Solve ∫ (from 0 to 1) of x^3-2x^2+2x

$$\int _ { 0 } ^ { 1 } x ^ { 3 } - 2 x ^ { 2 } + 2 x$$

Evaluate

$$\frac{7}{12}\approx 0.583333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

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

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}.

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

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.

$$\frac{1^{4}}{4}-\frac{2}{3}\times 1^{3}+1^{2}-\left(\frac{0^{4}}{4}-\frac{2}{3}\times 0^{3}+0^{2}\right)$$

Simplify.

$$\frac{7}{12}$$