Samacheer Kalvi Class 12 Maths Solutions Chapter 9 Applications of Integration Exercise 9.1

Question 1. Find an approximate value of \( \int_{1}^{1.5} x \, dx \) by applying the left-end rule with the partition \( \{1.1, 1.2, 1.3, 1.4, 1.5\} \)
Answer: To find the approximate value using the left-end rule, we first identify the given values:
\( a = 1 \), \( b = 1.5 \), \( n = 5 \) (number of subintervals), and \( f(x) = x \).
Now, we calculate the width of each subinterval, \( h \), also known as \( \Delta x \):
\( h = \Delta x = \frac{b-a}{n} = \frac{1.5-1}{5} = \frac{0.5}{5} = 0.1 \)
The left endpoints of the subintervals are \( x_0 = 1 \), \( x_1 = 1.1 \), \( x_2 = 1.2 \), \( x_3 = 1.3 \), \( x_4 = 1.4 \). We use these points to evaluate the function.
The left-end rule for the Riemann sum \( S \) is:
\( S = [f(x_0) + f(x_1) + f(x_2) + f(x_3) + f(x_4)] \Delta x \)
Substitute the values into the formula:
\( S = [f(1) + f(1.1) + f(1.2) + f(1.3) + f(1.4)] (0.1) \)
Since \( f(x) = x \), we simply use the values of \( x \):
\( S = [1 + 1.1 + 1.2 + 1.3 + 1.4] (0.1) \)
Add the numbers inside the brackets:
\( S = [6] (0.1) \)
Finally, multiply to get the approximate value:
\( S = 0.6 \). The left-end rule uses the start of each segment to estimate the area.
In simple words: First, find the width of each small section. Then, add up the function values at the left side of each section and multiply by the width. This gives an estimated area under the curve.

🎯 Exam Tip: Remember that for the left-end rule, you always use the starting point of each subinterval to calculate the height of the rectangle.

Question 2. Find an approximate value of \( \int_{1}^{1.5} x^2 \, dx \) by applying the right-end rule with the partition \( \{1.1, 1.2, 1.3, 1.4, 1.5\} \)
Answer: To find the approximate value using the right-end rule, we first list the given information:
\( a = 1 \), \( b = 1.5 \), \( n = 5 \) (number of subintervals), and \( f(x) = x^2 \).
First, calculate the width of each subinterval, \( h \), also known as \( \Delta x \):
\( h = \Delta x = \frac{b-a}{n} = \frac{1.5-1}{5} = \frac{0.5}{5} = 0.1 \)
The right endpoints of the subintervals are \( x_1 = 1.1 \), \( x_2 = 1.2 \), \( x_3 = 1.3 \), \( x_4 = 1.4 \), \( x_5 = 1.5 \). These are the points where we will evaluate the function.
The right-end rule for the Riemann sum \( S \) with equal width \( \Delta x \) is:
\( S = [f(x_1) + f(x_2) + f(x_3) + f(x_4) + f(x_5)] \Delta x \)
Substitute the values into the formula:
\( S = [f(1.1) + f(1.2) + f(1.3) + f(1.4) + f(1.5)](0.1) \)
Since \( f(x) = x^2 \), we square each right endpoint:
\( S = [ (1.1)^2 + (1.2)^2 + (1.3)^2 + (1.4)^2 + (1.5)^2 ] \times 0.1 \)
Calculate the squares:
\( S = (1.21 + 1.44 + 1.69 + 1.96 + 2.25) \times 0.1 \)
Add the numbers inside the parenthesis:
\( S = 8.55 \times 0.1 \)
Finally, multiply to get the approximate value:
\( S = 0.855 \). The right-end rule considers the end of each segment to determine the height.
In simple words: Find the width of each segment. Then, add up the function values at the right side of each segment and multiply by the width. This gives an estimate of the area under the curve.

🎯 Exam Tip: For the right-end rule, make sure to use the rightmost point of each subinterval to calculate the height of the rectangle, and adjust your sum accordingly.

Question 3. Find an approximate value of \( \int_{1}^{1.5} (2-x) \, dx \) by applying the mid-point rule with the partition \( \{1.1, 1.2, 1.3, 1.4, 1.5\} \)
Answer: To find the approximate value using the mid-point rule, let's identify the given information:
\( a = 1 \), \( b = 1.5 \), \( n = 5 \) (number of subintervals), and \( f(x) = 2-x \).
First, calculate the width of each subinterval, \( h \), which is \( \Delta x \):
\( h = \Delta x = \frac{b-a}{n} = \frac{1.5-1}{5} = \frac{0.5}{5} = 0.1 \)
The partition points are \( x_0 = 1 \), \( x_1 = 1.1 \), \( x_2 = 1.2 \), \( x_3 = 1.3 \), \( x_4 = 1.4 \), \( x_5 = 1.5 \).
Next, we find the midpoint of each subinterval and evaluate \( f(x) = 2-x \) at these midpoints:
For the first interval \( [1, 1.1] \): Midpoint \( \frac{1+1.1}{2} = 1.05 \).
\( f(1.05) = 2 - 1.05 = 0.95 \)
For the second interval \( [1.1, 1.2] \): Midpoint \( \frac{1.1+1.2}{2} = 1.15 \).
\( f(1.15) = 2 - 1.15 = 0.85 \)
For the third interval \( [1.2, 1.3] \): Midpoint \( \frac{1.2+1.3}{2} = 1.25 \).
\( f(1.25) = 2 - 1.25 = 0.75 \)
For the fourth interval \( [1.3, 1.4] \): Midpoint \( \frac{1.3+1.4}{2} = 1.35 \).
\( f(1.35) = 2 - 1.35 = 0.65 \)
For the fifth interval \( [1.4, 1.5] \): Midpoint \( \frac{1.4+1.5}{2} = 1.45 \).
\( f(1.45) = 2 - 1.45 = 0.55 \)
The mid-point rule for the Riemann sum \( S \) with equal width \( \Delta x \) is the sum of these function values multiplied by \( \Delta x \):
\( S = (0.95 + 0.85 + 0.75 + 0.65 + 0.55) \times 0.1 \)
Add the values inside the parenthesis:
\( S = 3.75 \times 0.1 \)
Finally, multiply to get the approximate value:
\( S = 0.375 \). The midpoint rule often provides a more accurate approximation than the left or right-end rules.
In simple words: Find the width of each section. Then, for each section, find the middle point. Add up the function values at these middle points and multiply by the width. This estimates the area under the curve.

🎯 Exam Tip: When using the midpoint rule, ensure you correctly calculate the midpoint for each subinterval before evaluating the function at that point.