Solve ∫ (from 0 to 1) of x^2(1-x)^3 wrt x

$$\int _ { 0 } ^ { 1 } x ^ { 2 } ( 1 - x ) ^ { 3 } d x$$

Evaluate

$$\frac{1}{60}\approx 0.016666667$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(1-x\right)^{3}.

$$\int _{0}^{1}x^{2}\left(1-3x+3x^{2}-x^{3}\right)\mathrm{d}x$$

Use the distributive property to multiply x^{2} by 1-3x+3x^{2}-x^{3}.

$$\int _{0}^{1}x^{2}-3x^{3}+3x^{4}-x^{5}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int x^{2}-3x^{3}+3x^{4}-x^{5}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int -3x^{3}\mathrm{d}x+\int 3x^{4}\mathrm{d}x+\int -x^{5}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x-3\int x^{3}\mathrm{d}x+3\int x^{4}\mathrm{d}x-\int x^{5}\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}-3\int x^{3}\mathrm{d}x+3\int x^{4}\mathrm{d}x-\int x^{5}\mathrm{d}x$$

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

$$\frac{x^{3}}{3}-\frac{3x^{4}}{4}+3\int x^{4}\mathrm{d}x-\int x^{5}\mathrm{d}x$$

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}. Multiply 3 times \frac{x^{5}}{5}.

$$\frac{x^{3}}{3}-\frac{3x^{4}}{4}+\frac{3x^{5}}{5}-\int x^{5}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}. Multiply -1 times \frac{x^{6}}{6}.

$$\frac{x^{3}}{3}-\frac{3x^{4}}{4}+\frac{3x^{5}}{5}-\frac{x^{6}}{6}$$

Simplify.

$$-\frac{x^{6}}{6}+\frac{3x^{5}}{5}-\frac{3x^{4}}{4}+\frac{x^{3}}{3}$$

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.

$$-\frac{1^{6}}{6}+\frac{3}{5}\times 1^{5}-\frac{3}{4}\times 1^{4}+\frac{1^{3}}{3}-\left(-\frac{0^{6}}{6}+\frac{3}{5}\times 0^{5}-\frac{3}{4}\times 0^{4}+\frac{0^{3}}{3}\right)$$

Simplify.

$$\frac{1}{60}$$