Q:

# 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.

Accepted Solution

A:
$$\bf \displaystyle \int_{-1}^{0}~(2x^4+8x)^3(4x^3+4)\cdot dx\\\\ -------------------------------\\\\ u=2x^4+8x\implies \cfrac{du}{dx}=8x^3+8\implies \cfrac{du}{2(4x^3+4)}=dx\\\\ -------------------------------\\\\ \displaystyle \int_{-1}^{0}~u^3\underline{(4x^3+4)}\cdot \cfrac{du}{2\underline{(4x^3+4)}}\implies \cfrac{1}{2}\int_{-1}^0~u^3\cdot du\\\\ -------------------------------\\\\$$

$$\bf \textit{now, let's change the bounds, using }u(x)\\\\ u(-1)=2(-1)^4+8(-1)\implies u(-1)=-6 \\\\\\ u(0)=2(0)^4+8(0)\implies u(0)=0\\\\ -------------------------------\\\\ \displaystyle \cfrac{1}{2}\int_{-6}^0~u^3\cdot du\implies \left.\cfrac{1}{2}\cdot \cfrac{u^4}{4} \right]_{-6}^0\implies \left. \cfrac{u^4}{8} \right]_{-6}^0 \\\\\\\ [0]~~-~~[162]\implies -162$$

notice, by changing the bounds using u(x), we do not need to substitute back the "u".