Solve ∫ (from 1 to 2) of x(x-1)(x-2) wrt x

$$\int_{ 1 }^{ 2 } x(x-1)(x-2) d x$$

Evaluate

$$-\frac{1}{4}=-0.25$$

Solution Steps

Use the distributive property to multiply x by x-1.

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

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

$$\int _{1}^{2}x^{3}-2x^{2}-x^{2}+2x\mathrm{d}x$$

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

$$\int _{1}^{2}x^{3}-3x^{2}+2x\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

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

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

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

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.

$$\frac{2^{4}}{4}-2^{3}+2^{2}-\left(\frac{1^{4}}{4}-1^{3}+1^{2}\right)$$

Simplify.

$$-\frac{1}{4}$$