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

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

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}.

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

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.

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

Simplify.

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