Solve ∫ (from 0 to 2) of x(1/12(8-x^3)) wrt x

$$\int_{ 0 }^{ 2 } x( \frac{ 1 }{ 12 } (8- { x }^{ 3 } )) d x$$

Evaluate

$$\frac{4}{5}=0.8$$

Solution Steps

Use the distributive property to multiply x\times \frac{1}{12} by 8-x^{3}.

$$\int _{0}^{2}8x\times \frac{1}{12}-\frac{1}{12}x^{4}\mathrm{d}x$$

Multiply 8 and \frac{1}{12} to get \frac{2}{3}.

$$\int _{0}^{2}\frac{2}{3}x-\frac{1}{12}x^{4}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{2x}{3}-\frac{x^{4}}{12}\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{2x}{3}\mathrm{d}x+\int -\frac{x^{4}}{12}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{2\int x\mathrm{d}x}{3}-\frac{\int x^{4}\mathrm{d}x}{12}$$

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 \frac{2}{3} times \frac{x^{2}}{2}.

$$\frac{x^{2}}{3}-\frac{\int x^{4}\mathrm{d}x}{12}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply -\frac{1}{12} times \frac{x^{5}}{5}.

$$\frac{x^{2}}{3}-\frac{x^{5}}{60}$$

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.

$$\frac{2^{2}}{3}-\frac{2^{5}}{60}-\left(\frac{0^{2}}{3}-\frac{0^{5}}{60}\right)$$

Simplify.

$$\frac{4}{5}$$