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

$\\int_{0}^{1}\\sqrt{x^{2}-2x + 1}dx$ \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, factor $x^{2}-2x + 1=(x - 1)^{2}$. So, $\sqrt{x^{2}-2x + 1}=\vert x - 1\vert$. For $x\in[0,1]$, $\vert x - 1\vert=1 - x$.

Step2: Calculate the definite - integral

We have $\int_{0}^{1}\sqrt{x^{2}-2x + 1}dx=\int_{0}^{1}(1 - x)dx$. Using the integral rules $\int(1 - x)dx=\int 1dx-\int xdx=x-\frac{x^{2}}{2}+C$. Evaluating the definite - integral: $\left[x-\frac{x^{2}}{2}\right]_{0}^{1}=(1-\frac{1^{2}}{2})-(0 - 0)$. $=1-\frac{1}{2}=\frac{1}{2}$.

Answer:

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