Question:

The area bounded by the curve \(x = y^2\), the y-axis, and the lines \(y = 3\) and \(y = 4\) is:

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When given a curve in the form \(x = f(y)\), and the limits are on the \(y\)-axis, integrate with respect to \(y\): \[ \text{Area} = \int_{y=a}^{b} x \, dy \] This avoids needing to invert the function.
  • \(\dfrac{74}{3} \text{ sq units}\)
  • \(\dfrac{37}{3} \text{ sq units}\)
  • \(74 \text{ sq units}\)
  • \(37 \text{ sq units}\)
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The Correct Option is B

Solution and Explanation

We are given the curve \(x = y^2\), and we need to find the area bounded by this curve, the \(y\)-axis, and the horizontal lines \(y = 3\) and \(y = 4\). Since we are integrating with respect to \(y\), the area \(A\) is given by: \[ A = \int_{y=3}^{4} x \, dy = \int_{3}^{4} y^2 \, dy \] Now compute the integral: \[ \int_{3}^{4} y^2 \, dy = \left[\frac{y^3}{3}\right]_{3}^{4} = \frac{4^3}{3} - \frac{3^3}{3} = \frac{64}{3} - \frac{27}{3} = \frac{37}{3} \] \[ \boxed{A = \frac{37}{3} \text{ sq units}} \]
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