Solve ∫ (from 2 to 3) of (1/2x-1/2)^2-(x-2) wrt x

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

Evaluate

$$\frac{1}{12}\approx 0.083333333$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{2}x-\frac{1}{2}\right)^{2}.

$$\int _{2}^{3}\frac{1}{4}x^{2}-\frac{1}{2}x+\frac{1}{4}-\left(x-2\right)\mathrm{d}x$$

To find the opposite of x-2, find the opposite of each term.

$$\int _{2}^{3}\frac{1}{4}x^{2}-\frac{1}{2}x+\frac{1}{4}-x+2\mathrm{d}x$$

Combine -\frac{1}{2}x and -x to get -\frac{3}{2}x.

$$\int _{2}^{3}\frac{1}{4}x^{2}-\frac{3}{2}x+\frac{1}{4}+2\mathrm{d}x$$

Add \frac{1}{4} and 2 to get \frac{9}{4}.

$$\int _{2}^{3}\frac{1}{4}x^{2}-\frac{3}{2}x+\frac{9}{4}\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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{1}{4} times \frac{x^{3}}{3}.

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

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 -\frac{3}{2} times \frac{x^{2}}{2}.

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

Find the integral of \frac{9}{4} using the table of common integrals rule \int a\mathrm{d}x=ax.

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

Simplify.

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

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.

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

Simplify.

$$\frac{1}{12}$$