Solve ∫ (from - 1 to 1) of (-x^2-2x+3) wrt x

$$\int _ { - 1 } ^ { 1 } ( - x ^ { 2 } - 2 x + 3 ) d x$$

Evaluate

$$\frac{16}{3}\approx 5.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

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

Find the integral of 3 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

$$\frac{16}{3}$$