Solve ∫ (from - 2 to 2) of x(x+3)(2x-1)

$$\int _ { - 2 } ^ { 2 } x ( x + 3 ) ( 2 x - 1 )$$

Evaluate

$$\frac{80}{3}\approx 26.666666667$$

Solution Steps

Use the distributive property to multiply x by x+3.

$$\int _{-2}^{2}\left(x^{2}+3x\right)\left(2x-1\right)\mathrm{d}x$$

Apply the distributive property by multiplying each term of x^{2}+3x by each term of 2x-1.

$$\int _{-2}^{2}2x^{3}-x^{2}+6x^{2}-3x\mathrm{d}x$$

Combine -x^{2} and 6x^{2} to get 5x^{2}.

$$\int _{-2}^{2}2x^{3}+5x^{2}-3x\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 2x^{3}+5x^{2}-3x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{3}\mathrm{d}x+\int 5x^{2}\mathrm{d}x+\int -3x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{3}\mathrm{d}x+5\int x^{2}\mathrm{d}x-3\int x\mathrm{d}x$$

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}.

$$\frac{x^{4}}{2}+5\int x^{2}\mathrm{d}x-3\int x\mathrm{d}x$$

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

$$\frac{x^{4}}{2}+\frac{5x^{3}}{3}-3\int x\mathrm{d}x$$

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

$$\frac{x^{4}}{2}+\frac{5x^{3}}{3}-\frac{3x^{2}}{2}$$

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.

$$\frac{2^{4}}{2}+\frac{5}{3}\times 2^{3}-\frac{3}{2}\times 2^{2}-\left(\frac{\left(-2\right)^{4}}{2}+\frac{5}{3}\left(-2\right)^{3}-\frac{3}{2}\left(-2\right)^{2}\right)$$

Simplify.

$$\frac{80}{3}$$