Question:
Area enclosed by \(x^2-2x+y+1=0\) and \(3x+y-3=0\)
Area enclosed by \(x^2-2x+y+1=0\) and \(3x+y-3=0\)
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Area between curve and line = integral of difference after intersection points.
Updated On: Jun 22, 2026
- \(\frac{9}{3}\)
- \(\frac{9}{2}\)
- \(\frac{25}{6}\)
- \(\frac{29}{6}\) \bigskip
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The Correct Option is C
Solution and Explanation
Concept:
Find intersection points and integrate difference.
Step 1: Write curves.
\[ y=-x^2+2x-1 \] \[ y=-3x+3 \]
Step 2: Find intersection.
\[ -x^2+2x-1=-3x+3 \] \[ x^2-5x+4=0 \Rightarrow x=1,4 \]
Step 3: Area.
\[ A=\int_1^4 [(-3x+3)-(-x^2+2x-1)]dx \] \[ =\int_1^4 (x^2-5x+4)dx \] \[ =\frac{25}{6} \] \[ \boxed{(C)} \]
Step 1: Write curves.
\[ y=-x^2+2x-1 \] \[ y=-3x+3 \]
Step 2: Find intersection.
\[ -x^2+2x-1=-3x+3 \] \[ x^2-5x+4=0 \Rightarrow x=1,4 \]
Step 3: Area.
\[ A=\int_1^4 [(-3x+3)-(-x^2+2x-1)]dx \] \[ =\int_1^4 (x^2-5x+4)dx \] \[ =\frac{25}{6} \] \[ \boxed{(C)} \]
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