Solve = ∫ (from - 3 to 2) of -x^2-x+6 wrt x

$$= \int _ { - 3 } ^ { 2 } - x ^ { 2 } - x + 6 d x$$

Evaluate

$$\frac{125}{6}\approx 20.833333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -x^{2}-x+6\mathrm{d}x$$

Integrate the sum term by term.

$$\int -x^{2}\mathrm{d}x+\int -x\mathrm{d}x+\int 6\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\int x^{2}\mathrm{d}x-\int x\mathrm{d}x+\int 6\mathrm{d}x$$

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}.

$$-\frac{x^{3}}{3}-\int x\mathrm{d}x+\int 6\mathrm{d}x$$

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}.

$$-\frac{x^{3}}{3}-\frac{x^{2}}{2}+\int 6\mathrm{d}x$$

Find the integral of 6 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-\frac{x^{3}}{3}-\frac{x^{2}}{2}+6x$$

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.

$$-\frac{2^{3}}{3}-\frac{2^{2}}{2}+6\times 2-\left(-\frac{\left(-3\right)^{3}}{3}-\frac{\left(-3\right)^{2}}{2}+6\left(-3\right)\right)$$

Simplify.

$$\frac{125}{6}$$