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

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

Evaluate

$$-\frac{3}{2}=-1.5$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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

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

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

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

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.

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

Simplify.

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