Q. 18 The definite integral of a funct... [FREE SOLUTION]

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Chapter 4: Q. 18 (page 352)

The definite integral of a function $f$ on an interval $[a, b]$ is defined as a limit of Riemann sums. How can it be that the sum of the areas of infinitely many rectangles that are each 鈥渋nfinitely thin鈥� is a finite number? On the one hand, shouldn鈥檛 it be infinite, since we are adding up infinitely many rectangles? On the other hand, shouldn鈥檛 it always be zero, since the width of each of the rectangles is approaching zero as $n 鈫� 鈭�$ ?

Short Answer

Expert verified

The sums of the areas of infinitely many rectangles which are infinitely thin is a finite number is proved.

Step by step solution

Step 1. Given information

The definite integral of a function $f$ on an interval $[a, b]$ is defined as a limit of Riemann sums.

Step 2. The above proof can be shown with an example.

Let the definite integral be,

${鈭�}_{2}^{5} (5 - x) d x$ .

The right sum defined for $n$ rectangles on $[a, b]$ is, ${鈭�}_{k = 1}^{n} f (x_{k}) 螖 x$ .

where, $螖 x = \frac{b - a}{n}, x_{k} = a + k 螖 x .$

The interval is, role="math" localid="1648713229251" $[2, 5]$ .

Now,

$鈭� x = \frac{5 - 2}{n} = \frac{3}{n}$

And,

$x_{k} = 2 + k (\frac{3}{n}) = 2 + \frac{3 k}{n}$

Step 3. The right sum is,

${鈭�}_{k = 1}^{n} (5 - (2 - \frac{3 k}{n})) (\frac{3}{n}) = (\frac{3}{n}) {鈭�}_{k = 1}^{n} (3 - \frac{3 k}{n}) = \frac{3}{n} {鈭�}_{k = 1}^{n} 3 - \frac{3}{n} {鈭�}_{k = 1}^{n} \frac{3 k}{n} = \frac{3}{n} (3 n) - (\frac{3}{n}) (\frac{3}{n}) (\frac{n (n + 1)}{2}) = \frac{3}{n} (3 n) - \frac{9}{n^{2}} (\frac{n (n + 1)}{2})$

Therefore, the right sum is, $\frac{3}{n} (3 n) - \frac{9}{n^{2}} (\frac{n (n + 1)}{2})$ .

Step 4.Now applying limit on infinite rectangles for infinitely thin rectangles,

${鈭�}_{2}^{5} (5 - x) d x = \lim_{n 鈫� 鈭�} (\frac{3}{n} (3 n) - \frac{9}{n^{2}} (\frac{n (n + 1)}{2})) = \lim_{n 鈫� 鈭�} (9 - \frac{9}{n^{2}} (\frac{n^{2} + n}{2})) = \lim_{n 鈫� 鈭�} (\frac{18 - 9 n^{2} - 9 n}{2 n^{2}}) = \frac{9}{2}$

The exact value is finite.

Therefore, the sums of the areas of infinitely many rectangles which are infinitely thin is a finite number.

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Short Answer

Step by step solution

Step 1. Given information

Step 2. The above proof can be shown with an example.

Step 3. The right sum is,

Step 4.Now applying limit on infinite rectangles for infinitely thin rectangles,

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