Question

The area of the region bounded by the line y = 2x + 1, x-axis and the ordinates x = -1 and x = 1 is

• A. \(\frac{9}{4}\)
• B. 2
• C. \(\frac{5}{2}\)
• D. 5

Answer 1

The Correct Option is C

We need to find the area of the region bounded by the line \(y = 2x + 1\), the x-axis, and the ordinates \(x = -1\) and \(x = 1\).

First, find where the line intersects the x-axis (y = 0):

\(0 = 2x + 1\)

\(2x = -1\)

\(x = -\frac{1}{2}\)

So the line crosses the x-axis at \(x = -\frac{1}{2}\). 

This means from x = -1 to x = -1/2, the line is below the x-axis, and from x = -1/2 to x = 1, the line is above the x-axis.

The area is given by the sum of the absolute values of the integrals from -1 to -1/2 and from -1/2 to 1:

Area = \(|\int_{-1}^{-1/2} (2x+1) \, dx| + \int_{-1/2}^{1} (2x+1) \, dx\)

The antiderivative of \(2x + 1\) is \(x^2 + x\).

\(\int_{-1}^{-1/2} (2x+1) \, dx = [x^2 + x]_{-1}^{-1/2} = ((\frac{-1}{2})^2 + (-\frac{1}{2})) - ((-1)^2 + (-1)) = (\frac{1}{4} - \frac{1}{2}) - (1-1) = -\frac{1}{4} - 0 = -\frac{1}{4}\)

So the absolute value of the first integral is \(|-\frac{1}{4}| = \frac{1}{4}\).

\(\int_{-1/2}^{1} (2x+1) \, dx = [x^2 + x]_{-1/2}^{1} = ((1)^2 + (1)) - ((\frac{-1}{2})^2 + (-\frac{1}{2})) = (1+1) - (\frac{1}{4} - \frac{1}{2}) = 2 - (-\frac{1}{4}) = 2 + \frac{1}{4} = \frac{8}{4} + \frac{1}{4} = \frac{9}{4}\)

The total area is \(\frac{1}{4} + \frac{9}{4} = \frac{10}{4} = \frac{5}{2}\)

Therefore, the correct option is (C) \(\frac{5}{2}\).