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

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

Evaluate

$$-\frac{1}{2}=-0.5$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 3x^{2}-x-1\mathrm{d}x$$

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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 -1 times \frac{x^{2}}{2}.

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

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

$$x^{3}-\frac{x^{2}}{2}-x$$

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.

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

Simplify.

$$-\frac{1}{2}$$