Question:

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

Updated On: Jul 7, 2022
  • $\frac{7}{3}$ Sq, unit
  • $\frac{4}{5}$ Sq, unit
  • $\frac{3}{4}$ Sq, unit
  • none of these
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The Correct Option is A

Solution and Explanation

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
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Concepts Used:

Area under Simple Curves

  • The area of the region bounded by the curve y = f (x), x-axis and the lines x = a and x = b (b > a) - given by the formula:
\[\text{Area}=\int_a^bydx=\int_a^bf(x)dx\]
  • The area of the region bounded by the curve x = φ (y), y-axis and the lines y = c, y = d - given by the formula:
\[\text{Area}=\int_c^dxdy=\int_c^d\phi(y)dy\]

Read More: Area under the curve formula