Solve ∫ (from 0 to 4) of (6-(4-sqrt{x})^2) wrt x

$$\int _ { 0 } ^ { 4 } ( 6 - ( 4 - \sqrt { x } ) ^ { 2 } ) d x$$

Evaluate

$$-\frac{16}{3}\approx -5.333333333$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-\sqrt{x}\right)^{2}.

$$\int _{0}^{4}6-\left(16-8\sqrt{x}+\left(\sqrt{x}\right)^{2}\right)\mathrm{d}x$$

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

$$\int _{0}^{4}6-\left(16-8\sqrt{x}+x\right)\mathrm{d}x$$

To find the opposite of 16-8\sqrt{x}+x, find the opposite of each term.

$$\int _{0}^{4}6-16+8\sqrt{x}-x\mathrm{d}x$$

Subtract 16 from 6 to get -10.

$$\int _{0}^{4}-10+8\sqrt{x}-x\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -10+8\sqrt{x}-x\mathrm{d}x$$

Integrate the sum term by term.

$$\int -10\mathrm{d}x+\int 8\sqrt{x}\mathrm{d}x+\int -x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int -10\mathrm{d}x+8\int \sqrt{x}\mathrm{d}x-\int x\mathrm{d}x$$

Find the integral of -10 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-10x+8\int \sqrt{x}\mathrm{d}x-\int x\mathrm{d}x$$

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. Multiply 8 times \frac{2x^{\frac{3}{2}}}{3}.

$$-10x+\frac{16x^{\frac{3}{2}}}{3}-\int x\mathrm{d}x$$

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}.

$$-10x+\frac{16x^{\frac{3}{2}}}{3}-\frac{x^{2}}{2}$$

Simplify.

$$-10x-\frac{x^{2}}{2}+\frac{16x^{\frac{3}{2}}}{3}$$

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.

$$-10\times 4-\frac{4^{2}}{2}+\frac{16}{3}\times 4^{\frac{3}{2}}-\left(-10\times 0-\frac{0^{2}}{2}+\frac{16}{3}\times 0^{\frac{3}{2}}\right)$$

Simplify.

$$-\frac{16}{3}$$