evaluate the integral from -1 to 0 of (2x^5+6x)^3(5x^4+3)dx using substitution and the antiderivative

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evaluate the integral from -1 to 0 of (2x^5+6x)^3(5x^4+3)dx using substitution and the antiderivative

MIT 18.01 Single Variable Calculus (OCW)
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  • phi
let \( u = (2x^5 +6x) \) and \( du = (10x^4+6) \ dx\) solving for dx we get \[ dx = \frac{du}{2(5x^4+3) } \] the lower limit x= -1 corresponds to u= -2-6= -8 the upper limit x=0 corresponds to u=0 substitute into your problem to get \[ \frac{1}{2} \int_{-8}^0 u^3 du\]

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