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

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

Evaluate

$$\frac{7}{60}\approx 0.116666667$$

Solution Steps

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

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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 -1 times \frac{x^{3}}{3}.

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

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 -1 times \frac{x^{4}}{4}.

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

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}.

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

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.

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

Simplify.

$$\frac{7}{60}$$