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

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

Evaluate

$$\frac{3}{10}=0.3$$

Solution Steps

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

$$\int _{1}^{2}x\left(8-12x+6x^{2}-x^{3}\right)\mathrm{d}x$$

Use the distributive property to multiply x by 8-12x+6x^{2}-x^{3}.

$$\int _{1}^{2}8x-12x^{2}+6x^{3}-x^{4}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 8x-12x^{2}+6x^{3}-x^{4}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 8x\mathrm{d}x+\int -12x^{2}\mathrm{d}x+\int 6x^{3}\mathrm{d}x+\int -x^{4}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$8\int x\mathrm{d}x-12\int x^{2}\mathrm{d}x+6\int x^{3}\mathrm{d}x-\int x^{4}\mathrm{d}x$$

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}. Multiply 8 times \frac{x^{2}}{2}.

$$4x^{2}-12\int x^{2}\mathrm{d}x+6\int x^{3}\mathrm{d}x-\int x^{4}\mathrm{d}x$$

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 -12 times \frac{x^{3}}{3}.

$$4x^{2}-4x^{3}+6\int x^{3}\mathrm{d}x-\int x^{4}\mathrm{d}x$$

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

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

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

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

Simplify.

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

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.

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

Simplify.

$$\frac{3}{10}$$