Solve ∫ (from - 1 to 0) of (2-3x-1/2x^2) wrt x=

$$\int _ { - 1 } ^ { 0 } ( 2 - 3 x - \frac { 1 } { 2 } x ^ { 2 } ) d x =$$

Evaluate

$$\frac{10}{3}\approx 3.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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 -3 times \frac{x^{2}}{2}.

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

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 -\frac{1}{2} times \frac{x^{3}}{3}.

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

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.

$$2\times 0-\frac{3}{2}\times 0^{2}-\frac{0^{3}}{6}-\left(2\left(-1\right)-\frac{3}{2}\left(-1\right)^{2}-\frac{\left(-1\right)^{3}}{6}\right)$$

Simplify.

$$\frac{10}{3}$$