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

$$\int _ { - 2 } ^ { 1 } ( 3 x ^ { 3 } - 2 x ^ { 2 } - 1 ) d x =$$

Evaluate

$$-\frac{81}{4}=-20.25$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 3x^{3}-2x^{2}-1\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x^{3}\mathrm{d}x+\int -2x^{2}\mathrm{d}x+\int -1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x^{3}\mathrm{d}x-2\int x^{2}\mathrm{d}x+\int -1\mathrm{d}x$$

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

$$\frac{3x^{4}}{4}-2\int x^{2}\mathrm{d}x+\int -1\mathrm{d}x$$

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 -2 times \frac{x^{3}}{3}.

$$\frac{3x^{4}}{4}-\frac{2x^{3}}{3}+\int -1\mathrm{d}x$$

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

$$-\frac{81}{4}$$