Solve ∫ _{1/2}^1(sqrt{2x-1})^2 wrt x

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

Evaluate

$$\frac{1}{4}=0.25$$

Solution Steps

Calculate \sqrt{2x-1} to the power of 2 and get 2x-1.

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

Evaluate the indefinite integral first.

$$\int 2x-1\mathrm{d}x$$

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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

$$x^{2}-x$$

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.

$$1^{2}-1-\left(\left(\frac{1}{2}\right)^{2}-\frac{1}{2}\right)$$

Simplify.

$$\frac{1}{4}$$