Question

The area in the first quadrant bounded by $y = 4x^2$ , x = 0, y =1 and y = 4 is

• A. $\frac{7}{3}$ Sq, unit
• B. $\frac{4}{5}$ Sq, unit
• C. $\frac{3}{4}$ Sq, unit
• D. none of these

Answer 1

Answer: (A)

$\frac{7}{3}$ Sq, unit

Answer 2

Given $y = 4x^2, x = 0, y = 1$ and $y = 4$ Now, $y = 4x^{2} \Rightarrow x^{2} = \frac{y}{4} \Rightarrow x = \frac{1}{2} \sqrt{y}$ Required area = $\int^{4}_{1} x dy = \frac{1}{2} \int^{4}_{1} \sqrt{y} dy$ $= \frac{1}{2} \left[\frac{y^{3/2}}{3/2}\right]^{4}_{1} = \frac{1}{3} \left[y^{3/2}\right]^{4}_{1}$ $= \frac{1}{3} \left[4^{3/2} - 1^{3/2}\right] = \frac{7}{3}$ s unit