4x^3+4)\cdot dx\\\\
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u=2x^4+8x\implies \cfrac{du}{dx}=8x^3+8\implies \cfrac{du}{2(4x^3+4)}=dx\\\\
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\displaystyle \int_{-1}^{0}~u^3\underline{(4x^3+4)}\cdot \cfrac{du}{2\underline{(4x^3+4)}}\implies \cfrac{1}{2}\int_{-1}^0~u^3\cdot du\\\\
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[/tex]
[tex]\bf \textit{now, let's change the bounds, using }u(x)\\\\
u(-1)=2(-1)^4+8(-1)\implies u(-1)=-6
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u(0)=2(0)^4+8(0)\implies u(0)=0\\\\
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\displaystyle \cfrac{1}{2}\int_{-6}^0~u^3\cdot du\implies \left.\cfrac{1}{2}\cdot \cfrac{u^4}{4} \right]_{-6}^0\implies \left. \cfrac{u^4}{8} \right]_{-6}^0
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[0]~~-~~[162]\implies -162[/tex]
notice, by changing the bounds using u(x), we do not need to substitute back the "u".
