Solve ∫ (from - 1 to 3) of [x+1/2]

$$\int _ { - 1 } ^ { 3 } [ x + \frac { 1 } { 2 } ]$$

Evaluate

$$6$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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}.

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

Find the integral of \frac{1}{2} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{2}+x}{2}$$

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.

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

Simplify.

$$6$$