Solve ∫ (from 0 to 1) of (x)(1-x)^2 wrt x

$$\int _ { 0 } ^ { 1 } ( x ) ( 1 - x ) ^ { 2 } d x$$

Evaluate

$$\frac{1}{12}\approx 0.083333333$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$\int _{0}^{1}x\left(1-2x+x^{2}\right)\mathrm{d}x$$

Use the distributive property to multiply x by 1-2x+x^{2}.

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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}.

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

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}.

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

Simplify.

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

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.

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

Simplify.

$$\frac{1}{12}$$