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Question

Mathematics Question on Integration

Let f:RRf:\R\rightarrow\R be defined by
f(x)={xif x1 x+2if x>1f(x) = \begin{cases} x & \text{if}\ x\leq1 \\\ -x+2 & \text{if}\ x\gt1\end{cases}
Then 02f(x)dx=\int^{2}_{0}f(x)dx=

A

π2\frac{\pi}{2}

B

1

C

2

D

4

E

π6\frac{\pi}{6}

Answer

1

Explanation

Solution

The correct option is (B): 1