Solve ∫ (from 0 to 2) of left(3x-4right)^3 wrt x

$$\int_{ 0 }^{ 2 } { \left(3x-4 \right) }^{ 3 } d x$$

Evaluate

$$-20$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(3x-4\right)^{3}.

$$\int _{0}^{2}27x^{3}-108x^{2}+144x-64\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 27x^{3}-108x^{2}+144x-64\mathrm{d}x$$

Integrate the sum term by term.

$$\int 27x^{3}\mathrm{d}x+\int -108x^{2}\mathrm{d}x+\int 144x\mathrm{d}x+\int -64\mathrm{d}x$$

Factor out the constant in each of the terms.

$$27\int x^{3}\mathrm{d}x-108\int x^{2}\mathrm{d}x+144\int x\mathrm{d}x+\int -64\mathrm{d}x$$

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

$$\frac{27x^{4}}{4}-108\int x^{2}\mathrm{d}x+144\int x\mathrm{d}x+\int -64\mathrm{d}x$$

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 -108 times \frac{x^{3}}{3}.

$$\frac{27x^{4}}{4}-36x^{3}+144\int x\mathrm{d}x+\int -64\mathrm{d}x$$

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

$$\frac{27x^{4}}{4}-36x^{3}+72x^{2}+\int -64\mathrm{d}x$$

Find the integral of -64 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{27x^{4}}{4}-36x^{3}+72x^{2}-64x$$

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.

$$\frac{27}{4}\times 2^{4}-36\times 2^{3}+72\times 2^{2}-64\times 2-\left(\frac{27}{4}\times 0^{4}-36\times 0^{3}+72\times 0^{2}-64\times 0\right)$$

Simplify.

$$-20$$