Solve ∫ (from 0 to 3) of (-12x-12x^2-6x^3) wrt x

$$\int_{ 0 }^{ 3 } (-12x-12 { x }^{ 2 } -6 { x }^{ 3 } ) d x$$

Evaluate

$$-\frac{567}{2}=-283.5$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -12x-12x^{2}-6x^{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -12x\mathrm{d}x+\int -12x^{2}\mathrm{d}x+\int -6x^{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-12\int x\mathrm{d}x-12\int x^{2}\mathrm{d}x-6\int x^{3}\mathrm{d}x$$

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}.

$$-6x^{2}-12\int x^{2}\mathrm{d}x-6\int x^{3}\mathrm{d}x$$

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 -12 times \frac{x^{3}}{3}.

$$-6x^{2}-4x^{3}-6\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply -6 times \frac{x^{4}}{4}.

$$-6x^{2}-4x^{3}-\frac{3x^{4}}{2}$$

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.

$$-6\times 3^{2}-4\times 3^{3}-\frac{3}{2}\times 3^{4}-\left(-6\times 0^{2}-4\times 0^{3}-\frac{3}{2}\times 0^{4}\right)$$

Simplify.

$$-\frac{567}{2}$$