Solve ∫ _{1/2}^1(2x-1)^2 wrt x

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

Evaluate

$$\frac{1}{6}\approx 0.166666667$$

Solution Steps

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

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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 -4 times \frac{x^{2}}{2}.

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

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$\frac{4}{3}\times 1^{3}-2\times 1^{2}+1-\left(\frac{4}{3}\times \left(\frac{1}{2}\right)^{3}-2\times \left(\frac{1}{2}\right)^{2}+\frac{1}{2}\right)$$

Simplify.

$$\frac{1}{6}$$