Solve ∫ (from 1 to 2) of 3x^2-3x-6

$$\int _ { 1 } ^ { 2 } 3 x ^ { 2 } - 3 x - 6$$

Evaluate

$$-\frac{7}{2}=-3.5$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 3x^{2}-3x-6\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x^{2}\mathrm{d}x+\int -3x\mathrm{d}x+\int -6\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x^{2}\mathrm{d}x-3\int x\mathrm{d}x+\int -6\mathrm{d}x$$

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

$$x^{3}-3\int x\mathrm{d}x+\int -6\mathrm{d}x$$

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}.

$$x^{3}-\frac{3x^{2}}{2}+\int -6\mathrm{d}x$$

Find the integral of -6 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

$$-\frac{7}{2}$$