Solve 2 pi ∫ (from 0 to 4) of (32-2x^2)(x-4) wrt x+2piint_-4^0(32-2x^2)(x-4)dx

$$2 \pi \int _ { 0 } ^ { 4 } ( 32 - 2 x ^ { 2 } ) ( x - 4 ) d x + 2 \pi \int _ { - 4 } ^ { 0 } ( 32 - 2 x ^ { 2 } ) ( x - 4 ) d x$$

Factor

$$-\frac{4096 \pi}{3} = -4289.321169701264$$

Solution Steps

Factor out 2.

$$2\left(\pi \int _{0}^{4}\left(32-2x^{2}\right)\left(x-4\right)\mathrm{d}x+\pi \int _{-4}^{0}\left(32-2x^{2}\right)\left(x-4\right)\mathrm{d}x\right)$$

Consider \pi \int _{0}^{4}\left(32-2x^{2}\right)\left(x-4\right)\mathrm{d}x+\pi \int _{-4}^{0}\left(32-2x^{2}\right)\left(x-4\right)\mathrm{d}x. Factor out \pi .

$$\pi \left(\int _{0}^{4}\left(32-2x^{2}\right)\left(x-4\right)\mathrm{d}x+\int _{-4}^{0}\left(32-2x^{2}\right)\left(x-4\right)\mathrm{d}x\right)$$

Rewrite the complete factored expression.

$$2\pi \left(\int _{0}^{4}\left(32-2x^{2}\right)\left(x-4\right)\mathrm{d}x+\int _{-4}^{0}\left(32-2x^{2}\right)\left(x-4\right)\mathrm{d}x\right)$$

Evaluate

$$-\frac{4096\pi }{3}\approx -4289.321169701$$