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

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

Evaluate

$$\frac{135}{2}=67.5$$

Solution Steps

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

$$\int _{0}^{3}\left(9x^{2}-12x+4\right)\left(2x-3\right)\mathrm{d}x$$

Use the distributive property to multiply 9x^{2}-12x+4 by 2x-3 and combine like terms.

$$\int _{0}^{3}18x^{3}-51x^{2}+44x-12\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 18x^{3}-51x^{2}+44x-12\mathrm{d}x$$

Integrate the sum term by term.

$$\int 18x^{3}\mathrm{d}x+\int -51x^{2}\mathrm{d}x+\int 44x\mathrm{d}x+\int -12\mathrm{d}x$$

Factor out the constant in each of the terms.

$$18\int x^{3}\mathrm{d}x-51\int x^{2}\mathrm{d}x+44\int x\mathrm{d}x+\int -12\mathrm{d}x$$

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

$$\frac{9x^{4}}{2}-51\int x^{2}\mathrm{d}x+44\int x\mathrm{d}x+\int -12\mathrm{d}x$$

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 -51 times \frac{x^{3}}{3}.

$$\frac{9x^{4}}{2}-17x^{3}+44\int x\mathrm{d}x+\int -12\mathrm{d}x$$

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

$$\frac{9x^{4}}{2}-17x^{3}+22x^{2}+\int -12\mathrm{d}x$$

Find the integral of -12 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{9x^{4}}{2}-17x^{3}+22x^{2}-12x$$

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.

$$22\times 3^{2}-12\times 3-17\times 3^{3}+\frac{9}{2}\times 3^{4}-\left(22\times 0^{2}-12\times 0-17\times 0^{3}+\frac{9}{2}\times 0^{4}\right)$$

Simplify.

$$\frac{135}{2}$$