Question

Area enclosed by curve \(y = 2x^2\) and lines \(x\geq 1, y\leq 4\) is ..... sq. units

• A. \(\frac{8\sqrt{2}-10}{3}\)
• B. \(\frac{8(\sqrt{2}-1)}{3}\)
• C. \(\frac{4\sqrt{2}-5}{3}\)
• D. \(\frac{4(\sqrt{2}-1)}{3}\)

Hint:

The curve meets \(y=4\) at \(x=\sqrt2\), so integrate \(4-2x^2\) from 1 to \(\sqrt2\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The region is bounded above by \(y=4\), below by the parabola \(y=2x^2\), and on the left by \(x=1\).

Step 2: Key Formula or Approach
\(2x^2=4\) gives \(x=\sqrt2\). So the limits are \(x=1\) to \(x=\sqrt2\).

Step 3: Detailed Explanation
\[ \text{Area}=\int_1^{\sqrt2}(4-2x^2)\,dx=\left[4x-\frac{2x^3}{3}\right]_1^{\sqrt2} \]
At \(\sqrt2\): \(4\sqrt2-\dfrac{2\cdot2\sqrt2}{3}=\dfrac{8\sqrt2}{3}\). At 1: \(4-\dfrac23=\dfrac{10}{3}\).
\[ \text{Area}=\frac{8\sqrt2-10}{3} \]

Final Answer:
The area is \(\dfrac{8\sqrt2-10}{3}\) square units, option (A). \[ \boxed{\dfrac{8\sqrt2-10}{3}\ \text{(A)}} \]