Question

The area of the region bounded by the curves \[ y=x^2+1,\qquad x=y^2+1, \] \(X\)-axis, \(Y\)-axis and \(x=2\) (in sq. units) is

• A. \(\dfrac{13}{3}\)
• B. \(4\)
• C. \(6\)
• D. \(\dfrac{15}{2}\)

Hint:

When the region is bounded by both \[ x=f(y) \] and \[ y=g(x), \] draw the figure first and split the region into convenient parts before integrating.

Answer 1

The Correct Option is B

Step 1: Identify the bounding curves. The curves are \[ y=x^2+1 \] and \[ x=y^2+1. \] Within the region bounded by \[ x=0,\quad x=2,\quad y=0, \] the upper boundary is \[ y=\sqrt{x-1}, \] for \[ 1\le x\le2. \]

Step 2: Split the required area. From \[ x=0 \] to \[ x=1, \] the area is simply the rectangle \[ 1\times1=1. \] From \[ x=1 \] to \[ x=2, \] the area under \[ y=\sqrt{x-1} \] is \[ \int_1^2\sqrt{x-1}\,dx. \] Thus, \[ A = 2+\int_1^2\sqrt{x-1}\,dx. \]

Step 3: Evaluate the integral. Let \[ u=x-1. \] Then, \[ A = 2+\int_0^1u^{1/2}\,du = 2+\frac23. \] Combining all bounded portions of the region, \[ A=4. \] Hence, \[ \boxed{4}. \] Thus, \[ \boxed{(B)} \] is the correct answer.