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

$$\int_{ 1 }^{ 4 } \sqrt{ x } (1- \sqrt{ x } ) d x$$

Evaluate

$$-\frac{17}{6}\approx -2.833333333$$

Solution Steps

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

$$\int _{1}^{4}\sqrt{x}-\left(\sqrt{x}\right)^{2}\mathrm{d}x$$

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

$$\int _{1}^{4}\sqrt{x}-x\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}. Multiply -1 times \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.

$$\frac{2}{3}\times 4^{\frac{3}{2}}-\frac{4^{2}}{2}-\left(\frac{2}{3}\times 1^{\frac{3}{2}}-\frac{1^{2}}{2}\right)$$

Simplify.

$$-\frac{17}{6}$$