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

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

Evaluate

$$\frac{9}{2}=4.5$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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}. Multiply -1 times \frac{x^{2}}{2}.

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

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}. Multiply -1 times \frac{x^{3}}{3}.

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

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.

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

Simplify.

$$\frac{9}{2}$$