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Step 1:

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The integral expression is $\"image\"$ on interval $\"image\"$ .

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The expression is $\"image\"$

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Consider $\"image\"$ .

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Using summation theorem :

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If $\"\"$ is integral on $\"\"$ , then $\"image\"$ .

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Where $\"image\"$ , $\"image\"$ and $\"image\"$ .

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The integral expression is $\"image\"$ .

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Step 2:

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Here $\"image\"$ and $\"image\"$ .

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$\"image\"$

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$\"image\"$

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

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$\"image\"$ .

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$\"image\"$ .

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Step 3:

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substitute $\"image\"$ values from 1 to 4.

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$\"image\"$

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$\"image\"$

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$\"image\"$

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$\"image\"$

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$\"image\"$

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Step 4:

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using mid point theorem Area = $\"image\"$ .

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$\"image\"$

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$\"image\"$ .

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Solution:

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Area of the function is $\"image\"$ .

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