Solve ∫ (from 0 to 3) of xleft(x-2right)left(x+3right) wrt x

$$\int_{ 0 }^{ 3 } x \left( x-2 \right) \left( x+3 \right) d x$$

Evaluate

$$\frac{9}{4}=2.25$$

Solution Steps

Use the distributive property to multiply x by x-2.

$$\int _{0}^{3}\left(x^{2}-2x\right)\left(x+3\right)\mathrm{d}x$$

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

$$\int _{0}^{3}x^{3}+3x^{2}-2x^{2}-6x\mathrm{d}x$$

Combine 3x^{2} and -2x^{2} to get x^{2}.

$$\int _{0}^{3}x^{3}+x^{2}-6x\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

$$\frac{x^{4}}{4}+\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}+\frac{x^{3}}{3}-6\int x\mathrm{d}x$$

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 -6 times \frac{x^{2}}{2}.

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

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.

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

Simplify.

$$\frac{9}{4}$$