Solve ∫ (from -4 to 1) of -x^3-x-12 wrt x

$$\int_{ -4 }^{ 1 } - { x }^{ 3 } -x-12 d x$$

Evaluate

$$\frac{45}{4}=11.25$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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 -1 times \frac{x^{2}}{2}.

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

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

$$-\frac{x^{4}}{4}-\frac{x^{2}}{2}-12x$$

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.

$$-\frac{1^{4}}{4}-\frac{1^{2}}{2}-12-\left(-\frac{\left(-4\right)^{4}}{4}-\frac{\left(-4\right)^{2}}{2}-12\left(-4\right)\right)$$

Simplify.

$$\frac{45}{4}$$