Solve ∫ (from 4 to 9) of sqrt{x}(1+sqrt{x}) wrt x=

$$\int _ { 4 } ^ { 9 } \sqrt { x } ( 1 + \sqrt { x } ) d x =$$

Evaluate

$$\frac{271}{6}\approx 45.166666667$$

Solution Steps

Use the distributive property to multiply \sqrt{x} by 1+\sqrt{x}.

$$\int _{4}^{9}\sqrt{x}+\left(\sqrt{x}\right)^{2}\mathrm{d}x$$

Calculate \sqrt{x} to the power of 2 and get x.

$$\int _{4}^{9}\sqrt{x}+x\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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.

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

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}.

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

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.

$$\frac{2}{3}\times 9^{\frac{3}{2}}+\frac{9^{2}}{2}-\left(\frac{2}{3}\times 4^{\frac{3}{2}}+\frac{4^{2}}{2}\right)$$

Simplify.

$$\frac{271}{6}$$