Solve ∫ (from 0 to 2) of (2x-5)^3 wrt x

$$\int _ { 0 } ^ { 2 } ( 2 x - 5 ) ^ { 3 } d x$$

Evaluate

$$-78$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(2x-5\right)^{3}.

$$\int _{0}^{2}8x^{3}-60x^{2}+150x-125\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 8x^{3}-60x^{2}+150x-125\mathrm{d}x$$

Integrate the sum term by term.

$$\int 8x^{3}\mathrm{d}x+\int -60x^{2}\mathrm{d}x+\int 150x\mathrm{d}x+\int -125\mathrm{d}x$$

Factor out the constant in each of the terms.

$$8\int x^{3}\mathrm{d}x-60\int x^{2}\mathrm{d}x+150\int x\mathrm{d}x+\int -125\mathrm{d}x$$

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}.

$$2x^{4}-60\int x^{2}\mathrm{d}x+150\int x\mathrm{d}x+\int -125\mathrm{d}x$$

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 -60 times \frac{x^{3}}{3}.

$$2x^{4}-20x^{3}+150\int x\mathrm{d}x+\int -125\mathrm{d}x$$

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

$$2x^{4}-20x^{3}+75x^{2}+\int -125\mathrm{d}x$$

Find the integral of -125 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2x^{4}-20x^{3}+75x^{2}-125x$$

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.

$$2\times 2^{4}-20\times 2^{3}+75\times 2^{2}-125\times 2-\left(2\times 0^{4}-20\times 0^{3}+75\times 0^{2}-125\times 0\right)$$

Simplify.

$$-78$$