Solve ∫ (x-2)^2(x) wrt x

$$\int ( x - 2 ) ^ { 2 } ( x ) d x$$

Evaluate

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

Solution Steps

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

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

Use the distributive property to multiply x^{2}-4x+4 by x.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.

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

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

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

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}.

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

Simplify.

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

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

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

Differentiate w.r.t. x

$$x\left(x-2\right)^{2}$$