Solve ∫ (from 0 to 2) of (4-2x)(4-x^2)

$$\int _ { 0 } ^ { 2 } ( 4 - 2 x ) ( 4 - x ^ { 2 } )$$

Evaluate

$$\frac{40}{3}\approx 13.333333333$$

Solution Steps

Use the distributive property to multiply 4-2x by 4-x^{2}.

$$\int _{0}^{2}16-4x^{2}-8x+2x^{3}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 16-4x^{2}-8x+2x^{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 16\mathrm{d}x+\int -4x^{2}\mathrm{d}x+\int -8x\mathrm{d}x+\int 2x^{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 16\mathrm{d}x-4\int x^{2}\mathrm{d}x-8\int x\mathrm{d}x+2\int x^{3}\mathrm{d}x$$

Find the integral of 16 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$16x-4\int x^{2}\mathrm{d}x-8\int x\mathrm{d}x+2\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -4 times \frac{x^{3}}{3}.

$$16x-\frac{4x^{3}}{3}-8\int x\mathrm{d}x+2\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply -8 times \frac{x^{2}}{2}.

$$16x-\frac{4x^{3}}{3}-4x^{2}+2\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply 2 times \frac{x^{4}}{4}.

$$16x-\frac{4x^{3}}{3}-4x^{2}+\frac{x^{4}}{2}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$16\times 2-\frac{4}{3}\times 2^{3}-4\times 2^{2}+\frac{2^{4}}{2}-\left(16\times 0-\frac{4}{3}\times 0^{3}-4\times 0^{2}+\frac{0^{4}}{2}\right)$$

Simplify.

$$\frac{40}{3}$$