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

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 $\frac{1}{2} \int_{-8}^0 u^3 du$