Find the area bounded by the curves y = -x² + 4x - 3

Find the area bounded by the curves $y = -x^2 + 4x - 3$ and $y = \frac{3}{2}x^2 - \frac{3}{2}x - 3$.

We will find where the parabolas cross and integrate the difference between them.

Step 1: Find intersection points

Setting the two equations equal:

\[-x^2 + 4x - 3 = \frac{3}{2}x^2 - \frac{3}{2}x - 3\]

\[-\frac{5}{2}x^2 + \frac{11}{2}x = 0\]

\[\frac{x}{2}(-5x + 11) = 0\]

Therefore: $x=0$ or $x=\frac{11}{5}$.

At $x=0$: $y=-3$, point $(0,-3)$. At $x=\frac{11}{5}$: $y=\frac{24}{25}$, point $\left(\frac{11}{5},\frac{24}{25}\right)$.

Step 2: Determine which parabola is above

At $x=1$: first parabola gives $y_1=-1+4-3=0$; second gives $y_2=\frac32-\frac32-3=-3$. Since $0>-3$, the first parabola $y=-x^2+4x-3$ is above the second.

Step 3: Set up the integral

\[A = \int_0^{11/5} \left[(-x^2 + 4x - 3) - \left(\frac{3}{2}x^2 - \frac{3}{2}x - 3\right)\right] dx = \int_0^{11/5} \left(-\frac{5}{2}x^2 + \frac{11}{2}x\right) dx\]

Step 4: Evaluate the integral

\[A = \frac{1}{2}\left[-\frac{5x^3}{3} + \frac{11x^2}{2}\right]_0^{11/5}\]

At $x=\frac{11}{5}$:

\[-\frac{5}{3}\cdot\frac{1331}{125} + \frac{11}{2}\cdot\frac{121}{25} = -\frac{1331}{75} + \frac{1331}{50} = \frac{1331}{150}\]

Therefore:

\[A = \frac{1}{2}\cdot\frac{1331}{150} = \frac{1331}{300}\]

\[\boxed{A = \frac{1331}{300} = \frac{11^3}{300} \text{ square units}}\]

Step 5: Numerical value

\[A \approx 4.437 \text{ square units}\]

Summary

\[\boxed{A = \frac{1331}{300} \approx 4.437 \text{ square units}}\]

Key insight: The difference of the two parabolas simplifies to $\frac{x}{2}(11-5x)$, which factors nicely and leads to the elegant result $\frac{11^3}{300}$.