Solve ∫ (from - 2 to 1) of (2x-6)^3

$$\int _ { - 2 } ^ { 1 } ( 2 x - 6 ) ^ { 3 }$$

Evaluate

$$-1218$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(2x-6\right)^{3}.

$$\int _{-2}^{1}8x^{3}-72x^{2}+216x-216\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 8x^{3}-72x^{2}+216x-216\mathrm{d}x$$

Integrate the sum term by term.

$$\int 8x^{3}\mathrm{d}x+\int -72x^{2}\mathrm{d}x+\int 216x\mathrm{d}x+\int -216\mathrm{d}x$$

Factor out the constant in each of the terms.

$$8\int x^{3}\mathrm{d}x-72\int x^{2}\mathrm{d}x+216\int x\mathrm{d}x+\int -216\mathrm{d}x$$

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

$$2x^{4}-72\int x^{2}\mathrm{d}x+216\int x\mathrm{d}x+\int -216\mathrm{d}x$$

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 -72 times \frac{x^{3}}{3}.

$$2x^{4}-24x^{3}+216\int x\mathrm{d}x+\int -216\mathrm{d}x$$

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

$$2x^{4}-24x^{3}+108x^{2}+\int -216\mathrm{d}x$$

Find the integral of -216 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2x^{4}-24x^{3}+108x^{2}-216x$$

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.

$$2\times 1^{4}-24\times 1^{3}+108\times 1^{2}-216-\left(2\left(-2\right)^{4}-24\left(-2\right)^{3}+108\left(-2\right)^{2}-216\left(-2\right)\right)$$

Simplify.

$$-1218$$