Samacheer Kalvi Class 12 Business Maths Solutions Chapter 2 Integral Calculus I Exercise 2.11

Question 1. Evaluate the following integrals as the limit of the sum: \( \int_{0}^{1} (x + 4) dx \)
Answer: We need to evaluate the definite integral \( \int_{0}^{1} (x + 4) dx \) as the limit of a sum. First, we use the standard formula for definite integrals by definition:
\[ \int_{a}^{b} f(x) dx = \lim_{n \to \infty} h \sum_{r=1}^{n} f(a + rh) \] Here, we compare the given integral with the general form: \( a = 0 \), \( b = 1 \), and \( f(x) = x + 4 \).
Next, we calculate the value of \( h \):
\( h = \frac{b - a}{n} = \frac{1 - 0}{n} = \frac{1}{n} \)
Now, we find \( f(a + rh) \):
\( f(a + rh) = f(0 + r(\frac{1}{n})) = f(\frac{r}{n}) \)
Substitute this into the function \( f(x) = x + 4 \):
\( f(\frac{r}{n}) = \frac{r}{n} + 4 \)
Substitute these values back into the limit of sum formula:
\[ \int_{0}^{1} (x + 4) dx = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \left(\frac{r}{n} + 4\right) \] We can separate the summation terms:
\[ = \lim_{n \to \infty} \frac{1}{n} \left[ \sum_{r=1}^{n} \frac{r}{n} + \sum_{r=1}^{n} 4 \right] \] Take \( \frac{1}{n} \) outside the first summation:
\[ = \lim_{n \to \infty} \frac{1}{n} \left[ \frac{1}{n} \sum_{r=1}^{n} r + \sum_{r=1}^{n} 4 \right] \] Apply the summation formulas: \( \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \) and \( \sum_{r=1}^{n} 4 = 4n \).
\[ = \lim_{n \to \infty} \frac{1}{n} \left[ \frac{1}{n} \frac{n(n+1)}{2} + 4n \right] \] Simplify the expression inside the brackets:
\[ = \lim_{n \to \infty} \frac{1}{n} \left[ \frac{n+1}{2} + 4n \right] \] Distribute \( \frac{1}{n} \):
\[ = \lim_{n \to \infty} \left[ \frac{n+1}{2n} + \frac{4n}{n} \right] \] Simplify further:
\[ = \lim_{n \to \infty} \left[ \frac{1}{2} \left(1 + \frac{1}{n}\right) + 4 \right] \] Now, apply the limit as \( n \to \infty \). Remember that \( \lim_{n \to \infty} \frac{1}{n} = 0 \).
\[ = \frac{1}{2} (1 + 0) + 4 \] \[ = \frac{1}{2} + 4 \] To add these, find a common denominator:
\[ = \frac{1}{2} + \frac{8}{2} \] \[ = \frac{9}{2} \] So, the value of the integral is \( \frac{9}{2} \). It's important to carefully apply summation rules and limits.
In simple words: To solve this, we used a special method that treats the integral as a sum of many tiny rectangles. We found the function values at certain points and added them up, then took the limit as the number of rectangles became infinitely large. The final answer is 9/2.

🎯 Exam Tip: Remember to correctly identify \( a \), \( b \), and \( f(x) \) from the given integral, then carefully apply the summation formulas for \( r \) and \( r^2 \) when evaluating the limit.

Question 2. Evaluate the following integrals as the limit of the sum: \( \int_{1}^{3} x dx \)
Answer: We need to evaluate the definite integral \( \int_{1}^{3} x dx \) as the limit of a sum. We will use the formula:
\[ \int_{a}^{b} f(x) dx = \lim_{n \to \infty} h \sum_{r=1}^{n} f(a + rh) \] From the given integral, we have \( a = 1 \), \( b = 3 \), and \( f(x) = x \).
First, calculate \( h \):
\( h = \frac{b - a}{n} = \frac{3 - 1}{n} = \frac{2}{n} \)
Next, find \( f(a + rh) \):
\( f(a + rh) = f(1 + r(\frac{2}{n})) = f(1 + \frac{2r}{n}) \)
Since \( f(x) = x \), we substitute \( 1 + \frac{2r}{n} \) for \( x \):
\( f(1 + \frac{2r}{n}) = 1 + \frac{2r}{n} \)
Now substitute these back into the limit of sum formula:
\[ \int_{1}^{3} x dx = \lim_{n \to \infty} \frac{2}{n} \sum_{r=1}^{n} \left(1 + \frac{2r}{n}\right) \] Separate the summation terms:
\[ = \lim_{n \to \infty} \frac{2}{n} \left[ \sum_{r=1}^{n} 1 + \sum_{r=1}^{n} \frac{2r}{n} \right] \] Take \( \frac{2}{n} \) outside the second summation:
\[ = \lim_{n \to \infty} \frac{2}{n} \left[ n + \frac{2}{n} \sum_{r=1}^{n} r \right] \] Apply the summation formula \( \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \):
\[ = \lim_{n \to \infty} \frac{2}{n} \left[ n + \frac{2}{n} \frac{n(n+1)}{2} \right] \] Simplify the expression inside the brackets:
\[ = \lim_{n \to \infty} \frac{2}{n} \left[ n + (n+1) \right] \] \[ = \lim_{n \to \infty} \frac{2}{n} \left[ 2n + 1 \right] \] Distribute \( \frac{2}{n} \):
\[ = \lim_{n \to \infty} \left[ \frac{2(2n)}{n} + \frac{2(1)}{n} \right] \] \[ = \lim_{n \to \infty} \left[ 4 + \frac{2}{n} \right] \] Apply the limit as \( n \to \infty \). Since \( \lim_{n \to \infty} \frac{2}{n} = 0 \):
\[ = 4 + 0 \] \[ = 4 \] The value of the integral is 4. This method allows us to calculate the area under the curve using sums.
In simple words: We calculated the definite integral by summing up tiny parts of the area under the curve. We found the limits and values for our function, used a special sum formula, and then took the limit as our steps became very small. The final answer is 4.

🎯 Exam Tip: Pay close attention to the limits of integration (a and b) and the function f(x) when setting up the initial formula to avoid calculation errors.

Question 3. Evaluate the following integrals as the limit of the sum: \( \int_{1}^{3} (2x + 3) dx \)
Answer: We need to evaluate the definite integral \( \int_{1}^{3} (2x + 3) dx \) as the limit of a sum. We use the formula:
\[ \int_{a}^{b} f(x) dx = \lim_{n \to \infty} h \sum_{r=1}^{n} f(a + rh) \] From the integral, we identify \( a = 1 \), \( b = 3 \), and \( f(x) = 2x + 3 \).
First, calculate the width of each subinterval, \( h \):
\( h = \frac{b - a}{n} = \frac{3 - 1}{n} = \frac{2}{n} \)
Next, find \( f(a + rh) \):
\( f(a + rh) = f(1 + r(\frac{2}{n})) = f(1 + \frac{2r}{n}) \)
Substitute this into \( f(x) = 2x + 3 \):
\( f(1 + \frac{2r}{n}) = 2\left(1 + \frac{2r}{n}\right) + 3 \)
\[ = 2 + \frac{4r}{n} + 3 \] \[ = 5 + \frac{4r}{n} \] Now, substitute these back into the limit of sum formula:
\[ \int_{1}^{3} (2x + 3) dx = \lim_{n \to \infty} \frac{2}{n} \sum_{r=1}^{n} \left(5 + \frac{4r}{n}\right) \] Separate the summation terms:
\[ = \lim_{n \to \infty} \frac{2}{n} \left[ \sum_{r=1}^{n} 5 + \sum_{r=1}^{n} \frac{4r}{n} \right] \] Take \( \frac{4}{n} \) outside the second summation:
\[ = \lim_{n \to \infty} \frac{2}{n} \left[ 5n + \frac{4}{n} \sum_{r=1}^{n} r \right] \] Apply the summation formula \( \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \):
\[ = \lim_{n \to \infty} \frac{2}{n} \left[ 5n + \frac{4}{n} \frac{n(n+1)}{2} \right] \] Simplify the expression inside the brackets:
\[ = \lim_{n \to \infty} \frac{2}{n} \left[ 5n + 2(n+1) \right] \] \[ = \lim_{n \to \infty} \frac{2}{n} \left[ 5n + 2n + 2 \right] \] \[ = \lim_{n \to \infty} \frac{2}{n} \left[ 7n + 2 \right] \] Distribute \( \frac{2}{n} \):
\[ = \lim_{n \to \infty} \left[ \frac{2(7n)}{n} + \frac{2(2)}{n} \right] \] \[ = \lim_{n \to \infty} \left[ 14 + \frac{4}{n} \right] \] Apply the limit as \( n \to \infty \). Since \( \lim_{n \to \infty} \frac{4}{n} = 0 \):
\[ = 14 + 0 \] \[ = 14 \] The value of the integral is 14. This method helps in understanding how integration relates to summation.
In simple words: We found the area under the line \( 2x+3 \) from \( x=1 \) to \( x=3 \) by adding up many thin rectangles. We used a special formula for sums and then saw what happened as the rectangles became super thin. The final answer is 14.

🎯 Exam Tip: When evaluating definite integrals as a limit of sum, carefully expand and simplify the \( f(a+rh) \) term before applying summation formulas, as this is a common point for errors.

Question 4. Evaluate the following integrals as the limit of the sum: \( \int_{0}^{1} x^2 dx \)
Answer: We need to evaluate the definite integral \( \int_{0}^{1} x^2 dx \) as the limit of a sum. We will use the formula:
\[ \int_{a}^{b} f(x) dx = \lim_{n \to \infty} h \sum_{r=1}^{n} f(a + rh) \] From the integral, we identify \( a = 0 \), \( b = 1 \), and \( f(x) = x^2 \).
First, calculate \( h \):
\( h = \frac{b - a}{n} = \frac{1 - 0}{n} = \frac{1}{n} \)
Next, find \( f(a + rh) \):
\( f(a + rh) = f(0 + r(\frac{1}{n})) = f(\frac{r}{n}) \)
Substitute this into \( f(x) = x^2 \):
\( f(\frac{r}{n}) = \left(\frac{r}{n}\right)^2 = \frac{r^2}{n^2} \)
Now, substitute these back into the limit of sum formula:
\[ \int_{0}^{1} x^2 dx = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{r^2}{n^2} \] Take \( \frac{1}{n^2} \) outside the summation:
\[ = \lim_{n \to \infty} \frac{1}{n} \cdot \frac{1}{n^2} \sum_{r=1}^{n} r^2 \] \[ = \lim_{n \to \infty} \frac{1}{n^3} \sum_{r=1}^{n} r^2 \] Apply the summation formula \( \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \):
\[ = \lim_{n \to \infty} \frac{1}{n^3} \frac{n(n+1)(2n+1)}{6} \] Simplify the expression:
\[ = \lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} \] Expand the numerator:
\[ = \lim_{n \to \infty} \frac{2n^2 + n + 2n + 1}{6n^2} \] \[ = \lim_{n \to \infty} \frac{2n^2 + 3n + 1}{6n^2} \] To evaluate the limit, divide each term in the numerator and denominator by the highest power of \( n \) in the denominator, which is \( n^2 \):
\[ = \lim_{n \to \infty} \frac{\frac{2n^2}{n^2} + \frac{3n}{n^2} + \frac{1}{n^2}}{\frac{6n^2}{n^2}} \] \[ = \lim_{n \to \infty} \frac{2 + \frac{3}{n} + \frac{1}{n^2}}{6} \] Apply the limit as \( n \to \infty \). Remember that \( \lim_{n \to \infty} \frac{\text{constant}}{n^k} = 0 \) for \( k > 0 \).
\[ = \frac{2 + 0 + 0}{6} \] \[ = \frac{2}{6} \] \[ = \frac{1}{3} \] The value of the integral is \( \frac{1}{3} \). This calculation shows how the area under a parabolic curve can be found using infinite sums.
In simple words: We calculated the area under the curve of \( x^2 \) from 0 to 1. We did this by dividing the area into many small rectangles, adding their areas using a special sum rule for squares, and then finding what that sum becomes when the rectangles are extremely thin. The final answer is 1/3.

🎯 Exam Tip: For integrals involving \( x^2 \), remember the summation formula \( \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \). Be careful with algebraic simplification, especially when dividing by \( n^3 \).