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If, by value, you mean the area under the curve from 0 to 1, the solution would be 1/2 sqrt(π) • e • rf(1) + 1. This is simply found by removing the parentheses, taking the derivative, and then solving from there as usual.

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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TopNewestIf, by value, you mean the area under the curve from 0 to 1, the solution would be 1/2 sqrt(π) • e • rf(1) + 1. This is simply found by removing the parentheses, taking the derivative, and then solving from there as usual.

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Oh can you explain more.

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Yes - I'll post a new discussion called JEE 1981 Int Calc with the information enclosed. I need my LaTeX!

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@RISHU Jaar It's in the new section.

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Read Error function.

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@Pi Han Goh , thank you , its a jee 1981 question but didn't knew the maths involved was so higher.

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Then you can only evaluate it using numerical methods, like trapezoidal rule or Simpson's rule.

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Trapezium Rule

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@Rishabh Cool and others please give a proof !

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