Question

The area of the region bounded by the curves \[ y=2^x,\qquad y^2=4x \] and the lines \[ x=\frac12,\qquad x=1 \] is

• A. \(\dfrac{2-\sqrt2}{\log2}-\dfrac{4-\sqrt2}{3}\)
• B. \(\dfrac{4-\sqrt2}{3}-\dfrac{2-\sqrt2}{\log2}\)
• C. \(\dfrac{4+\sqrt2}{3}-\dfrac{2-\sqrt2}{\log2}\)
• D. \(\dfrac{2-\sqrt2}{\log2}-\dfrac{4+\sqrt2}{3}\)

Hint:

When finding the area between two curves, \[ \boxed{ \text{Area} = \int_a^b (\text{Upper curve}-\text{Lower curve})\,dx. } \] Always determine which curve lies above the other before integrating.

Answer 1

The Correct Option is B

Step 1: Identify the upper and lower curves. The curves are \[ y=2^x \] and \[ y^2=4x \quad\Rightarrow\quad y=2\sqrt{x}, \] since \(y>0\) in the given interval. For \[ \frac12\le x\le1, \] we have \[ 2\sqrt{x}>2^x. \] Hence, \[ \text{Area} = \int_{1/2}^{1} \left(2\sqrt{x}-2^x\right)\,dx. \]

Step 2: Integrate each function. Now, \[ \int2\sqrt{x}\,dx = 2\cdot\frac23x^{3/2} = \frac43x^{3/2}, \] and \[ \int2^x\,dx = \frac{2^x}{\log2}. \] Therefore, \[ \text{Area} = \left[ \frac43x^{3/2} - \frac{2^x}{\log2} \right]_{1/2}^{1}. \]

Step 3: Substitute the limits. At \[ x=1, \] \[ \frac43(1)^{3/2} - \frac2{\log2} = \frac43-\frac2{\log2}. \] At \[ x=\frac12, \] \[ \frac43 \left(\frac12\right)^{3/2} - \frac{\sqrt2}{\log2} = \frac{\sqrt2}{3} - \frac{\sqrt2}{\log2}. \] Hence, \[ \text{Area} = \left( \frac43-\frac2{\log2} \right) - \left( \frac{\sqrt2}{3} - \frac{\sqrt2}{\log2} \right). \] Simplifying, \[ \boxed{ \text{Area} = \frac{4-\sqrt2}{3} - \frac{2-\sqrt2}{\log2}. } \] Therefore, the correct option is \(\boxed{(B)}\).