Solve ∫ (from 0 to 4) of (sqrt{x}+1) wrt x

$$\int _ { 0 } ^ { 4 } ( \sqrt { x } + 1 ) d x$$

Evaluate

$$\frac{28}{3}\approx 9.333333333$$

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.

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

Simplify.

$$\frac{28}{3}$$