RBSE Solutions Class 12 Maths Chapter 10 Definite Integral Exercise 10.1

Find the value of definite integrals, as limit of sum (by first principle)

Question 1. \( \int_{3}^{5} (x-2)dx \)
Answer: To find the definite integral \( \int_{3}^{5} (x-2)dx \) as the limit of a sum, we use the formula: \( \int_{a}^{b} f(x)dx = \lim_{h \to 0} h[f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)] \) where \( nh = b-a \). Here, \( a=3 \), \( b=5 \), and \( f(x)=x-2 \). First, find \( nh \): \( nh = 5-3 = 2 \). Now, let's find the terms of the sum: \( f(a) = f(3) = 3-2 = 1 \) \( f(a+h) = f(3+h) = (3+h)-2 = 1+h \) \( f(a+2h) = f(3+2h)-2 = (3+2h)-2 = 1+2h \) ... \( f(a+(n-1)h) = f(3+(n-1)h)-2 = (3+(n-1)h)-2 = 1+(n-1)h \) Substitute these into the formula: \( \int_{3}^{5} (x-2)dx = \lim_{h \to 0} h[f(3+h) + f(3+2h) + ... + f(3+nh)] \) Note: The source uses `f(3+nh)` which implies a sum up to \( n \) terms, not \( n-1 \). We will follow the source's summation up to \( n \) terms. \( \implies \lim_{h \to 0} h[(3+h-2) + (3+2h-2) + (3+3h-2) + ... + (3+nh-2)] \) \( \implies \lim_{h \to 0} h[(1+h) + (1+2h) + (1+3h) + ... + (1+nh)] \) Group the 1s and the terms with h: \( \implies \lim_{h \to 0} h[(1+1+1+...+1) + (h+2h+3h+...+nh)] \) There are \( n \) terms of 1, and \( h \) is a common factor in the second part: \( \implies \lim_{h \to 0} h[n + h(1+2+3+...+n)] \) The sum of the first \( n \) natural numbers is \( \frac{n(n+1)}{2} \): \( \implies \lim_{h \to 0} h[n + h \frac{n(n+1)}{2}] \) Multiply by \( h \): \( \implies \lim_{h \to 0} [nh + h^2 \frac{n(n+1)}{2}] \) We know \( nh = 2 \). Replace \( n \) with \( \frac{2}{h} \): \( \implies \lim_{h \to 0} [nh + h^2 \frac{\frac{2}{h}(\frac{2}{h}+1)}{2}] \) \( \implies \lim_{h \to 0} [nh + h^2 \frac{2(2+h)}{2h^2}] \) \( \implies \lim_{h \to 0} [nh + \frac{2(2+h)}{2}] \) Now, substitute \( nh=2 \) and let \( h \to 0 \): \( \implies 2 + \frac{2(2+0)}{2} \) \( \implies 2 + \frac{2 \times 2}{2} \) \( \implies 2 + 2 = 4 \) Using the fundamental theorem of calculus, \( \int_{3}^{5} (x-2)dx = \left[ \frac{x^2}{2} - 2x \right]_{3}^{5} = \left( \frac{5^2}{2} - 2(5) \right) - \left( \frac{3^2}{2} - 2(3) \right) = \left( \frac{25}{2} - 10 \right) - \left( \frac{9}{2} - 6 \right) = \left( \frac{25-20}{2} \right) - \left( \frac{9-12}{2} \right) = \frac{5}{2} - \left( -\frac{3}{2} \right) = \frac{5+3}{2} = \frac{8}{2} = 4 \). This confirms our result.
In simple words: To solve this, we used a special method that adds up many tiny rectangles under the curve of the function. As these rectangles become infinitely thin, their sum gives us the exact area, which is the value of the definite integral. The final answer is 4.

🎯 Exam Tip: Remember to correctly identify \( a \), \( b \), \( f(x) \), and \( nh = b-a \) before expanding the sum. Be careful with algebraic simplifications and limits.

Question 2. \( \int_{0}^{2} x^{2}dx \)
Answer: To find the definite integral \( \int_{0}^{2} x^{2}dx \) as the limit of a sum, we use the formula: \( \int_{a}^{b} f(x)dx = \lim_{h \to 0} h[f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)] \) where \( nh = b-a \). Here, \( a=0 \), \( b=2 \), and \( f(x)=x^2 \). First, find \( nh \): \( nh = 2-0 = 2 \). Now, let's find the terms of the sum: \( f(a) = f(0) = 0^2 = 0 \) \( f(a+h) = f(h) = h^2 \) \( f(a+2h) = f(2h) = (2h)^2 \) ... \( f(a+(n-1)h) = f((n-1)h) = ((n-1)h)^2 \) Substitute these into the formula: \( \int_{0}^{2} x^{2}dx = \lim_{h \to 0} h[f(0) + f(h) + f(2h) + ... + f((n-1)h)] \) \( \implies \lim_{h \to 0} h[0^2 + h^2 + (2h)^2 + ... + ((n-1)h)^2] \) \( \implies \lim_{h \to 0} h[h^2 (1^2 + 2^2 + ... + (n-1)^2)] \) The sum of squares of the first \( (n-1) \) natural numbers is \( \frac{(n-1)n(2n-1)}{6} \). The problem's solution uses \( n \) up to \( n \) term. Let's adapt based on source's solution steps, which seems to use \( \frac{n(n+1)(2n+1)}{6} \) (sum of first \( n \) squares). Let's follow the solution given (which seems to simplify directly to a form where \( nh \) is substituted). The source arrives at: \( \lim_{h \to 0} \frac{1}{3} (b-a)^3 \) which is the result for \( \int x^2 dx \) from \( a \) to \( b \). The integral is \( \frac{b^3-a^3}{3} \). So, \( \int_{0}^{2} x^{2}dx = \frac{2^3-0^3}{3} = \frac{8}{3} \). The detailed steps provided seem to be a general derivation for the formula of \( \int_{a}^{b} x^2 dx \). Let's use the standard formula for the sum of squares up to \( (n-1) \): \( \sum_{k=1}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6} \) So, \( \lim_{h \to 0} h[h^2 \frac{(n-1)n(2n-1)}{6}] \) \( \implies \lim_{h \to 0} \frac{h^3}{6} (n-1)n(2n-1) \) Rearrange to use \( nh \): \( \implies \lim_{h \to 0} \frac{1}{6} (nh-h)(nh)(2nh-h) \) Substitute \( nh = 2 \) and let \( h \to 0 \): \( \implies \frac{1}{6} (2-0)(2)(2(2)-0) \) \( \implies \frac{1}{6} (2)(2)(4) \) \( \implies \frac{16}{6} = \frac{8}{3} \) This is the expected result. The enriching sentence can highlight the power rule for integration.
In simple words: This problem involves finding the area under the curve of \( x^2 \) from 0 to 2 using the method of summing many small rectangles. We use a specific formula for the sum of squares, and as the width of the rectangles gets smaller and smaller, the sum approaches the exact area. The result is \( \frac{8}{3} \).

🎯 Exam Tip: When dealing with \( x^2 \) integrals as a limit of sum, remember the formula for the sum of squares \( \sum_{k=1}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6} \) and make sure to correctly substitute \( nh \) values and take the limit.

Question 3. \( \int_{1}^{3} (x^{2}+5x)dx \)
Answer: To find the definite integral \( \int_{1}^{3} (x^{2}+5x)dx \) as the limit of a sum, we can split it into two integrals: \( \int_{1}^{3} (x^{2}+5x)dx = \int_{1}^{3} x^{2}dx + \int_{1}^{3} 5xdx \) Let \( I_1 = \int_{1}^{3} x^{2}dx \) and \( I_2 = \int_{1}^{3} 5xdx \). For \( I_1 = \int_{1}^{3} x^{2}dx \): Here, \( a=1 \), \( b=3 \), \( f(x)=x^2 \). So \( nh = b-a = 3-1 = 2 \). \( I_1 = \lim_{h \to 0} h[f(a) + f(a+h) + ... + f(a+(n-1)h)] \) \( \implies \lim_{h \to 0} h[1^2 + (1+h)^2 + (1+2h)^2 + ... + (1+(n-1)h)^2] \) \( \implies \lim_{h \to 0} h[ (1 + 1^2h + 2\cdot1\cdot h) + (1 + (2h)^2 + 2\cdot1\cdot2h) + ... + (1 + ((n-1)h)^2 + 2\cdot1\cdot(n-1)h) ] \) Expand each term: \( (1+kh)^2 = 1 + 2kh + k^2h^2 \). Sum over \( k=0 \) to \( n-1 \): \( \implies \lim_{h \to 0} h \left[ \sum_{k=0}^{n-1} 1 + \sum_{k=0}^{n-1} 2kh + \sum_{k=0}^{n-1} k^2h^2 \right] \) \( \implies \lim_{h \to 0} h \left[ n + 2h \sum_{k=0}^{n-1} k + h^2 \sum_{k=0}^{n-1} k^2 \right] \) Using \( \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2} \) and \( \sum_{k=0}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6} \): \( \implies \lim_{h \to 0} h \left[ n + 2h \frac{(n-1)n}{2} + h^2 \frac{(n-1)n(2n-1)}{6} \right] \) Multiply by \( h \): \( \implies \lim_{h \to 0} \left[ nh + h^2(n-1)n + h^3 \frac{(n-1)n(2n-1)}{6} \right] \) Substitute \( nh=2 \): \( \implies \lim_{h \to 0} \left[ nh + (nh-h)nh + \frac{(nh-h)nh(2nh-h)}{6} \right] \) \( \implies \lim_{h \to 0} \left[ 2 + (2-h)2 + \frac{(2-h)2(4-h)}{6} \right] \) Let \( h \to 0 \): \( \implies 2 + (2-0)2 + \frac{(2-0)2(4-0)}{6} \) \( \implies 2 + 4 + \frac{2 \cdot 2 \cdot 4}{6} \) \( \implies 6 + \frac{16}{6} = 6 + \frac{8}{3} = \frac{18+8}{3} = \frac{26}{3} \) So, \( I_1 = \frac{26}{3} \). For \( I_2 = \int_{1}^{3} 5xdx \): Here, \( a=1 \), \( b=3 \), \( f(x)=5x \). So \( nh = 2 \). \( I_2 = \lim_{h \to 0} h[5(1) + 5(1+h) + ... + 5(1+(n-1)h)] \) \( \implies \lim_{h \to 0} 5h[1 + (1+h) + (1+2h) + ... + (1+(n-1)h)] \) \( \implies \lim_{h \to 0} 5h[(1+1+...+1) + (h+2h+...+(n-1)h)] \) \( \implies \lim_{h \to 0} 5h[n + h(1+2+...+(n-1))] \) \( \implies \lim_{h \to 0} 5h[n + h \frac{(n-1)n}{2}] \) Multiply by \( h \): \( \implies \lim_{h \to 0} 5[nh + h^2 \frac{(n-1)n}{2}] \) Substitute \( nh=2 \): \( \implies \lim_{h \to 0} 5[nh + \frac{(nh-h)nh}{2}] \) \( \implies \lim_{h \to 0} 5[2 + \frac{(2-h)2}{2}] \) Let \( h \to 0 \): \( \implies 5[2 + \frac{(2-0)2}{2}] \) \( \implies 5[2 + 2] = 5 \times 4 = 20 \) So, \( I_2 = 20 \). Finally, \( \int_{1}^{3} (x^{2}+5x)dx = I_1 + I_2 = \frac{26}{3} + 20 = \frac{26+60}{3} = \frac{86}{3} \). Using the fundamental theorem of calculus, \( \int_{1}^{3} (x^2+5x)dx = \left[ \frac{x^3}{3} + \frac{5x^2}{2} \right]_{1}^{3} = \left( \frac{3^3}{3} + \frac{5 \cdot 3^2}{2} \right) - \left( \frac{1^3}{3} + \frac{5 \cdot 1^2}{2} \right) = \left( 9 + \frac{45}{2} \right) - \left( \frac{1}{3} + \frac{5}{2} \right) = \left( \frac{18+45}{2} \right) - \left( \frac{2+15}{6} \right) = \frac{63}{2} - \frac{17}{6} = \frac{189-17}{6} = \frac{172}{6} = \frac{86}{3} \). The calculation matches the result.
In simple words: We broke this integral into two parts, one for \( x^2 \) and one for \( 5x \), then solved each part separately using the limit of sums method. After finding the value for each part, we added them together to get the total area. This method helps us find the exact area under a combined curve.

🎯 Exam Tip: When the integrand is a sum of functions, you can often split the integral into a sum of integrals. Remember the formulas for sums of natural numbers and squares, and always substitute \( nh \) carefully.

Question 4. \( \int_{a}^{b} e^{x}dx \)
Answer: To find the definite integral \( \int_{a}^{b} e^{x}dx \) as the limit of a sum, we use the formula: \( \int_{a}^{b} f(x)dx = \lim_{h \to 0} h[f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)] \) where \( nh = b-a \). Here, \( f(x)=e^x \). So the sum is: \( \lim_{h \to 0} h[e^a + e^{a+h} + e^{a+2h} + ... + e^{a+(n-1)h}] \) We can factor out \( e^a \): \( \implies \lim_{h \to 0} h e^a[1 + e^h + e^{2h} + ... + e^{(n-1)h}] \) The terms inside the bracket form a geometric progression with first term 1, common ratio \( e^h \), and \( n \) terms. The sum of a geometric series is \( S_n = \frac{r^n-1}{r-1} \). Here, \( r = e^h \), so the sum is \( \frac{(e^h)^n-1}{e^h-1} = \frac{e^{nh}-1}{e^h-1} \). Substitute this back into the limit: \( \implies \lim_{h \to 0} h e^a \frac{e^{nh}-1}{e^h-1} \) Rearrange the terms: \( \implies e^a (e^{nh}-1) \lim_{h \to 0} \frac{h}{e^h-1} \) We know a standard limit: \( \lim_{x \to 0} \frac{x}{e^x-1} = 1 \). So, \( \lim_{h \to 0} \frac{h}{e^h-1} = 1 \). Substitute \( nh = b-a \): \( \implies e^a (e^{b-a}-1) \times 1 \) \( \implies e^a e^{b-a} - e^a \) \( \implies e^{a+b-a} - e^a \) \( \implies e^b - e^a \) This result is consistent with the fundamental theorem of calculus: \( \int_{a}^{b} e^{x}dx = [e^x]_{a}^{b} = e^b - e^a \). The formula works universally.
In simple words: For this problem, we used the sum of a geometric series to find the area under the exponential curve \( e^x \) between points \( a \) and \( b \). As the width of the small rectangles goes to zero, the sum becomes exactly \( e^b - e^a \). This shows how exponential functions integrate in a straightforward way.

🎯 Exam Tip: When dealing with exponential functions, remember that the sum will form a geometric progression. Carefully apply the sum formula and the standard limit for \( \frac{x}{e^x-1} \) to arrive at the correct result.

Question 5. \( \int_{0}^{2} (x+4)dx \)
Answer: To find the definite integral \( \int_{0}^{2} (x+4)dx \) as the limit of a sum, we use the formula: \( \int_{a}^{b} f(x)dx = \lim_{h \to 0} h[f(a) + f(a+h) + f(a+2h) + ... + f(a+(n-1)h)] \) where \( nh = b-a \). Here, \( a=0 \), \( b=2 \), and \( f(x)=x+4 \). First, find \( nh \): \( nh = 2-0 = 2 \). Now, let's find the terms of the sum: \( f(a) = f(0) = 0+4 = 4 \) \( f(a+h) = f(h) = h+4 \) \( f(a+2h) = f(2h) = 2h+4 \) ... \( f(a+(n-1)h) = f((n-1)h) = (n-1)h+4 \) Substitute these into the formula: \( \int_{0}^{2} (x+4)dx = \lim_{h \to 0} h[f(0) + f(h) + f(2h) + ... + f((n-1)h)] \) \( \implies \lim_{h \to 0} h[(0+4) + (h+4) + (2h+4) + ... + ((n-1)h+4)] \) Group the constant terms and the terms with \( h \): \( \implies \lim_{h \to 0} h[(4+4+4+...+4) + (0+h+2h+...+(n-1)h)] \) There are \( n \) terms of 4. Factor out \( h \) from the second part: \( \implies \lim_{h \to 0} h[4n + h(0+1+2+...+(n-1))] \) The sum of the first \( (n-1) \) natural numbers is \( \frac{(n-1)n}{2} \): \( \implies \lim_{h \to 0} h[4n + h \frac{(n-1)n}{2}] \) Multiply by \( h \): \( \implies \lim_{h \to 0} [4nh + h^2 \frac{(n-1)n}{2}] \) Substitute \( nh = 2 \): \( \implies \lim_{h \to 0} [4nh + \frac{(nh-h)nh}{2}] \) \( \implies \lim_{h \to 0} [4(2) + \frac{(2-h)2}{2}] \) Let \( h \to 0 \): \( \implies 8 + \frac{(2-0)2}{2} \) \( \implies 8 + \frac{4}{2} \) \( \implies 8 + 2 = 10 \) This result matches the fundamental theorem of calculus: \( \int_{0}^{2} (x+4)dx = \left[ \frac{x^2}{2} + 4x \right]_{0}^{2} = \left( \frac{2^2}{2} + 4(2) \right) - \left( \frac{0^2}{2} + 4(0) \right) = (2+8) - (0) = 10 \). This method confirms the area.
In simple words: We found the area under the line \( y=x+4 \) from 0 to 2 by adding up infinitely thin rectangles. By using the limit of a sum and some simple algebra, we found that the total area is 10. This is a basic example of how integration helps find areas.

🎯 Exam Tip: When integrating linear functions using the limit of sums, remember to separate the constant terms from the terms involving \( h \). The sum of an arithmetic progression will be useful here.

Question 6. \( \int_{1}^{2} (2x^{2}+5)dx \)
Answer: To find the definite integral \( \int_{1}^{2} (2x^{2}+5)dx \) as the limit of a sum, we can split it into two integrals: \( \int_{1}^{2} (2x^{2}+5)dx = \int_{1}^{2} 2x^{2}dx + \int_{1}^{2} 5dx \) Let \( I_1 = \int_{1}^{2} 2x^{2}dx \) and \( I_2 = \int_{1}^{2} 5dx \). For \( I_1 = \int_{1}^{2} 2x^{2}dx \): Here, \( a=1 \), \( b=2 \), \( f(x)=2x^2 \). So \( nh = b-a = 2-1 = 1 \). \( I_1 = \lim_{h \to 0} h[2(1)^2 + 2(1+h)^2 + ... + 2(1+(n-1)h)^2] \) \( \implies \lim_{h \to 0} 2h[1^2 + (1+h)^2 + ... + (1+(n-1)h)^2] \) Expand each term: \( (1+kh)^2 = 1 + 2kh + k^2h^2 \). Sum over \( k=0 \) to \( n-1 \): \( \implies \lim_{h \to 0} 2h \left[ \sum_{k=0}^{n-1} 1 + \sum_{k=0}^{n-1} 2kh + \sum_{k=0}^{n-1} k^2h^2 \right] \) \( \implies \lim_{h \to 0} 2h \left[ n + 2h \frac{(n-1)n}{2} + h^2 \frac{(n-1)n(2n-1)}{6} \right] \) Multiply by \( 2h \): \( \implies \lim_{h \to 0} \left[ 2nh + 2h^2(n-1)n + 2h^3 \frac{(n-1)n(2n-1)}{6} \right] \) Substitute \( nh=1 \): \( \implies \lim_{h \to 0} \left[ 2nh + 2(nh-h)nh + \frac{2(nh-h)nh(2nh-h)}{6} \right] \) \( \implies \lim_{h \to 0} \left[ 2(1) + 2(1-h)(1) + \frac{2(1-h)(1)(2(1)-h)}{6} \right] \) Let \( h \to 0 \): \( \implies 2 + 2(1-0)(1) + \frac{2(1-0)(1)(2-0)}{6} \) \( \implies 2 + 2 + \frac{2 \cdot 1 \cdot 1 \cdot 2}{6} \) \( \implies 4 + \frac{4}{6} = 4 + \frac{2}{3} = \frac{12+2}{3} = \frac{14}{3} \) So, \( I_1 = \frac{14}{3} \). For \( I_2 = \int_{1}^{2} 5dx \): Here, \( a=1 \), \( b=2 \), \( f(x)=5 \). So \( nh = 1 \). \( I_2 = \lim_{h \to 0} h[f(1) + f(1+h) + ... + f(1+(n-1)h)] \) \( \implies \lim_{h \to 0} h[5 + 5 + ... + 5] \) (n times) \( \implies \lim_{h \to 0} h[5n] \) \( \implies \lim_{h \to 0} 5(nh) \) Substitute \( nh=1 \): \( \implies 5(1) = 5 \) So, \( I_2 = 5 \). Finally, \( \int_{1}^{2} (2x^{2}+5)dx = I_1 + I_2 = \frac{14}{3} + 5 = \frac{14+15}{3} = \frac{29}{3} \). Using the fundamental theorem of calculus, \( \int_{1}^{2} (2x^2+5)dx = \left[ \frac{2x^3}{3} + 5x \right]_{1}^{2} = \left( \frac{2(2)^3}{3} + 5(2) \right) - \left( \frac{2(1)^3}{3} + 5(1) \right) = \left( \frac{16}{3} + 10 \right) - \left( \frac{2}{3} + 5 \right) = \left( \frac{16+30}{3} \right) - \left( \frac{2+15}{3} \right) = \frac{46}{3} - \frac{17}{3} = \frac{29}{3} \). The calculation is correct.
In simple words: This problem asks us to find the area under the curve of \( 2x^2+5 \) between 1 and 2 by adding up many tiny rectangles. We split the problem into two parts, one for \( 2x^2 \) and one for 5. After calculating each part using the limit of sums, we combined them to get the final area of \( \frac{29}{3} \).

🎯 Exam Tip: Always break down complex integrands into simpler parts. Pay close attention to the constants in the function and the limits of integration. Ensure to correctly apply the sum formulas for \( k \) and \( k^2 \).