Solve ∫ (from 0 to 1) of sqrt{x}-1 wrt x=

$$\int_{ 0 }^{ 1 } \sqrt{ x } -1 d x =$$

Evaluate

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

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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.

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

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

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

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.

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

Simplify.

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