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Integrals - Definite integrals as a limit of a sum

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The definite integral ∫abf(x)dx\int_{a}^{b} f(x) dx is geometrically interpreted as the signed area of the region bounded by the graph of the function y=f(x)y = f(x), the x-axis, and the vertical lines x=ax = a and x=bx = b. Visually, this is the space between the curve and the horizontal axis within the specified boundaries.

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The interval [a,b][a, b] is divided into nn equal sub-intervals, each of width hh. As nn becomes very large, the width hh approaches zero, which is expressed by the relation h=b−anh = \frac{b-a}{n} or nh=b−anh = b-a. This process is visually represented by slicing the area under the curve into extremely thin vertical strips.

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The area under the curve is approximated by the sum of the areas of these thin vertical rectangles. For each sub-interval, the height of the rectangle is determined by the value of the function at a specific point within that interval, such as the right endpoint f(a+rh)f(a+rh).

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A Riemann Sum is the mathematical expression of this rectangular approximation: Sn=h[f(a+h)+f(a+2h)+⋯+f(a+nh)]S_n = h[f(a+h) + f(a+2h) + \dots + f(a+nh)]. Visually, as nn increases, the jagged 'staircase' formed by the tops of the rectangles aligns more closely with the smooth curve of the function.

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The definite integral is defined as the limit of this sum as n→∞n \to \infty (or h→0h \to 0). If this limit exists, the function f(x)f(x) is said to be integrable over the interval [a,b][a, b]. For CBSE purposes, we generally assume f(x)f(x) is continuous on [a,b][a, b] to ensure the limit exists.

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Evaluating an integral as a limit of a sum involves substituting the function into the summation formula, simplifying using standard series sum formulas (like the sum of natural numbers or squares), and then applying the limit as hh tends to zero while keeping nhnh as a constant.

📐Formulae

∫abf(x)dx=lim⁡h→0h∑r=1nf(a+rh)\int_{a}^{b} f(x) dx = \lim_{h \to 0} h \sum_{r=1}^{n} f(a+rh), where nh=b−anh = b-a

∑r=1nr=1+2+3+⋯+n=n(n+1)2\sum_{r=1}^{n} r = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}

∑r=1nr2=12+22+32+⋯+n2=n(n+1)(2n+1)6\sum_{r=1}^{n} r^2 = 1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}

∑r=1nr3=[n(n+1)2]2\sum_{r=1}^{n} r^3 = [\frac{n(n+1)}{2}]^2

a+ar+ar2+⋯+arn−1=a(rn−1)r−1a + ar + ar^2 + \dots + ar^{n-1} = \frac{a(r^n - 1)}{r-1} (Sum of a Geometric Progression)

lim⁡h→0eh−1h=1\lim_{h \to 0} \frac{e^h - 1}{h} = 1

💡Examples

Problem 1:

Evaluate ∫02(x+1)dx\int_{0}^{2} (x+1) dx as a limit of a sum.

Solution:

  1. Identify parameters: a=0,b=2,f(x)=x+1a=0, b=2, f(x)=x+1. \n2. Calculate nhnh: nh=b−a=2−0=2nh = b-a = 2-0 = 2. \n3. Set up the sum: Sn=h∑r=1nf(a+rh)=h∑r=1nf(0+rh)=h∑r=1n(rh+1)S_n = h \sum_{r=1}^{n} f(a+rh) = h \sum_{r=1}^{n} f(0+rh) = h \sum_{r=1}^{n} (rh+1). \n4. Expand the summation: Sn=h[h∑r=1nr+∑r=1n1]=h[hn(n+1)2+n]S_n = h [h \sum_{r=1}^{n} r + \sum_{r=1}^{n} 1] = h [h \frac{n(n+1)}{2} + n]. \n5. Distribute hh: Sn=(nh)(nh+h)2+nhS_n = \frac{(nh)(nh+h)}{2} + nh. \n6. Apply the limit h→0h \to 0 and substitute nh=2nh=2: lim⁡h→0[2(2+h)2+2]=2(2)2+2=2+2=4\lim_{h \to 0} [\frac{2(2+h)}{2} + 2] = \frac{2(2)}{2} + 2 = 2 + 2 = 4.

Explanation:

This approach converts the definite integral into a sum of nn rectangles. We use the formula for the sum of the first nn natural numbers and then evaluate the limit as the rectangle width hh approaches zero.

Problem 2:

Evaluate ∫02exdx\int_{0}^{2} e^x dx as a limit of a sum.

Solution:

  1. Identify parameters: a=0,b=2,f(x)=ex,nh=2a=0, b=2, f(x)=e^x, nh = 2. \n2. Set up the sum: Sn=h∑r=1nf(0+rh)=h[eh+e2h+⋯+enh]S_n = h \sum_{r=1}^{n} f(0+rh) = h [e^h + e^{2h} + \dots + e^{nh}]. \n3. This is a Geometric Progression with first term A=ehA=e^h, common ratio R=ehR=e^h, and nn terms. \n4. Sum of GP: Sn=heh((eh)n−1)eh−1=heh(enh−1)eh−1S_n = h \frac{e^h((e^h)^n - 1)}{e^h - 1} = h \frac{e^h(e^{nh} - 1)}{e^h - 1}. \n5. Rearrange for limit: Sn=eh(enh−1)⋅heh−1S_n = e^h(e^{nh} - 1) \cdot \frac{h}{e^h - 1}. \n6. Substitute nh=2nh=2 and apply lim⁡h→0\lim_{h \to 0}: lim⁡h→0eh(e2−1)⋅1eh−1h\lim_{h \to 0} e^h(e^2 - 1) \cdot \frac{1}{\frac{e^h - 1}{h}}. \n7. Since lim⁡h→0eh=1\lim_{h \to 0} e^h = 1 and lim⁡h→0eh−1h=1\lim_{h \to 0} \frac{e^h - 1}{h} = 1, the result is 1⋅(e2−1)⋅1=e2−11 \cdot (e^2 - 1) \cdot 1 = e^2 - 1.

Explanation:

For exponential functions, the Riemann sum results in a Geometric Progression. The evaluation requires the standard limit lim⁡h→0eh−1h=1\lim_{h \to 0} \frac{e^h-1}{h} = 1 to simplify the expression.