Solve ∫ (from 0 to 6) of sqrt{x}-(6-x)

$$\int _ { 0 } ^ { 6 } \sqrt { x } - ( 6 - x )$$

Evaluate

$$4\sqrt{6}-18\approx -8.202041029$$

Solution Steps

To find the opposite of 6-x, find the opposite of each term.

$$\int _{0}^{6}\sqrt{x}-6-\left(-x\right)\mathrm{d}x$$

The opposite of -x is x.

$$\int _{0}^{6}\sqrt{x}-6+x\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \sqrt{x}-6+x\mathrm{d}x$$

Integrate the sum term by term.

$$\int \sqrt{x}\mathrm{d}x+\int -6\mathrm{d}x+\int x\mathrm{d}x$$

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.

$$\frac{2x^{\frac{3}{2}}}{3}+\int -6\mathrm{d}x+\int x\mathrm{d}x$$

Find the integral of -6 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{2x^{\frac{3}{2}}}{3}-6x+\int x\mathrm{d}x$$

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}.

$$\frac{2x^{\frac{3}{2}}}{3}-6x+\frac{x^{2}}{2}$$

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.

$$\frac{2}{3}\times 6^{\frac{3}{2}}-6\times 6+\frac{6^{2}}{2}-\left(\frac{2}{3}\times 0^{\frac{3}{2}}-6\times 0+\frac{0^{2}}{2}\right)$$

Simplify.

$$4\sqrt{6}-18$$