$\\int_{0}^{1}\\sqrt{x^{2}-2x + 1}dx$\na -1\nb $-\\frac{1}{2}$\nc $\\frac{1}{2}$\nd 1\ne none of the above

$\\int_{0}^{1}\\sqrt{x^{2}-2x + 1}dx$\na -1\nb $-\\frac{1}{2}$\nc $\\frac{1}{2}$\nd 1\ne none of the above

$\\int_{0}^{1}\\sqrt{x^{2}-2x + 1}dx$\na -1\nb $-\\frac{1}{2}$\nc $\\frac{1}{2}$\nd 1\ne none of the above

Answer

Explanation:

Step1: Simplify the integrand

First, simplify $\sqrt{x^{2}-2x + 1}$. Since $x^{2}-2x + 1=(x - 1)^{2}$, then $\sqrt{x^{2}-2x + 1}=\vert x - 1\vert$. For $x\in[0,1]$, $\vert x - 1\vert=1 - x$.

Step2: Calculate the definite - integral

We know that $\int_{0}^{1}\sqrt{x^{2}-2x + 1}dx=\int_{0}^{1}(1 - x)dx$. According to the integral formula $\int(1 - x)dx=x-\frac{1}{2}x^{2}+C$. Then $\int_{0}^{1}(1 - x)dx=\left[x-\frac{1}{2}x^{2}\right]_{0}^{1}$.

Step3: Evaluate the definite - integral

Substitute the upper and lower limits: $\left(1-\frac{1}{2}\times1^{2}\right)-\left(0-\frac{1}{2}\times0^{2}\right)=1-\frac{1}{2}=\frac{1}{2}$.

Answer:

C. $\frac{1}{2}$