Mathematics, 15.07.2019 00:50 hipstersale4913
Evaluate exactly the value of the integral from negative 1 to 0 of the product of the cube of the quantity 2 times x to the 4th power plus 8 times x and 4 times x to the 3rd power plus 4, dx. your work must include the use of substitution and the antiderivative.
Answers: 1
![\bf \textit{now, let's change the bounds, using }u(x)\\\\ u(-1)=2(-1)^4+8(-1)\implies u(-1)=-6 \\\\\\ u(0)=2(0)^4+8(0)\implies u(0)=0\\\\ -------------------------------\\\\ \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 \\\\\\\ [0]~~-~~[162]\implies -162](/tpl/images/0090/5938/1b3b7.png)
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Evaluate exactly the value of the integral from negative 1 to 0 of the product of the cube of the qu...
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