Solve ∫ (from - 2 to 1) of (x^2-6x+12) wrt x

$$\int _ { - 2 } ^ { 1 } ( x ^ { 2 } - 6 x + 12 ) d x$$

Evaluate

$$48$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{2}-6x+12\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int -6x\mathrm{d}x+\int 12\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x+\int 12\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}-6\int x\mathrm{d}x+\int 12\mathrm{d}x$$

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 -6 times \frac{x^{2}}{2}.

$$\frac{x^{3}}{3}-3x^{2}+\int 12\mathrm{d}x$$

Find the integral of 12 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}-3x^{2}+12x$$

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.

$$\frac{1^{3}}{3}-3\times 1^{2}+12\times 1-\left(\frac{\left(-2\right)^{3}}{3}-3\left(-2\right)^{2}+12\left(-2\right)\right)$$

Simplify.

$$48$$