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

$$\int_{ 2 }^{ 4 } (x+ \sqrt{ x } ) d x$$

Evaluate

$$\frac{34-4\sqrt{2}}{3}\approx 9.44771525$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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}.

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

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.

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

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.

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

Simplify.

$$\frac{34-4\sqrt{2}}{3}$$