Solve ∫ (from 0 to 4) of [4-(x-2)^2] wrt x

$$\int _ { 0 } ^ { 4 } [ 4 - ( x - 2 ) ^ { 2 } ] d x$$

Evaluate

$$\frac{32}{3}\approx 10.666666667$$

Solution Steps

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

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

To find the opposite of x^{2}-4x+4, find the opposite of each term.

$$\int _{0}^{4}4-x^{2}+4x-4\mathrm{d}x$$

Subtract 4 from 4 to get 0.

$$\int _{0}^{4}-x^{2}+4x\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -1 times \frac{x^{3}}{3}.

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

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 4 times \frac{x^{2}}{2}.

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

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.

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

Simplify.

$$\frac{32}{3}$$