Question:
The area of the region bounded by the curves
\[
y=x^2-3x+3
\]
\[
y=2x^2-1
\]
is
The area of the region bounded by the curves
\[
y=x^2-3x+3
\]
\[
y=2x^2-1
\]
is
Show Hint
Always verify which curve lies above by checking one point inside the interval.
Updated On: Jun 15, 2026
- \(\frac{403}{6}\)
- \(27\)
- \(19\)
- \(\frac{125}{6}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
Area bounded between two curves is
\[
A=\int_a^b(y_{upper}-y_{lower})dx
\]
where limits are intersection points.
Step 1: Find points of intersection.
Equating equations \[ x^2-3x+3=2x^2-1 \] \[ x^2+3x-4=0 \] \[ (x+4)(x-1)=0 \] Thus \[ x=-4,\qquad x=1 \]
Step 2: Determine upper curve.
Check at \[ x=0 \] First curve \[ =3 \] Second curve \[ =-1 \] So upper curve: \[ x^2-3x+3 \] Area \[ A= \int_{-4}^{1} [(x^2-3x+3)-(2x^2-1)]dx \] \[ = \int_{-4}^{1} (-x^2-3x+4)dx \]
Step 3: Integrate.
\[ A= \left[ -\frac{x^3}{3} -\frac{3x^2}{2} +4x \right]_{-4}^{1} \] Substituting limits, \[ A=\frac{125}{6} \] Hence \[ \boxed{\frac{125}{6}} \]
Step 1: Find points of intersection.
Equating equations \[ x^2-3x+3=2x^2-1 \] \[ x^2+3x-4=0 \] \[ (x+4)(x-1)=0 \] Thus \[ x=-4,\qquad x=1 \]
Step 2: Determine upper curve.
Check at \[ x=0 \] First curve \[ =3 \] Second curve \[ =-1 \] So upper curve: \[ x^2-3x+3 \] Area \[ A= \int_{-4}^{1} [(x^2-3x+3)-(2x^2-1)]dx \] \[ = \int_{-4}^{1} (-x^2-3x+4)dx \]
Step 3: Integrate.
\[ A= \left[ -\frac{x^3}{3} -\frac{3x^2}{2} +4x \right]_{-4}^{1} \] Substituting limits, \[ A=\frac{125}{6} \] Hence \[ \boxed{\frac{125}{6}} \]
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