Question
Download Solution PDFFor the data,
\(\begin{array}{*{20}{c}} {x:}\\ {f\left( x \right):} \end{array}\begin{array}{*{20}{c}} 0\\ 5 \end{array}\begin{array}{*{20}{c}} 1\\ 2 \end{array}\begin{array}{*{20}{c}} 2\\ 1 \end{array}\begin{array}{*{20}{c}} 3\\ 3 \end{array}\begin{array}{*{20}{c}} 4\\ 7 \end{array}\)
the value of \(\mathop \smallint \nolimits_0^4 2f\left( x \right)dx\) will be:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Trapezoidal rule is given by:
\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{2}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_{\rm{n}}} + 2\left( {{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}{\rm{\;}} \ldots } \right)} \right]\)
\({\rm{Number\;of\;intervals(n)}} = \frac{{{\rm{b}} - {\rm{a}}}}{{\rm{h}}}{\rm{\;}}\)
where b is the upper limit, a is the lower limit, h is the step size.
Calculation:
From question
|
x |
0 |
1 |
2 |
3 |
4 |
|
f(x) |
5 |
2 |
1 |
3 |
7 |
| 2f(x) | 10 | 4 | 2 | 6 | 14 |
\(I = \mathop \smallint \nolimits_0^4 2.f\left( x \right).dx\)
From the formula:
\(\mathop \smallint \nolimits_0^4 f\left( x \right)dx = \frac{h}{2}\left[ {\left( {{y_0} + {y_4}} \right) + 2\left( {{y_1} + {y_2} + {y_3}} \right)} \right]\;\)
\(= \frac{{1}}{2}\left[ {\left( {5 + 7} \right) + 2\left( {2 + 1 + 3} \right)} \right]\)
∴ \(I = \frac{{24}}{2} = 12\)
Here we calculated for f(x), but asked for 2 f(x)
So the answer comes out to be 2 × 12 = 24
Last updated on Jul 16, 2025
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