Question:
Let \(f:\R\rightarrow\R\) be defined by
\(f(x) = \begin{cases} x & \text{if}\ x\leq1 \\ -x+2 & \text{if}\ x\gt1\end{cases}\)
Then \(\int^{2}_{0}f(x)dx=\)
Let \(f:\R\rightarrow\R\) be defined by
\(f(x) = \begin{cases} x & \text{if}\ x\leq1 \\ -x+2 & \text{if}\ x\gt1\end{cases}\)
Then \(\int^{2}_{0}f(x)dx=\)
\(f(x) = \begin{cases} x & \text{if}\ x\leq1 \\ -x+2 & \text{if}\ x\gt1\end{cases}\)
Then \(\int^{2}_{0}f(x)dx=\)
Updated On: Jun 10, 2024
- \(\frac{\pi}{2}\)
- 1
- 2
- 4
- \(\frac{\pi}{6}\)
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The Correct Option is B
Solution and Explanation
The correct option is (B): 1
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