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

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

Evaluate

$$\frac{3}{20}=0.15$$

Solution Steps

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

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

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

$$\int _{0}^{1}2x^{4}-7x^{3}+9x^{2}-5x+1\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 2x^{4}-7x^{3}+9x^{2}-5x+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{4}\mathrm{d}x+\int -7x^{3}\mathrm{d}x+\int 9x^{2}\mathrm{d}x+\int -5x\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{4}\mathrm{d}x-7\int x^{3}\mathrm{d}x+9\int x^{2}\mathrm{d}x-5\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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 2 times \frac{x^{5}}{5}.

$$\frac{2x^{5}}{5}-7\int x^{3}\mathrm{d}x+9\int x^{2}\mathrm{d}x-5\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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 -7 times \frac{x^{4}}{4}.

$$\frac{2x^{5}}{5}-\frac{7x^{4}}{4}+9\int x^{2}\mathrm{d}x-5\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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

$$\frac{2x^{5}}{5}-\frac{7x^{4}}{4}+3x^{3}-5\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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 -5 times \frac{x^{2}}{2}.

$$\frac{2x^{5}}{5}-\frac{7x^{4}}{4}+3x^{3}-\frac{5x^{2}}{2}+\int 1\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{2x^{5}}{5}-\frac{7x^{4}}{4}+3x^{3}-\frac{5x^{2}}{2}+x$$

Simplify.

$$-\frac{5x^{2}}{2}+x+3x^{3}-\frac{7x^{4}}{4}+\frac{2x^{5}}{5}$$

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.

$$-\frac{5}{2}\times 1^{2}+1+3\times 1^{3}-\frac{7}{4}\times 1^{4}+\frac{2}{5}\times 1^{5}-\left(-\frac{5}{2}\times 0^{2}+0+3\times 0^{3}-\frac{7}{4}\times 0^{4}+\frac{2}{5}\times 0^{5}\right)$$

Simplify.

$$\frac{3}{20}$$