Question

The area of the region bounded by the curves \(x^2=4y\), the line x = 3 and x axis is:

• A. \(\frac{3}{4}\)
• B. \(\frac{7}{4}\)
• C. \(\frac{9}{4}\)
• D. \(\frac{18}{4}\)

Answer 1

The Correct Option is C

To find the area of the region bounded by the curve \(x^2=4y\), the line \(x=3\), and the x-axis, we first analyze the curve and the given line boundaries. The parabola \(x^2=4y\) can be rewritten as \(y=\frac{x^2}{4}\). This parabola opens upwards and meets the x-axis when \(y=0\), i.e., \(x^2=0\) or \(x=0\).

We are interested in the region from \(x=0\) to \(x=3\) along the parabola and down to the x-axis, as specified by the bounding \(x=3\).

The area can be computed using the definite integral of the function \(y=\frac{x^2}{4}\) over the interval from \(x=0\) to \(x=3\).

The area \(A\) is:

\[ A = \int_{0}^{3} \frac{x^2}{4} \, dx \]

Calculating the integral:

\[ A = \frac{1}{4} \int_{0}^{3} x^2 \, dx = \frac{1}{4} \left[\frac{x^3}{3}\right]_{0}^{3} \]

\[ = \frac{1}{4} \left(\frac{3^3}{3} - \frac{0^3}{3}\right) \]

\[ = \frac{1}{4} \left(\frac{27}{3}\right) \]

\[ = \frac{1}{4} \times 9 \]

\[ = \frac{9}{4} \]

Therefore, the area of the region bounded by the curves is \(\frac{9}{4}\).