Solve ∫ (from -1 to 2) of x^3-3x^2-4 wrt x

$$\int_{ -1 }^{ 2 } { x }^{ 3 } -3 { x }^{ 2 } -4 d x$$

Evaluate

$$-\frac{69}{4}=-17.25$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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}.

$$\frac{x^{4}}{4}-x^{3}+\int -4\mathrm{d}x$$

Find the integral of -4 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$\frac{2^{4}}{4}-2^{3}-4\times 2-\left(\frac{\left(-1\right)^{4}}{4}-\left(-1\right)^{3}-4\left(-1\right)\right)$$

Simplify.

$$-\frac{69}{4}$$