Solve ∫ (from - 2 to 1) of (-x^2+4-x-2)

$$\int _ { - 2 } ^ { 1 } ( - x ^ { 2 } + 4 - x - 2 )$$

Evaluate

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

Solution Steps

Evaluate the indefinite integral first.

$$\int -x^{2}+4-x-2\mathrm{d}x$$

Integrate the sum term by term.

$$\int -x^{2}\mathrm{d}x+\int 4\mathrm{d}x+\int -x\mathrm{d}x+\int -2\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\int x^{2}\mathrm{d}x+\int 4\mathrm{d}x-\int x\mathrm{d}x+\int -2\mathrm{d}x$$

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

$$-\frac{x^{3}}{3}+\int 4\mathrm{d}x-\int x\mathrm{d}x+\int -2\mathrm{d}x$$

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-\frac{x^{3}}{3}+4x-\int x\mathrm{d}x+\int -2\mathrm{d}x$$

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

$$-\frac{x^{3}}{3}+4x-\frac{x^{2}}{2}+\int -2\mathrm{d}x$$

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

$$-\frac{x^{3}}{3}+4x-\frac{x^{2}}{2}-2x$$

Simplify.

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

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.

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

Simplify.

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