Evaluate the definite integral by the limit definition. Step 1 J4 5 dx To find the definite integral 5 dx by the limit definition, divide the interval [4, 9] into n subintervals. Step 2 Then the width of each interval is b-a Ax = Note that ||A||- → n J4 5 n 9-4 n O as n→ 00.
Evaluate the definite integral by the limit definition. Step 1 J4 5 dx To find the definite integral 5 dx by the limit definition, divide the interval [4, 9] into n subintervals. Step 2 Then the width of each interval is b-a Ax = Note that ||A||- → n J4 5 n 9-4 n O as n→ 00.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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Question
![Evaluate the definite integral by the limit definition.
\[
\int_{4}^{9} 5 \, dx
\]
**Step 1**
To find the definite integral \(\int_{4}^{9} 5 \, dx\) by the limit definition, divide the interval \([4, 9]\) into \(n\) subintervals.
The width of each interval is
\[
\Delta x = \frac{b-a}{n} = \frac{9-4}{n} = \frac{5}{n}.
\]
Note that \(\|\Delta\| \rightarrow 0\) as \(n \rightarrow \infty\).
**Step 2**
Choose \(c_i\) as the right endpoint of each subinterval. Then
\[
c_i = a + i(\Delta x) = 4 + i\left(\frac{5}{n}\right).
\]
So the definite integral is given by
\[
\int_{4}^{9} 5 \, dx = \lim_{\|\Delta\| \to 0} \sum_{i=1}^{n} f(c_i) \Delta x
\]
\[
= \lim_{n \to \infty} \sum_{i=1}^{n} f(c_i) \Delta x
\]
\[
= \lim_{n \to \infty} \sum_{i=1}^{n} f\left(4 + \frac{5i}{n}\right) \left(\frac{5}{n}\right)
\]
\[
= \lim_{n \to \infty} \sum_{i=1}^{n} \left(5\right)\left(\frac{5}{n}\right)
\]
\[
= \lim_{n \to \infty} \left( \frac{5}{n} \right) \sum_{i=1}^{n} 5
\]
\[
= \lim_{n \to \infty} \left(\frac{25}{n}\right) \sum_{i=1}^{n} 1
\]
\[
= \lim_{n \to \infty} 25
\]
\[
= 25.
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Transcribed Image Text:Evaluate the definite integral by the limit definition.
\[
\int_{4}^{9} 5 \, dx
\]
**Step 1**
To find the definite integral \(\int_{4}^{9} 5 \, dx\) by the limit definition, divide the interval \([4, 9]\) into \(n\) subintervals.
The width of each interval is
\[
\Delta x = \frac{b-a}{n} = \frac{9-4}{n} = \frac{5}{n}.
\]
Note that \(\|\Delta\| \rightarrow 0\) as \(n \rightarrow \infty\).
**Step 2**
Choose \(c_i\) as the right endpoint of each subinterval. Then
\[
c_i = a + i(\Delta x) = 4 + i\left(\frac{5}{n}\right).
\]
So the definite integral is given by
\[
\int_{4}^{9} 5 \, dx = \lim_{\|\Delta\| \to 0} \sum_{i=1}^{n} f(c_i) \Delta x
\]
\[
= \lim_{n \to \infty} \sum_{i=1}^{n} f(c_i) \Delta x
\]
\[
= \lim_{n \to \infty} \sum_{i=1}^{n} f\left(4 + \frac{5i}{n}\right) \left(\frac{5}{n}\right)
\]
\[
= \lim_{n \to \infty} \sum_{i=1}^{n} \left(5\right)\left(\frac{5}{n}\right)
\]
\[
= \lim_{n \to \infty} \left( \frac{5}{n} \right) \sum_{i=1}^{n} 5
\]
\[
= \lim_{n \to \infty} \left(\frac{25}{n}\right) \sum_{i=1}^{n} 1
\]
\[
= \lim_{n \to \infty} 25
\]
\[
= 25.
\]
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