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

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

Evaluate

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

Solution Steps

Multiply -1 and -1 to get 1.

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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}.

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

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}.

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

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.

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

Simplify.

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