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

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

Evaluate

$$\frac{1}{20}=0.05$$

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}.

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

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

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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

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

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

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

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}. Multiply -1 times \frac{x^{5}}{5}.

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

Simplify.

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

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.

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

Simplify.

$$\frac{1}{20}$$