Question

The area bounded by the curve $y = x^2$, the normal at (1, 1) and the x-axis is :

• A. $\frac{4}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. None

Answer 1

Answer: (A)

$\frac{4}{3}$

Answer 2

$y = x^{2} \Rightarrow \frac{dy}{dx} = 2 x \Rightarrow \left(\frac{dy}{dx}\right)_{\left(1,1\right)} = 2$ $\therefore$ Slope of tangent at (1,1) = 2 $\therefore$ Slope of normal at (1,1) = $- \frac{1}{2}$ Equation to the normal at (1, 1) is $y - 1 = - \frac{1}{2} (x -1) \, \Rightarrow \, x + 2y - 3 = 0$ The normal intersects x-axis of (3,0) the required area is shown as shaded region. Required area A = area (OANO)+area (NABN) $A = \int^{1}_{0} x^{2}dx + \int^{3}_{1} \frac{1}{2}\left(3-x\right)dx$ $= \left[\frac{x^{3}}{3}\right]^{1}_{0} + \frac{1}{2} \left[3x- \frac{x^{2}}{2}\right]^{3}_{1} = \frac{4}{3}$