Solve ∫ (from -2 to 0) of -x^2-1 wrt x

$$\int_{ -2 }^{ 0 } - { x }^{ 2 } -1 d x$$

Evaluate

$$-\frac{14}{3}\approx -4.666666667$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

$$\int -x^{2}\mathrm{d}x+\int -1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\int x^{2}\mathrm{d}x+\int -1\mathrm{d}x$$

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}.

$$-\frac{x^{3}}{3}+\int -1\mathrm{d}x$$

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$-\frac{0^{3}}{3}-0-\left(-\frac{\left(-2\right)^{3}}{3}-\left(-2\right)\right)$$

Simplify.

$$-\frac{14}{3}$$