Solve ∫ (from 0 to 1) of {left(1-sqrt{x}right)}^2 wrt x

$$\int_{ 0 }^{ 1 } { \left(1- \sqrt{ x } \right) }^{ 2 } d x$$

Evaluate

$$\frac{1}{6}\approx 0.166666667$$

Solution Steps

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

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

Calculate \sqrt{x} to the power of 2 and get x.

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

Rewrite \sqrt{x} as x^{\frac{1}{2}}. Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{1}{2}}\mathrm{d}x with \frac{x^{\frac{3}{2}}}{\frac{3}{2}}. Simplify. Multiply -2 times \frac{2x^{\frac{3}{2}}}{3}.

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

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}.

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

Simplify.

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

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.

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

Simplify.

$$\frac{1}{6}$$