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

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

Evaluate

$$-\frac{3}{4}=-0.75$$

Solution Steps

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

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

Evaluate the indefinite integral first.

$$\int x^{3}-1\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int -1\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}+\int -1\mathrm{d}x$$

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

$$\frac{x^{4}}{4}-x$$

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.

$$\frac{1^{4}}{4}-1-\left(\frac{0^{4}}{4}-0\right)$$

Simplify.

$$-\frac{3}{4}$$