Question:
The value of the integral
\[ \int_{0.2}^{1.4} (\sin x - \log_e x + e^x) \, dx \]
using Simpson's three-eighth rule, by taking interval size \( h = 0.2 \), is:
The value of the integral
\[ \int_{0.2}^{1.4} (\sin x - \log_e x + e^x) \, dx \]
using Simpson's three-eighth rule, by taking interval size \( h = 0.2 \), is:
\[ \int_{0.2}^{1.4} (\sin x - \log_e x + e^x) \, dx \]
using Simpson's three-eighth rule, by taking interval size \( h = 0.2 \), is:
Show Hint
Simpson’s three-eighth rule is a numerical integration technique that is particularly effective when dealing with irregular intervals and provides better accuracy for certain functions
Updated On: Jan 17, 2026
- 4.05
- 4.052985
- 4.502985
- 4.052985
Show Solution
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The Correct Option is B
Solution and Explanation
To apply Simpson’s three-eighth rule we use the given function values at the specified intervals and calculate the weighted sum The result of the integral is approximately 4.052985
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