Step 1:

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The function is $\"\"$ and the interval is $\"\"$ .

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

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Definition of integral :

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

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Where $\"\"$ .

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Let number of sub intervals be $\"\"$

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

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Divide this interval into $\"\"$ equal width subintervals, each of which has a width of $\"\"$ .

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$\"\"$

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

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$\"\"$ .

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Step 2:

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$\"\"$

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$\"\"$

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Sum of first $\"image\"$ natural numbers is $\"image\"$ .

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$\"\"$

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$\"\"$

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Area of the region is between $\"\"$ and $\"\"$ -axis is 3.

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Step 3:

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Graph the function.

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$\"\"$

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Area of the region is 3 sq-units.

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Solution:

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Area of the region is 3 sq-units.