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

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

Evaluate

$$14$$

Solution Steps

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

$$\int _{0}^{2}\frac{9}{4}x^{2}-12x+16\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{9x^{2}}{4}-12x+16\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{9x^{2}}{4}\mathrm{d}x+\int -12x\mathrm{d}x+\int 16\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{9\int x^{2}\mathrm{d}x}{4}-12\int x\mathrm{d}x+\int 16\mathrm{d}x$$

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 \frac{9}{4} times \frac{x^{3}}{3}.

$$\frac{3x^{3}}{4}-12\int x\mathrm{d}x+\int 16\mathrm{d}x$$

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 -12 times \frac{x^{2}}{2}.

$$\frac{3x^{3}}{4}-6x^{2}+\int 16\mathrm{d}x$$

Find the integral of 16 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{3x^{3}}{4}-6x^{2}+16x$$

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.

$$\frac{3}{4}\times 2^{3}-6\times 2^{2}+16\times 2-\left(\frac{3}{4}\times 0^{3}-6\times 0^{2}+16\times 0\right)$$

Simplify.

$$14$$