Question

The area of the region bounded by the line y = x +1, and the lines x = 3 and x = 5 is

• A. \(\frac{11}{2}\) sq. units
• B. 10 sq. units
• C. 7 sq. units
• D. \(\frac{7}{2}\) sq. units

Hint:

When calculating the area between a curve and vertical lines, set up the integral of the function between the limits of integration. Then, evaluate the integral by substituting the limits into the antiderivative. This process will give you the area of the region.

Answer 1

The Correct Option is B

The correct answer is: (B): 10 sq. units.

We are tasked with finding the area of the region bounded by the line \( y = x + 1 \) and the vertical lines \( x = 3 \) and \( x = 5 \).

Step 1: Set up the integral

We know that the area under the curve between two vertical lines is given by the integral of the function between the limits of the lines. The function is \( y = x + 1 \), and the limits are \( x = 3 \) and \( x = 5 \). Therefore, we set up the integral for the area as follows:

\( A = \int_{3}^{5} (x + 1) \, dx \)

Step 2: Compute the integral

Now, we compute the integral:

\( \int (x + 1) \, dx = \frac{x^2}{2} + x \)

Step 3: Evaluate the integral

Substitute the limits \( x = 5 \) and \( x = 3 \) into the integral:

\( A = \left[ \frac{x^2}{2} + x \right]_{3}^{5} \)

Evaluating at \( x = 5 \) and \( x = 3 \):

\( A = \left( \frac{5^2}{2} + 5 \right) - \left( \frac{3^2}{2} + 3 \right) \)

\( A = \left( \frac{25}{2} + 5 \right) - \left( \frac{9}{2} + 3 \right) \)

\( A = \left( \frac{25}{2} + \frac{10}{2} \right) - \left( \frac{9}{2} + \frac{6}{2} \right) \)

\( A = \frac{35}{2} - \frac{15}{2} = \frac{20}{2} = 10 \)

Conclusion:
The area of the region is \( 10 \) square units, so the correct answer is (B): 10 sq. units.