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

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

Evaluate

$$\frac{1}{3}\approx 0.333333333$$

Solution Steps

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

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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}.

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

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}.

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

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.

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

Simplify.

$$\frac{1}{3}$$