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

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

Evaluate

$$\frac{16}{3}\approx 5.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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}.

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

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}.

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

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

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

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.

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

Simplify.

$$\frac{16}{3}$$