Question:

The area of the region bounded by the line 2y = 5x +7, x-axis and the lines x - 1 and x -3 is :

Updated On: May 11, 2025
  • 15
  • 17
  • 16
  • 19
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The Correct Option is B

Solution and Explanation

To find the area of the region bounded by the line, the x-axis, and the vertical lines, we start by expressing the line equation in slope-intercept form: \(2y = 5x + 7\) simplifies to \(y = \frac{5}{2}x + \frac{7}{2}\).
Next, identify the x-intercepts of the line by setting \(y = 0\):
\(0 = \frac{5}{2}x + \frac{7}{2}\)
Solving for \(x\):
\(-\frac{7}{2} = \frac{5}{2}x\)
Multiplying both sides by \(\frac{2}{5}\), we get:
\(x = -\frac{7}{5}\)
The region is bounded by the vertical lines \(x=1\) and \(x=3\), so the x-intercepts within this range are unnecessary as the boundary is dictated by these verticals.
We calculate the area bounded by the x-axis (from above) and the line (from below):
The definite integral gives us the exact area under \(y = \frac{5}{2}x + \frac{7}{2}\) between \(x=1\) and \(x=3\):
\(\int_1^3 \left(\frac{5}{2}x + \frac{7}{2}\right)dx\)
Integrating, we have:
\(\left[\frac{5}{4}x^2 + \frac{7}{2}x\right]_1^3\)
Evaluating at the bounds:
\(\left(\frac{5}{4}(3)^2 + \frac{7}{2}(3)\right) - \left(\frac{5}{4}(1)^2 + \frac{7}{2}(1)\right)\)
Simplifying each term, we have:
\(=\left(\frac{45}{4} + \frac{21}{2}\right) - \left(\frac{5}{4} + \frac{7}{2}\right)\)
\(=\left(\frac{45}{4} + \frac{42}{4}\right) - \left(\frac{5}{4} + \frac{14}{4}\right)\)
\(=\left(\frac{87}{4}\right) - \left(\frac{19}{4}\right)\)
\(=\frac{68}{4} = 17\)
Thus, the area of the region is 17.
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