Solve ∫ (from 0 to 4) of (sqrt{x}-2)^2

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

Evaluate

$$\frac{8}{3}\approx 2.666666667$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{x}-2\right)^{2}.

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

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

$$\int _{0}^{4}x-4\sqrt{x}+4\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int x-4\sqrt{x}+4\mathrm{d}x$$

Integrate the sum term by term.

$$\int x\mathrm{d}x+\int -4\sqrt{x}\mathrm{d}x+\int 4\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x\mathrm{d}x-4\int \sqrt{x}\mathrm{d}x+\int 4\mathrm{d}x$$

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}.

$$\frac{x^{2}}{2}-4\int \sqrt{x}\mathrm{d}x+\int 4\mathrm{d}x$$

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. Multiply -4 times \frac{2x^{\frac{3}{2}}}{3}.

$$\frac{x^{2}}{2}-\frac{8x^{\frac{3}{2}}}{3}+\int 4\mathrm{d}x$$

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{2}}{2}-\frac{8x^{\frac{3}{2}}}{3}+4x$$

Simplify.

$$4x-\frac{8x^{\frac{3}{2}}}{3}+\frac{x^{2}}{2}$$

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.

$$4\times 4-\frac{8}{3}\times 4^{\frac{3}{2}}+\frac{4^{2}}{2}-\left(4\times 0-\frac{8}{3}\times 0^{\frac{3}{2}}+\frac{0^{2}}{2}\right)$$

Simplify.

$$\frac{8}{3}$$