Solve ∫ (from 0 to 1) of x^2left(1-xright)^2 wrt x

$$\int_{ 0 }^{ 1 } { x }^{ 2 } { \left(1-x \right) }^{ 2 } d x$$

Evaluate

$$\frac{1}{30}\approx 0.033333333$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$\int _{0}^{1}x^{2}\left(1-2x+x^{2}\right)\mathrm{d}x$$

Use the distributive property to multiply x^{2} by 1-2x+x^{2}.

$$\int _{0}^{1}x^{2}-2x^{3}+x^{4}\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}.

$$\frac{x^{3}}{3}-\frac{x^{4}}{2}+\frac{x^{5}}{5}$$

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.

$$\frac{1^{5}}{5}-\frac{1^{4}}{2}+\frac{1^{3}}{3}-\left(\frac{0^{5}}{5}-\frac{0^{4}}{2}+\frac{0^{3}}{3}\right)$$

Simplify.

$$\frac{1}{30}$$