Samacheer Kalvi Class 12 Maths Solutions Chapter 9 Applications of Integration Exercise 9.2

Question 1. Evaluate the following integrals as the limits of sums:
(i) \( \int_{0}^{1} (5x + 4)dx \)
(ii) \( \int_{1}^{2} (4x^2 - 1)dx \)
Answer:
(i) We need to evaluate the integral \( \int_{0}^{1} (5x + 4)dx \) using the limit of sums.
Here, the function is \( f(x) = 5x + 4 \), the lower limit is \( a = 0 \), and the upper limit is \( b = 1 \).
We use the formula: \( \int_a^b f(x)dx = \lim_{n \to \infty} \frac{b-a}{n} \sum_{r=1}^n f(a + \frac{r(b-a)}{n}) \).
Applying the values:
\( \int_0^1 (5x+4)dx = \lim_{n \to \infty} \frac{1-0}{n} \sum_{r=1}^n f(0 + \frac{r(1-0)}{n}) \)
\( = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^n f(\frac{r}{n}) \)
\( = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^n (5(\frac{r}{n}) + 4) \)
\( = \lim_{n \to \infty} \frac{1}{n} [\sum_{r=1}^n \frac{5r}{n} + \sum_{r=1}^n 4] \)
\( = \lim_{n \to \infty} \frac{1}{n} [\frac{5}{n} \sum_{r=1}^n r + 4n] \)
We know that \( \sum_{r=1}^n r = \frac{n(n+1)}{2} \). Substitute this into the expression:
\( = \lim_{n \to \infty} \frac{1}{n} [\frac{5}{n} \frac{n(n+1)}{2} + 4n] \)
\( = \lim_{n \to \infty} \frac{1}{n} [\frac{5(n+1)}{2} + 4n] \)
Now, we divide each term inside the bracket by \( n \):
\( = \lim_{n \to \infty} [\frac{5(n+1)}{2n} + \frac{4n}{n}] \)
\( = \lim_{n \to \infty} [\frac{5n+5}{2n} + 4] \)
\( = \lim_{n \to \infty} [\frac{5}{2} + \frac{5}{2n} + 4] \)
As \( n \to \infty \), the term \( \frac{5}{2n} \) approaches 0.
\( = \frac{5}{2} + 0 + 4 \)
\( = \frac{5+8}{2} \)
\( = \frac{13}{2} \)

(ii) We need to evaluate the integral \( \int_{1}^{2} (4x^2 - 1)dx \) using the limit of sums.
Here, the function is \( f(x) = 4x^2 - 1 \), the lower limit is \( a = 1 \), and the upper limit is \( b = 2 \).
Using the same formula for the limit of sums:
\( \int_a^b f(x)dx = \lim_{n \to \infty} \frac{b-a}{n} \sum_{r=1}^n f(a + \frac{r(b-a)}{n}) \)
Substitute the values of \( a \) and \( b \):
\( \int_1^2 (4x^2 - 1)dx = \lim_{n \to \infty} \frac{2-1}{n} \sum_{r=1}^n f(1 + \frac{r(2-1)}{n}) \)
\( = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^n f(1 + \frac{r}{n}) \)
\( = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^n [4(1 + \frac{r}{n})^2 - 1] \)
Expand the square term \( (1 + \frac{r}{n})^2 \):
\( = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^n [4(1 + \frac{2r}{n} + \frac{r^2}{n^2}) - 1] \)
\( = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^n [4 + \frac{8r}{n} + \frac{4r^2}{n^2} - 1] \)
\( = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^n [3 + \frac{8r}{n} + \frac{4r^2}{n^2}] \)
Separate the summation terms:
\( = \lim_{n \to \infty} \frac{1}{n} [\sum_{r=1}^n 3 + \frac{8}{n} \sum_{r=1}^n r + \frac{4}{n^2} \sum_{r=1}^n r^2] \)
We know the summation formulas: \( \sum_{r=1}^n 3 = 3n \), \( \sum_{r=1}^n r = \frac{n(n+1)}{2} \), and \( \sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6} \).
Substitute these formulas:
\( = \lim_{n \to \infty} \frac{1}{n} [3n + \frac{8}{n} \frac{n(n+1)}{2} + \frac{4}{n^2} \frac{n(n+1)(2n+1)}{6}] \)
Simplify the terms:
\( = \lim_{n \to \infty} [3 + \frac{8(n+1)}{2n} + \frac{4(n+1)(2n+1)}{6n^2}] \)
\( = \lim_{n \to \infty} [3 + 4(\frac{n+1}{n}) + \frac{2}{3}(\frac{n+1}{n})(\frac{2n+1}{n})] \)
\( = \lim_{n \to \infty} [3 + 4(1 + \frac{1}{n}) + \frac{2}{3}(1 + \frac{1}{n})(2 + \frac{1}{n})] \)
As \( n \to \infty \), the terms with \( \frac{1}{n} \) approach 0.
\( = 3 + 4(1 + 0) + \frac{2}{3}(1 + 0)(2 + 0) \)
\( = 3 + 4(1) + \frac{2}{3}(1)(2) \)
\( = 3 + 4 + \frac{4}{3} \)
\( = 7 + \frac{4}{3} \)
\( = \frac{21+4}{3} \)
\( = \frac{25}{3} \)
The definite integral represents the area under the curve of the function between the given limits.
In simple words: To find the exact value of an integral, we can imagine splitting the area under the curve into many tiny rectangles. We add up the areas of all these rectangles and then imagine making the rectangles infinitely thin. This method gives us the precise area, which is the value of the integral.

🎯 Exam Tip: Remember the standard summation formulas for \( \sum r \), \( \sum r^2 \), and \( \sum r^3 \) as they are crucial when evaluating integrals using the limit of sums method. Pay close attention to the algebraic simplification steps and the limit evaluation.