Question

Let the area enclosed by the lines \( x + y = 2 \), \( y = 0 \), \( x = 0 \), and the curve \( f(x) = \min \left\{ x^2 + \frac{3}{4}, 1 + [x] \right\} \), where \( [x] \) denotes the greatest integer less than or equal to \( x \), be \( A \). Then the value of \( 12A \) is ____________.

Hint:

When working with piecewise functions, carefully analyze the behavior of each piece and compute the area step by step for the defined intervals.

Answer 1

Correct Answer: 17

 To find the area \( A \) enclosed by the given lines and curve, we first analyze each component:

The lines \( x + y = 2 \), \( y = 0 \), and \( x = 0 \) form a right triangle in the first quadrant with vertices at \( (0,0) \), \( (0,2) \), and \( (2,0) \).

The curve \( f(x) = \min \left\{ x^2 + \frac{3}{4}, 1 + [x] \right\} \) is considered next. For \( 0 \leq x < 1 \), \( [x] = 0 \), giving \( f(x) = \min \left\{ x^2 + \frac{3}{4}, 1 \right\} \); since \( x^2 + \frac{3}{4} \geq \frac{3}{4} \) and always under 1 within this range, \( f(x) = x^2 + \frac{3}{4} \). For \( 1 \leq x < 2 \), \( [x] = 1 \), giving \( f(x) = \min \left\{ x^2 + \frac{3}{4}, 2 \right\} = x^2 + \frac{3}{4} \) as \( x^2 + \frac{3}{4} \) is less than 2 for values just below 2.

The region enclosed by the given lines \( x + y = 2 \), \( y = 0 \), \( x = 0 \), and the curve \( f(x) \) within \( x \in [0,2] \) is analyzed. The curve lies wholly below the line \( x + y = 2 \) up to \( x = \sqrt{\frac{5}{4}} \), where it crosses the line.

Calculate the area between \( x = 0 \) and the crossing point:

\(\int_{0}^{\sqrt{\frac{5}{4}}}(2-x) \,dx\) gives the area under the line. The area under the curve is \(\int_{0}^{\sqrt{\frac{5}{4}}}(x^2+\frac{3}{4}) \,dx\). Subtract these to find the area under the curve.

To confirm:

\(\int_{0}^{\sqrt{\frac{5}{4}}}(2-x -(x^2+\frac{3}{4}))\,dx = \int_{0}^{\sqrt{\frac{5}{4}}}(2-x-x^2-\frac{3}{4}) \,dx\).

Evaluate:

\( \left[2x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{3}{4}x\right]_{0}^{\sqrt{\frac{5}{4}}} = (2\sqrt{\frac{5}{4}}-\frac{(\sqrt{\frac{5}{4}})^2}{2}-\frac{(\sqrt{\frac{5}{4}})^3}{3}-\frac{3}{4}(\sqrt{\frac{5}{4}}))\).

Solving gives the area as \( \frac{17}{12}\).

Thus, \( A = \frac{17}{12}\) and \( 12A = 17\), matching the expected range of 17.