Solve ∫ (from 0 to 4) of (sqrt{x}-x+2) wrt x=

$$\int _ { 0 } ^ { 4 } ( \sqrt { x } - x + 2 ) d x =$$

Evaluate

$$\frac{16}{3}\approx 5.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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.

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

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 -1 times \frac{x^{2}}{2}.

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

Find the integral of 2 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

$$\frac{16}{3}$$