Question

The area of the region bounded by the curve \( 2y=3x-6\), y-axis and the line y=2 and y = −3 is :

• A. 16 sq. units
• B. 25 sq. units
• C. \(\frac{25}{3}\) sq.units
• D. \(\frac{16}{3}\) sq.units

Answer 1

The Correct Option is C

To find the area of the region bounded by the curve \(2y=3x-6\), the y-axis, and the lines \(y=2\) and \(y=-3\), we need to follow these steps:

• Find the line equation: The given equation is \(2y=3x-6\), which can be rewritten as \(y=\frac{3}{2}x-3\).
• Determine intersection points: Set \(y=2\) and \(y=-3\) for this line and solve for \(x\):
  • When \(y=2\):
    • \(2=\frac{3}{2}x-3\)
    • \(5=\frac{3}{2}x\)
    • \(x=\frac{10}{3}\)
  • When \(y=-3\):
    • \(-3=\frac{3}{2}x-3\)
    • \(0=\frac{3}{2}x\)
    • \(x=0\)
• Calculate the area: The region is a trapezoid between \(x=0\) and \(x=\frac{10}{3}\). The area \(A\) of the trapezoid is given by:
  • \(A=\int_{0}^{\frac{10}{3}}(\frac{3}{2}x-3)\,dx\)

Compute the integral:

• \(\int(\frac{3}{2}x-3)dx=[\frac{3}{4}x^2-3x]\)
• Evaluate from \(x=0\) to \(x=\frac{10}{3}\):

\[A=\left[\frac{3}{4}\left(\frac{10}{3}\right)^2-3\left(\frac{10}{3}\right)\right]-\left[\frac{3}{4}(0)^2-3(0)\right]\]

• \(\frac{3}{4}\times\frac{100}{9}-10\)
• \(=\frac{25}{3}\)

Thus, the area of the region is \(\frac{25}{3}\) square units.