$\"\"$

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The function is $\"\"$

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The function is defined on the interval $\"\"$ .

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Find $\"\"$ .

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Using integral theorem :

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If $\"\"$ is integrable on $\"\"$ , then $\"\"$ .

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Where width $\"\"$ and $\"\"$ .

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The integral expression is $\"\"$ .

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In this case $\"\"$ and $\"\"$ .

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The number of subintervals are $\"\"$ .

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Substitute $\"\"$ in $\"\"$ .

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$\"\"$ .

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Substitute $\"\"$ and $\"\"$ in $\"\"$ .

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$\"\"$ .

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$\"\"$

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$\"\"$

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$\"\"$

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The series is in harmonic, so does not converge.

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The function is not continuous at $\"\"$ .

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$\"\"$ does not exist.

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$\"\"$

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$\"\"$ does not exist.