Question

The area of the region given by \(A =\left\{( x , y ): x ^2 \leq y \leq \min \{ x +2,4-3 x \}\right\}\) is :

• A. \(\frac{31}{8}\)
• B. \(\frac{17}{6}\)
• C. \(\frac{19}{6}\)
• D. \(\frac{27}{8}\)

Answer 1

The Correct Option is B

To find the area of the region given by \(A = \{(x, y): x^2 \leq y \leq \min\{x+2, 4-3x\}\}\), we need to follow these steps:

1. The inequality \(x^2 \leq y\) represents the region above the parabola \(y = x^2\).
2. The inequality \(y \leq \min\{x+2, 4-3x\}\) means \(y\) is below either the line \(y = x+2\) or \(y = 4-3x\), whichever is smaller at that \(x\)-value.
3. First, find the intersection points of the lines \(y = x+2\) and \(y = 4-3x\):

x + 2 = 4 - 3x \\ 4x = 2 \\ x = \frac{1}{2}

When \(x = \frac{1}{2}\), \(y = x+2 = \frac{1}{2} + 2 = \frac{5}{2}\).

4. The lines intersect at \(\left(\frac{1}{2}, \frac{5}{2}\right)\). Let us now consider the intersection of each line with the parabola \(y = x^2\).
5. For the line \(y = x + 2\):

x^2 = x + 2 \\ x^2 - x - 2 = 0 \\ (x-2)(x+1) = 0 \\ x = 2, \, x = -1

6. For the line \(y = 4 - 3x\):

x^2 = 4 - 3x \\ x^2 + 3x - 4 = 0 \\ (x+4)(x-1) = 0 \\ x = 1, \, x = -4

7. We now need to evaluate the area of the region bounded by \(x^2\), \(x+2\), and \(4-3x\) from \(-1\) to \(2\) based on the x-intervals.
8. For \(-1 \leq x \leq \frac{1}{2}\), the upper boundary is \(x+2\) from \(-1\) to \(\frac{1}{2}\).

\text{Area}_1 = \int_{-1}^{\frac{1}{2}} ((x+2) - x^2) \, dx

9. For \(\frac{1}{2} \leq x \leq 1\), the upper boundary switches to \(4-3x\).

\text{Area}_2 = \int_{\frac{1}{2}}^{1} ((4-3x) - x^2) \, dx

10. Finally, for \(1 \leq x \leq 2\), the upper boundary is again \(x+2\).

\text{Area}_3 = \int_{1}^{2} ((x+2) - x^2) \, dx

11. Calculate each integral:

For \(\text{Area}_1\):

\int_{-1}^{\frac{1}{2}} [(x+2) - x^2] \, dx = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{\frac{1}{2}}

For \(\text{Area}_2\):

\int_{\frac{1}{2}}^{1} [(4-3x) - x^2] \, dx = \left[ 4x - \frac{3x^2}{2} - \frac{x^3}{3} \right]_{\frac{1}{2}}^{1}

For \(\text{Area}_3\):

\int_{1}^{2} [(x+2) - x^2] \, dx = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{1}^{2}

12. Adding the calculated areas gives the total area:

A = \text{Area}_1 + \text{Area}_2 + \text{Area}_3 = \frac{17}{6}

Thus, the correct answer is \(\frac{17}{6}\).