GSEB Class 12 Maths Solutions Chapter 7 સંકલન Exercise 7.8

Find the value of the following definite integrals in the form of the sum of limits.

Question 1. \( \int_a^bxdx \)
Answer: According to the definition:
\( \int_a^b f(x)dx = \lim_{n \to \infty} h [f(a) + f(a + h) + f(a + 2h) + ... + f(a + (n-1)h)] \)
Here, \( f(x) = x \) and \( h = \frac{b-a}{n} \implies nh = b-a \).
So, \( \int_a^b x dx = \lim_{n \to \infty} h[a + (a + h) + (a + 2h) + ... + (a + (n-1)h)] \)
\( = \lim_{n \to \infty} h [(a + a + a + ... n \text{ વખત}) + h (1 + 2 + ... + (n-1))] \)
\( = \lim_{n \to \infty} h [an + h \frac{(n-1)n}{2}] \) (Since \( \Sigma n = \frac{n(n+1)}{2} \))
\( = \lim_{n \to \infty} [anh + \frac{h^2n^2(1 - \frac{1}{n})}{2}] \)
\( = \lim_{n \to \infty} [a(nh) + \frac{(nh)^2(1 - \frac{1}{n})}{2}] \)
(Since \( nh = b-a \))
\( = a(b-a) + \frac{(b-a)^2}{2} \lim_{n \to \infty} (1 - \frac{1}{n}) \)
(Since \( n \to \infty \implies \frac{1}{n} \to 0 \))
\( = a(b-a) + \frac{(b-a)^2(1-0)}{2} \)
\( = a(b-a) + \frac{(b-a)^2}{2} \)
\( = \frac{2a(b-a) + (b-a)^2}{2} \)
\( = \frac{(b-a)(2a + b - a)}{2} \)
\( = \frac{(b-a)(b+a)}{2} \)
\( = \frac{b^2 - a^2}{2} \)
In simple words: We find the exact value of the integral by using a specific rule. We replace the function and "h" with their values, then simplify the expression using sums of series. After taking the limit as "n" gets very large, we end up with the formula \( \frac{b^2 - a^2}{2} \). This is a basic way to solve definite integrals using limits.

Exam Tip: Remember the formula for the sum of the first \( n-1 \) natural numbers and the limit definition of a definite integral. Be careful with algebraic simplifications.

Question 2. \( \int_0^5(x + 1)dx \)
Answer: Here \( a = 0, b = 5 \) and \( f(x) = x + 1 \).
\( h = \frac{b-a}{n} = \frac{5-0}{n} = \frac{5}{n} \implies nh = 5 \).
\( f(a) = f(0) = 0 + 1 = 1 \)
\( f(a+h) = f(0+h) = f(h) = h+1 \)
\( f(a+2h) = f(0+2h) = f(2h) = 2h+1 \)
...
\( f(a+(n-1)h) = f(0+(n-1)h) = f((n-1)h) = (n-1)h+1 \)
According to the definition:
\( \int_a^b f(x) dx = \lim_{n \to \infty} h[f(a) + f(a + h) + f(a + 2h) + ... + f(a + (n-1)h)] \)
\( \int_0^5 (x+1)dx = \lim_{n \to \infty} h[1 + (h+1) + (2h+1) + ... + ((n-1)h+1)] \)
\( = \lim_{n \to \infty} h[(1+1+1+...n \text{ વખત}) + (h+2h+3h+...+ (n-1)h)] \)
\( = \lim_{n \to \infty} h[n + h(1+2+3+...+(n-1))] \)
\( = \lim_{n \to \infty} h[n + h \frac{(n-1)n}{2}] \) (Since \( \Sigma n = \frac{n(n+1)}{2} \))
\( = \lim_{n \to \infty} [nh + \frac{h^2n^2(1 - \frac{1}{n})}{2}] \)
\( = \lim_{n \to \infty} [nh + \frac{(nh)^2(1 - \frac{1}{n})}{2}] \)
(Since \( nh = 5 \))
\( = 5 + \frac{5^2}{2} \lim_{n \to \infty} (1 - \frac{1}{n}) \)
(Since \( n \to \infty \implies \frac{1}{n} \to 0 \))
\( = 5 + \frac{25}{2}(1-0) \)
\( = 5 + \frac{25}{2} = \frac{10+25}{2} = \frac{35}{2} \)
In simple words: We used the limit definition to find the value of the integral. We put in the values of "a", "b", and "f(x)". Then we used the formula for the sum of natural numbers, and as "n" approached infinity, the term \( \frac{1}{n} \) went to zero, leaving us with \( \frac{35}{2} \).

Exam Tip: For definite integrals of polynomials, you can often check your answer using standard integration techniques. Make sure to correctly identify 'a', 'b', and 'f(x)'.

Question 3. \( \int_2^3x² dx \)
Answer: Here \( a = 2, b = 3 \) and \( f(x) = x^2 \).
\( h = \frac{b-a}{n} = \frac{3-2}{n} = \frac{1}{n} \implies nh = 1 \).
\( f(a) = f(2) = 2^2 = 4 \)
\( f(a+h) = f(2+h) = (2+h)^2 \)
\( f(a+2h) = f(2+2h) = (2+2h)^2 \)
...
\( f(a+(n-1)h) = f(2+(n-1)h) = (2+(n-1)h)^2 \)
According to the definition:
\( \int_a^b f(x) dx = \lim_{n \to \infty} h[f(a) + f(a + h) + f(a + 2h) + ... + f(a + (n-1)h)] \)
\( \int_2^3 x^2 dx = \lim_{n \to \infty} h[4 + (2+h)^2 + (2+2h)^2 + ... + (2+(n-1)h)^2] \)
\( = \lim_{n \to \infty} h[(4+4+4+...n \text{ વખત}) + 4h(1+2+3+...+(n-1)) + h^2(1^2+2^2+...+(n-1)^2)] \)
\( = \lim_{n \to \infty} h[4n + 4h \frac{(n-1)n}{2} + h^2 \frac{(n-1)n(2n-1)}{6}] \) (Using formulas for \( \Sigma n \) and \( \Sigma n^2 \))
\( = \lim_{n \to \infty} [4nh + \frac{4h^2n^2(1 - \frac{1}{n})}{2} + \frac{h^3n^3(1 - \frac{1}{n})(2 - \frac{1}{n})}{6}] \)
\( = \lim_{n \to \infty} [4(nh) + \frac{4(nh)^2(1 - \frac{1}{n})}{2} + \frac{(nh)^3(1 - \frac{1}{n})(2 - \frac{1}{n})}{6}] \)
(Since \( nh = 1 \))
\( = 4(1) + \frac{4(1)^2}{2} \lim_{n \to \infty} (1 - \frac{1}{n}) + \frac{(1)^3}{6} \lim_{n \to \infty} (1 - \frac{1}{n})(2 - \frac{1}{n}) \)
(Since \( n \to \infty \implies \frac{1}{n} \to 0 \))
\( = 4+2(1-0) + \frac{1}{6}(1-0)(2-0) \)
\( = 4+2 + \frac{2}{6} \)
\( = 6 + \frac{1}{3} = \frac{18+1}{3} = \frac{19}{3} \)
In simple words: For this integral, we substitute 'a' as 2, 'b' as 3, and \( f(x) \) as \( x^2 \). We then use the limit of sums, applying the sum formulas for natural numbers and their squares. By letting 'n' approach infinity, terms with \( \frac{1}{n} \) become zero, which gives us the final answer \( \frac{19}{3} \).

Exam Tip: When dealing with \( x^2 \) in the integral, remember to use the formula for the sum of squares, \( \Sigma n^2 = \frac{n(n+1)(2n+1)}{6} \). Careful handling of \( h^2n^2 \) as \( (nh)^2 \) is essential.

Question 4. \( \int_1^4(x² - x)dx \)
Answer: Here \( a = 1, b = 4 \) and \( f(x) = x^2 - x \).
\( h = \frac{b-a}{n} = \frac{4-1}{n} = \frac{3}{n} \implies nh = 3 \).
\( f(a) = f(1) = (1)^2 - 1 = 0 \)
\( f(a+h) = f(1+h) = (1+h)^2 - (1+h) = 1+2h+h^2 - 1 - h = h+h^2 \)
\( f(a+2h) = f(1+2h) = (1+2h)^2 - (1+2h) = 1+4h+4h^2 - 1 - 2h = 2h+4h^2 \)
\( f(a+3h) = f(1+3h) = (1+3h)^2 - (1+3h) = 1+6h+9h^2 - 1 - 3h = 3h+9h^2 \)
...
\( f(a+(n-1)h) = f(1+(n-1)h) = (1+(n-1)h)^2 - (1+(n-1)h) = (n-1)h + (n-1)^2h^2 \)
According to the definition:
\( \int_a^b f(x) dx = \lim_{n \to \infty} h[f(a) + f(a + h) + f(a + 2h) + f(a + 3h) + ... + f(a + (n-1)h)] \)
\( \int_1^4 (x^2-x)dx = \lim_{n \to \infty} h[0 + (h+h^2) + (2h+4h^2) + (3h+9h^2) + ... + ((n-1)h + (n-1)^2h^2)] \)
\( = \lim_{n \to \infty} h[h(1+2+3+...+(n-1)) + h^2(1^2+2^2+3^2+...+(n-1)^2)] \)
\( = \lim_{n \to \infty} h[h \frac{(n-1)n}{2} + h^2 \frac{(n-1)n(2n-1)}{6}] \) (Using formulas for \( \Sigma n \) and \( \Sigma n^2 \))
\( = \lim_{n \to \infty} [nh \frac{(nh)(1 - \frac{1}{n})}{2} + (nh)^2 \frac{(1 - \frac{1}{n})(2 - \frac{1}{n})}{6}] \)
(Since \( nh = 3 \))
\( = \lim_{n \to \infty} [3 \frac{3(1 - \frac{1}{n})}{2} + 3^2 \frac{(1 - \frac{1}{n})(2 - \frac{1}{n})}{6}] \)
(Since \( n \to \infty \implies \frac{1}{n} \to 0 \))
\( = \frac{9}{2}(1-0) + \frac{9}{6}(1-0)(2-0) \)
\( = \frac{9}{2} + \frac{9}{6}(2) \)
\( = \frac{9}{2} + 3 = \frac{9+6}{2} = \frac{15}{2} \)
In simple words: For this integral, we set 'a' to 1, 'b' to 4, and \( f(x) \) to \( x^2 - x \). We then use the limit definition, substituting the function values at different points. By applying the sum formulas for 'n' and \( n^2 \), and taking the limit as 'n' approaches infinity, the terms with \( \frac{1}{n} \) vanish, leading to the final result of \( \frac{15}{2} \).

Exam Tip: Expanding \( (1+kh)^2 \) carefully and grouping terms with h and \( h^2 \) is crucial. Remember to substitute \( nh \) with its value and take the limit correctly.

Question 5. \( \int_{-1}^1e^x dx \)
Answer: Here \( a = -1, b = 1 \) and \( f(x) = e^x \).
\( h = \frac{b-a}{n} = \frac{1-(-1)}{n} = \frac{2}{n} \implies nh = 2 \).
\( f(a) = f(-1) = e^{-1} \)
\( f(a+h) = f(-1+h) = e^{-1+h} \)
\( f(a+2h) = f(-1+2h) = e^{-1+2h} \)
\( f(a+3h) = f(-1+3h) = e^{-1+3h} \)
...
\( f(a+(n-1)h) = f(-1+(n-1)h) = e^{-1+(n-1)h} \)
According to the definition:
\( \int_a^b f(x)dx = \lim_{n \to \infty} h[f(a) + f(a + h) + f(a + 2h) + f(a + 3h) + ... + f(a + (n-1)h)] \)
\( \int_{-1}^1 e^x dx = \lim_{n \to \infty} h[e^{-1} + e^{-1+h} + e^{-1+2h} + e^{-1+3h} + ... + e^{-1+(n-1)h}] \)
\( = \lim_{h \to 0} h[e^{-1}(1 + e^h + e^{2h} + e^{3h} + ... + e^{(n-1)h})] \) (Since \( n \to \infty \implies h \to 0 \))
This is a geometric series with first term \( A = 1 \), common ratio \( R = e^h \), and number of terms \( N = n \).
The sum of the series is \( S_n = A \frac{(R^N - 1)}{(R - 1)} = 1 \cdot \frac{((e^h)^n - 1)}{(e^h - 1)} = \frac{(e^{nh} - 1)}{(e^h - 1)} \).
\( = \lim_{h \to 0} h \cdot e^{-1} \frac{(e^{nh} - 1)}{(e^h - 1)} \)
\( = e^{-1} (e^{nh} - 1) \lim_{h \to 0} \frac{h}{(e^h - 1)} \)
We know that \( \lim_{x \to 0} \frac{x}{e^x - 1} = \frac{1}{\log_e e} = 1 \).
(Since \( nh = 2 \))
\( = e^{-1} (e^2 - 1) \cdot 1 \)
\( = \frac{1}{e}(e^2 - 1) \)
\( = e - \frac{1}{e} \)
In simple words: We find the integral of \( e^x \) from -1 to 1 using the limit of sums. We identify the terms as a geometric series and use its sum formula. After taking the limit as 'h' goes to zero and using the fact that \( \lim_{x \to 0} \frac{x}{e^x - 1} = 1 \), the expression simplifies to \( e - \frac{1}{e} \).

Exam Tip: Recognize the exponential series as a geometric progression. The limit \( \lim_{x \to 0} \frac{x}{e^x - 1} = 1 \) is a standard result that simplifies such problems greatly.

Question 6. \( \int_0^4(x + e^{2x})dx \)
Answer: Here \( a = 0, b = 4 \) and \( f(x) = x + e^{2x} \).
\( h = \frac{b-a}{n} = \frac{4-0}{n} = \frac{4}{n} \implies nh = 4 \).
\( f(a) = f(0) = 0 + e^0 = 1 \)
\( f(a+h) = f(0+h) = f(h) = h + e^{2h} \)
\( f(a+2h) = f(0+2h) = f(2h) = 2h + e^{4h} \)
\( f(a+3h) = f(0+3h) = f(3h) = 3h + e^{6h} \)
...
\( f(a+(n-1)h) = f(0+(n-1)h) = f((n-1)h) = (n-1)h + e^{2(n-1)h} \)
According to the definition:
\( \int_a^b f(x)dx = \lim_{n \to \infty} h[f(a) + f(a + h) + f(a + 2h) + ... + f(a + (n-1)h)] \)
\( \int_0^4 (x+e^{2x})dx = \lim_{n \to \infty} h[1 + (h+e^{2h}) + (2h+e^{4h}) + (3h+e^{6h}) + ... + ((n-1)h + e^{2(n-1)h})] \)
\( = \lim_{n \to \infty} h[(1+e^{2h}+e^{4h}+e^{6h}+...+e^{2(n-1)h}) + (h+2h+3h+...+(n-1)h)] \)
\( = \lim_{h \to 0} h[(e^0+e^{2h}+e^{4h}+e^{6h}+...+e^{2(n-1)h}) + h(1+2+3+...+(n-1))] \)
\( = \lim_{h \to 0} h[\frac{e^{2nh}-1}{e^{2h}-1} + h\frac{(n-1)n}{2}] \) (Using sum formulas for geometric and arithmetic series)
\( = \lim_{h \to 0} [\frac{h(e^{2nh}-1)}{e^{2h}-1} + \frac{h^2n^2(1-\frac{1}{n})}{2}] \)
\( = \lim_{h \to 0} [\frac{h}{e^{2h}-1} \cdot (e^{2nh}-1) + \frac{(nh)^2(1-\frac{1}{n})}{2}] \)
(Since \( n \to \infty \implies h \to 0 \) and \( nh = 4 \))
\( = \frac{1}{\log_e e^2} (e^{2(4)}-1) + \frac{4^2}{2} \lim_{n \to \infty} (1-\frac{1}{n}) \)
\( = \frac{1}{2} (e^8-1) + \frac{16}{2} (1-0) \)
\( = \frac{e^8-1}{2} + 8 \)
\( = \frac{e^8-1+16}{2} \)
\( = \frac{e^8+15}{2} \)
In simple words: We separate the integral into two parts, one for "x" and one for \( e^{2x} \). For "x", we use the arithmetic series sum, and for \( e^{2x} \), we use the geometric series sum. After applying the limit as 'n' approaches infinity (or 'h' approaches zero) and substituting \( nh = 4 \), we arrive at the final answer \( \frac{e^8+15}{2} \).

Exam Tip: When the integrand is a sum of functions, you can handle each part separately in the limit definition. Remember to correctly identify the common ratio and the limit for \( \frac{h}{e^{kh}-1} \).

Friends, the following trigonometric functions can be obtained with the help of the definition of definite integrals.

Question 1. \( \int_0^\pi \sin x dx \)
Answer: Here \( a = 0, b = \pi \) and \( f(x) = \sin x \).
\( h = \frac{b-a}{n} = \frac{\pi-0}{n} = \frac{\pi}{n} \implies nh = \pi \).
\( f(a) = f(0) = \sin 0 = 0 \)
\( f(a+h) = f(0+h) = \sin h \)
\( f(a+2h) = f(0+2h) = \sin 2h \)
\( f(a+3h) = f(0+3h) = \sin 3h \)
...
\( f(a+(n-1)h) = f(0+(n-1)h) = \sin (n-1)h \)
According to the definition:
\( \int_a^b f(x)dx = \lim_{n \to \infty} h[f(a) + f(a + h) + f(a + 2h) + f(a + 3h) + ... + f(a + (n-1)h)] \)
\( \int_0^\pi \sin x dx = \lim_{h \to 0} h[0 + \sin h + \sin 2h + \sin 3h + ... + \sin(n-1)h] \)
We use the formula for the sum of sines: \( \sin \alpha + \sin(\alpha+\beta) + ... + \sin(\alpha+(n-1)\beta) = \frac{\sin \frac{n\beta}{2}}{\sin \frac{\beta}{2}} \sin(\alpha+\frac{(n-1)\beta}{2}) \).
Here, \( \alpha = h, \beta = h \).
Sum \( S = \frac{\sin \frac{nh}{2}}{\sin \frac{h}{2}} \sin(h+\frac{(n-1)h}{2}) = \frac{\sin \frac{nh}{2}}{\sin \frac{h}{2}} \sin(\frac{h+nh-h}{2}) = \frac{\sin \frac{nh}{2}}{\sin \frac{h}{2}} \sin(\frac{nh}{2}) \).
Since \( nh = \pi \), this becomes \( \frac{\sin \frac{\pi}{2}}{\sin \frac{h}{2}} \sin(\frac{\pi}{2}) = \frac{1}{\sin \frac{h}{2}} \cdot 1 = \frac{1}{\sin \frac{h}{2}} \).
Now, \( \int_0^\pi \sin x dx = \lim_{h \to 0} h \cdot \frac{\sin \frac{nh}{2}}{\sin \frac{h}{2}} \sin(\frac{nh}{2}) \)
\( = \lim_{h \to 0} h \cdot \frac{\sin \frac{\pi}{2}}{\sin \frac{h}{2}} \sin(\frac{\pi}{2}) \)
\( = \lim_{h \to 0} h \cdot \frac{1}{\sin \frac{h}{2}} \)
\( = \lim_{h \to 0} \frac{2 \cdot \frac{h}{2}}{\sin \frac{h}{2}} \)
\( = 2 \lim_{h \to 0} \frac{\frac{h}{2}}{\sin \frac{h}{2}} \)
We know that \( \lim_{\theta \to 0} \frac{\theta}{\sin \theta} = 1 \).
\( = 2 \cdot 1 = 2 \).
In simple words: We calculate the definite integral of \( \sin x \) from 0 to \( \pi \) using the limit of sums. We identify the series of sine terms and apply its specific summation formula. By substituting \( nh = \pi \) and using the known limit for \( \frac{\theta}{\sin \theta} \), the entire expression simplifies to 2.

Exam Tip: For integrals of trigonometric functions, recalling the sum of sine or cosine series is key. Pay close attention to the arguments of the sine functions in the summation formula and the limit evaluation.