Solve ∫ (from - 2 to 6) of (12x+4x-x^2) wrt x

$$\int _ { - 2 } ^ { 6 } ( 12 x + 4 x - x ^ { 2 } ) d x$$

Evaluate

$$\frac{544}{3}\approx 181.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 12x+4x-x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 12x\mathrm{d}x+\int 4x\mathrm{d}x+\int -x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$12\int x\mathrm{d}x+4\int x\mathrm{d}x-\int x^{2}\mathrm{d}x$$

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

$$6x^{2}+4\int x\mathrm{d}x-\int x^{2}\mathrm{d}x$$

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

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

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

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

Simplify.

$$8x^{2}-\frac{x^{3}}{3}$$

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.

$$8\times 6^{2}-\frac{6^{3}}{3}-\left(8\left(-2\right)^{2}-\frac{\left(-2\right)^{3}}{3}\right)$$

Simplify.

$$\frac{544}{3}$$