GSEB Class 12 Maths Solutions Chapter 7 Integrals Exercise 7.8

Evaluate The Following Definite Integrals As Limits Of The Sum:

Question 1. \( \int_{a}^{b} x dx \)
Answer: The general formula for a definite integral as a limit of sum is given by: \[ \int f(x)dx = \lim_{h \to 0} h[f(a) + f(a + h) + ... + f(a+n-1h)] \] where \( h = \frac{b-a}{n} \).
Here, for \( \int_{a}^{b} x dx \), we have \( f(x) = x \).
So, we get the following values: \( f(a) = a \) \( f(a + h) = a + h \) \( f(a + 2h) = a + 2h \) ... \( f(a + n-1h) = a + (n - 1)h \)
Now, we substitute these into the formula: \[ \int_{a}^{b} x dx = \lim_{h \to 0} h[a + (a + h)+(a + 2h) + ... + (a + n - 1)h] \] \[ = \lim_{h \to 0} h[na + h(1+2+...+n-1)] \] We use the sum of the first \( n-1 \) natural numbers formula: \( 1+2+...+n-1 = \frac{(n-1)n}{2} \). \[ = \lim_{h \to 0} h \left[na + h \frac{(n-1)n}{2} \right] \] \[ = \lim_{h \to 0} \left[nha + h^2 \frac{n(n-1)}{2} \right] \] \[ = \lim_{h \to 0} \left[nha + \frac{nh(nh-h)}{2} \right] \] We know that \( nh = b - a \). Substitute this into the expression: \[ = \lim_{h \to 0} \left[(b-a)a + \frac{(b-a)(b-a-h)}{2} \right] \] As \( h \to 0 \), the expression becomes: \[ = (b-a)a + \frac{(b-a)(b-a-0)}{2} \] \[ = (b-a)a + \frac{(b-a)^2}{2} \] \[ = (b-a) \left( a + \frac{b-a}{2} \right) \] \[ = (b-a) \left( \frac{2a + b-a}{2} \right) \] \[ = (b-a) \left( \frac{b+a}{2} \right) \] \[ = \frac{b^2 - a^2}{2} \] Therefore, the definite integral of \( x \) from \( a \) to \( b \) is \( \frac{b^2 - a^2}{2} \).
In simple words: Whenever you evaluate an integral using the limit of sums, you divide the area under the curve into many tiny rectangles. Summing the areas of these rectangles and taking a limit as their width becomes extremely small helps to find the exact area.

Exam Tip: Remember the core formula for the limit of a sum and carefully substitute \( f(x) \) and \( h = \frac{b-a}{n} \), then simplify the sum of series correctly.

Question 2. \( \int_{5}^{0} (x + 1) dx \)
Answer: We need to evaluate the definite integral \( \int_{5}^{0} (x + 1) dx \) as a limit of the sum. Comparing this with \( \int_{a}^{b} f(x) dx \), we have \( f(x) = x + 1 \). The given limits are \( a = 5 \) and \( b = 0 \). For applying the limit of sum formula in the usual way, we consider \( \int_{0}^{5} (x+1) dx \) and then reverse the sign. For \( \int_{0}^{5} (x+1) dx \), we have \( a = 0 \) and \( b = 5 \). Thus, \( nh = b - a = 5 - 0 = 5 \).
Now, let's find the function values: \( f(a) = f(0) = 0 + 1 = 1 \) \( f(a + h) = f(h) = h + 1 \) \( f(a + 2h) = f(2h) = 2h + 1 \) ... \( f(a + n-1h) = f((n-1)h) = (n-1)h + 1 \)
By definition, the integral is: \[ \int_{a}^{b} f(x) dx = \lim_{h \to 0} h[f(a) + f(a + h) + f(a + 2h) + ... + f(a + n - 1h)] \] Substituting the values: \[ \int_{0}^{5} (x + 1) dx = \lim_{h \to 0} h[1 + (1+h) + (1 + 2h) + ... + (1 + (n - 1)h)] \] \[ = \lim_{h \to 0} h[(1+1+1+... \text{ to n terms}) + (h+2h+...+(n-1)h)] \] \[ = \lim_{h \to 0} h \left[ n \cdot 1 + h(1+2+3+...+(n-1)) \right] \] Using the sum of the first \( n-1 \) natural numbers formula: \( 1+2+...+n-1 = \frac{(n-1)n}{2} \). \[ = \lim_{h \to 0} h \left[ n + h \frac{(n-1)n}{2} \right] \] \[ = \lim_{h \to 0} \left[ nh + h^2 \frac{n(n-1)}{2} \right] \] \[ = \lim_{h \to 0} \left[ nh + \frac{nh(nh-h)}{2} \right] \] We know \( nh = 5 \). Substitute this value: \[ = \lim_{h \to 0} \left[ 5 + \frac{5(5-h)}{2} \right] \] As \( h \to 0 \), the expression becomes: \[ = 5 + \frac{5(5-0)}{2} \] \[ = 5 + \frac{25}{2} \] \[ = \frac{10 + 25}{2} = \frac{35}{2} \] This is the value for \( \int_{0}^{5} (x+1) dx \). Since the original question was \( \int_{5}^{0} (x+1) dx \), we must use the property \( \int_{a}^{b} f(x) dx = - \int_{b}^{a} f(x) dx \). Therefore, \( \int_{5}^{0} (x + 1) dx = - \frac{35}{2} \).
In simple words: First, we treat the integral as \( \int_{0}^{5} \) and apply the limit of sum method. We find the sum of \( f(x) \) values multiplied by \( h \). After calculating the result, we then flip the sign because the original question had the limits reversed compared to the standard formula application.

Exam Tip: Be careful with the order of limits. If the lower limit is greater than the upper limit, calculate the integral with the standard order and then multiply the result by \( -1 \).

Question 3. \( \int_{-3}^{2} x^2 dx \)
Answer: We need to evaluate the definite integral \( \int_{-3}^{2} x^2 dx \) as a limit of the sum. Let \( I = \int_{-3}^{2} x^2 dx \). Comparing this with \( \int_{a}^{b} f(x) dx \), we have \( f(x) = x^2 \), \( a = -3 \), and \( b = 2 \). Thus, \( nh = b - a = 2 - (-3) = 5 \).
Now, let's find the function values: \( f(a) = f(-3) = (-3)^2 = 9 \) \( f(a + h) = f(-3 + h) = (-3 + h)^2 \) \( f(a + 2h) = f(-3 + 2h) = (-3 + 2h)^2 \) ... \( f(a + (n-1)h) = f(-3 + (n-1)h) = (-3 + (n-1)h)^2 \)
By definition, the integral is: \[ I = \lim_{h \to 0} h[f(a) + f(a + h) + ... + f(a + (n-1)h)] \] \[ = \lim_{h \to 0} h[ (-3)^2 + (-3+h)^2 + (-3+2h)^2 + ... + (-3+(n-1)h)^2 ] \] Expanding each term \( (-3+kh)^2 = 9 - 6kh + k^2h^2 \) (for \( k=0, 1, ..., n-1 \)): \[ = \lim_{h \to 0} h \sum_{k=0}^{n-1} (9 - 6kh + k^2h^2) \] \[ = \lim_{h \to 0} h \left[ 9n - 6h \sum_{k=0}^{n-1} k + h^2 \sum_{k=0}^{n-1} k^2 \right] \] Using the sum of series formulas: \( \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2} \) \( \sum_{k=0}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6} \) Substitute these into the expression: \[ = \lim_{h \to 0} h \left[ 9n - 6h \frac{n(n-1)}{2} + h^2 \frac{n(n-1)(2n-1)}{6} \right] \] Distribute \( h \) inside the bracket: \[ = \lim_{h \to 0} \left[ 9nh - 3h^2 n(n-1) + h^3 \frac{n(n-1)(2n-1)}{6} \right] \] Substitute \( nh = 5 \): \[ = \lim_{h \to 0} \left[ 9(5) - \frac{3(nh)(nh-h)}{1} + \frac{(nh)(nh-h)(2nh-h)}{6} \right] \] \[ = \lim_{h \to 0} \left[ 45 - 3(5)(5-h) + \frac{5(5-h)(10-h)}{6} \right] \] As \( h \to 0 \): \[ = 45 - 3(5)(5-0) + \frac{5(5-0)(10-0)}{6} \] \[ = 45 - 75 + \frac{250}{6} \] \[ = -30 + \frac{125}{3} \] \[ = \frac{-90 + 125}{3} = \frac{35}{3} \] Therefore, \( \int_{-3}^{2} x^2 dx = \frac{35}{3} \).
In simple words: To find the definite integral of \( x^2 \) from -3 to 2 using the limit of sums, we break the range into small parts. We use the formula involving the sum of the first \( n \) numbers and the sum of the first \( n \) squares. Then, we let the width of these parts become zero to get the exact answer.

Exam Tip: Remember the formulas for sums of powers of natural numbers (\( \sum k \) and \( \sum k^2 \)) as they are crucial for solving definite integrals by the limit of sum method.

Question 4. \( \int_{-4}^{1} (x^2 - x) dx \)
Answer: We need to evaluate the definite integral \( \int_{-4}^{1} (x^2 - x) dx \) as a limit of the sum. Comparing this with \( \int_{a}^{b} f(x) dx \), we have \( f(x) = x^2 - x \), \( a = -4 \), and \( b = 1 \). Thus, \( nh = b - a = 1 - (-4) = 5 \).
Now, let's find the function values \( f(a+kh) \): \( f(a) = f(-4) = (-4)^2 - (-4) = 16 + 4 = 20 \) \( f(a+h) = f(-4+h) = (-4+h)^2 - (-4+h) = (16 - 8h + h^2) - (-4+h) = 20 - 9h + h^2 \) \( f(a+2h) = f(-4+2h) = (-4+2h)^2 - (-4+2h) = (16 - 16h + 4h^2) - (-4+2h) = 20 - 18h + 4h^2 \) In general, \( f(a+kh) = 20 - 9kh + k^2h^2 \) for \( k=0, 1, ..., n-1 \).
By definition, the integral is: \[ I = \lim_{h \to 0} h[f(a) + f(a + h) + ... + f(a + (n-1)h)] \] \[ = \lim_{h \to 0} h \sum_{k=0}^{n-1} (20 - 9kh + k^2h^2) \] \[ = \lim_{h \to 0} h \left[ 20n - 9h \sum_{k=0}^{n-1} k + h^2 \sum_{k=0}^{n-1} k^2 \right] \] Using the sum of series formulas: \( \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2} \) \( \sum_{k=0}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6} \) Substitute these formulas into the expression: \[ = \lim_{h \to 0} h \left[ 20n - 9h \frac{n(n-1)}{2} + h^2 \frac{n(n-1)(2n-1)}{6} \right] \] Distribute \( h \) inside the bracket: \[ = \lim_{h \to 0} \left[ 20nh - 9h^2 \frac{n(n-1)}{2} + h^3 \frac{n(n-1)(2n-1)}{6} \right] \] Now, substitute \( nh = 5 \): \[ = \lim_{h \to 0} \left[ 20(5) - \frac{9(nh)(nh-h)}{2} + \frac{(nh)(nh-h)(2nh-h)}{6} \right] \] \[ = \lim_{h \to 0} \left[ 100 - \frac{9(5)(5-h)}{2} + \frac{5(5-h)(10-h)}{6} \right] \] As \( h \to 0 \), we evaluate the limit: \[ = 100 - \frac{9(5)(5)}{2} + \frac{5(5)(10)}{6} \] \[ = 100 - \frac{225}{2} + \frac{250}{6} \] \[ = 100 - 112.5 + \frac{125}{3} \] \[ = \frac{600}{6} - \frac{675}{6} + \frac{250}{6} \] \[ = \frac{600 - 675 + 250}{6} \] \[ = \frac{175}{6} \] Therefore, \( \int_{-4}^{1} (x^2 - x) dx = \frac{175}{6} \).
In simple words: For this integral, we first identify \( f(x) \) and the limits \( a \) and \( b \). Then, we use the limit of sums formula, finding the function value at each step. We apply the summation formulas for powers of integers and then take the limit as \( h \) approaches zero to get the exact area.

Exam Tip: Pay close attention to negative signs in the limits of integration and in the function \( f(x) \) itself, as they can easily lead to calculation errors.

Question 5. \( \int_{1}^{-1} e^x dx \)
Answer: We need to evaluate the definite integral \( \int_{1}^{-1} e^x dx \) as a limit of the sum. Comparing this with \( \int_{a}^{b} f(x) dx \), we have \( f(x) = e^x \). The given limits are \( a = 1 \) and \( b = -1 \). For application of the limit of sum formula, we generally use the smaller limit as \( a \) and larger as \( b \), so let's first evaluate \( \int_{-1}^{1} e^x dx \) and then reverse the sign for the original question. For \( \int_{-1}^{1} e^x dx \), we have \( a = -1 \) and \( b = 1 \). Thus, \( nh = b - a = 1 - (-1) = 2 \).
Now, let's find the function values: \( f(a) = f(-1) = e^{-1} \) \( f(a + h) = f(-1 + h) = e^{-1+h} \) \( f(a + 2h) = f(-1 + 2h) = e^{-1+2h} \) ... \( f(a + n-1h) = f(-1 + (n-1)h) = e^{-1+(n-1)h} \)
By definition, the integral is: \[ \int_{-1}^{1} e^x dx = \lim_{h \to 0} h[f(a) + f(a + h) + ... + f(a + (n-1)h)] \] \[ = \lim_{h \to 0} h[e^{-1} + e^{-1+h} + e^{-1+2h} + ... + e^{-1+(n-1)h}] \] Factor out \( e^{-1} \): \[ = \lim_{h \to 0} h \cdot e^{-1} [1 + e^h + e^{2h} + ... + e^{(n-1)h}] \] The terms inside the bracket form a geometric progression with first term \( A = 1 \), common ratio \( R = e^h \), and \( n \) terms. The sum is \( A \frac{R^n - 1}{R - 1} = 1 \cdot \frac{(e^h)^n - 1}{e^h - 1} = \frac{e^{nh} - 1}{e^h - 1} \). \[ = \lim_{h \to 0} h \cdot e^{-1} \frac{e^{nh} - 1}{e^h - 1} \] Rearrange to use a standard limit form: \[ = e^{-1} (e^{nh} - 1) \lim_{h \to 0} \frac{h}{e^h - 1} \] We know that \( \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \), so \( \lim_{h \to 0} \frac{h}{e^h - 1} = 1 \). Substitute \( nh = 2 \): \[ = e^{-1} (e^2 - 1) \cdot 1 \] \[ = e^{-1} (e^2 - 1) \] \[ = e - e^{-1} \] This is the value for \( \int_{-1}^{1} e^x dx \). Since the original question was \( \int_{1}^{-1} e^x dx \), we apply the property \( \int_{a}^{b} f(x) dx = - \int_{b}^{a} f(x) dx \). Therefore, \( \int_{1}^{-1} e^x dx = - (e - e^{-1}) = e^{-1} - e \).
In simple words: To evaluate an integral with exponential functions using the limit of sums, we recognize the sum as a geometric series. We use the geometric series sum formula, then apply the limit as \( h \) goes to zero. Remember to adjust the sign if the integration limits are reversed.

Exam Tip: For integrals involving exponential functions, identifying the geometric series and applying its sum formula, along with the limit \( \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \), is key.

Question 6. \( \int_{4}^{0} (x + e^{2x}) dx \)
Answer: We need to evaluate the definite integral \( \int_{4}^{0} (x + e^{2x}) dx \) as a limit of the sum. Comparing this with \( \int_{a}^{b} f(x) dx \), we have \( f(x) = x + e^{2x} \). The given limits are \( a = 4 \) and \( b = 0 \). For applying the limit of sum formula, we first evaluate \( \int_{0}^{4} (x + e^{2x}) dx \) and then reverse the sign. For \( \int_{0}^{4} (x + e^{2x}) dx \), we have \( a = 0 \) and \( b = 4 \). Thus, \( nh = b - a = 4 - 0 = 4 \).
Now, let's find the function values \( f(a+kh) \): \( f(a) = f(0) = 0 + e^{2 \cdot 0} = e^0 = 1 \) \( f(a+h) = f(h) = h + e^{2h} \) \( f(a+2h) = f(2h) = 2h + e^{4h} \) ... \( f(a+(n-1)h) = f((n-1)h) = (n-1)h + e^{2(n-1)h} \)
By definition, the integral is: \[ I = \lim_{h \to 0} h[f(a) + f(a + h) + ... + f(a + (n-1)h)] \] \[ = \lim_{h \to 0} h[1 + (h + e^{2h}) + (2h + e^{4h}) + ... + ((n-1)h + e^{2(n-1)h})] \] Group the linear terms and exponential terms separately: \[ = \lim_{h \to 0} h \left[ (h + 2h + ... + (n-1)h) + (1 + e^{2h} + e^{4h} + ... + e^{2(n-1)h}) \right] \] \[ = \lim_{h \to 0} h \left[ h(1 + 2 + ... + (n-1)) + (1 + e^{2h} + (e^{2h})^2 + ... + (e^{2h})^{n-1}) \right] \] Use the sum of the first \( (n-1) \) natural numbers formula \( \frac{(n-1)n}{2} \) and the sum of a geometric series \( \frac{R^n - 1}{R - 1} \) with \( R = e^{2h} \): \[ = \lim_{h \to 0} h \left[ h \frac{n(n-1)}{2} + \frac{(e^{2h})^n - 1}{e^{2h} - 1} \right] \] \[ = \lim_{h \to 0} h \left[ h \frac{n(n-1)}{2} + \frac{e^{2nh} - 1}{e^{2h} - 1} \right] \] Distribute \( h \) inside the bracket: \[ = \lim_{h \to 0} \left[ h^2 \frac{n(n-1)}{2} + h \frac{e^{2nh} - 1}{e^{2h} - 1} \right] \] Rearrange terms for substitution and limit evaluation: \[ = \lim_{h \to 0} \left[ \frac{(nh)(nh-h)}{2} + (e^{2nh} - 1) \frac{h}{e^{2h} - 1} \right] \] We know \( nh = 4 \). Substitute this value: \[ = \lim_{h \to 0} \left[ \frac{4(4-h)}{2} + (e^{2 \cdot 4} - 1) \frac{1}{2} \cdot \frac{2h}{e^{2h} - 1} \right] \] As \( h \to 0 \), \( \frac{2h}{e^{2h} - 1} \to 1 \). \[ = \frac{4(4-0)}{2} + (e^8 - 1) \frac{1}{2} \cdot 1 \] \[ = \frac{16}{2} + \frac{e^8 - 1}{2} \] \[ = 8 + \frac{e^8 - 1}{2} \] \[ = \frac{16 + e^8 - 1}{2} \] \[ = \frac{15 + e^8}{2} \] This is the value for \( \int_{0}^{4} (x + e^{2x}) dx \). Since the original question was \( \int_{4}^{0} (x + e^{2x}) dx \), we apply the property \( \int_{a}^{b} f(x) dx = - \int_{b}^{a} f(x) dx \). Therefore, \( \int_{4}^{0} (x + e^{2x}) dx = - \frac{15 + e^8}{2} \).
In simple words: This problem involves two types of sums: an arithmetic progression for \( x \) terms and a geometric progression for \( e^{2x} \) terms. We find the sum for each part separately, take the limits as \( h \) approaches zero, and then combine them. Remember to adjust the sign if the limits are reversed.

Exam Tip: When \( f(x) \) is a sum of multiple terms, calculate the limit of sum for each term separately and then combine them. Also, use the correct limit property for exponential functions.