Question

The area of the region bounded by the curves \[ y=x^2-3x+3 \] \[ y=2x^2-1 \] is

• A. \(\frac{403}{6}\)
• B. \(27\)
• C. \(19\)
• D. \(\frac{125}{6}\)

Hint:

Always verify which curve lies above by checking one point inside the interval.

Answer 1

The Correct Option is D

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 \[ \boxed{\frac{125}{6}} \]