$\"\"$

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The curves are $\"\"$ and $\"\"$ .

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Find the intersection points, by equating two curves.

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$\"\"$

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$\"\"$

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$\"\"$ .

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The doman of the logarithm functions is $\"\"$ , hence point of intersection of the two curves is $\"\"$

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Therefore, the area bounded between $\"\"$ and $\"\"$ .

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Graph:

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Graph the curves $\"\"$ and $\"\"$ .

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$\"\"$ .

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Observe the graph:

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The upper curve is $\"\"$ .

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The lower curve is $\"\"$ .

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$\"\"$

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Definite integral as area of the region:

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If $\"image\"$ is continuous and non-negative on the closed interval $\"\"$ , then the area of the region bounded by the graph $\"image\"$ , the $\"image\"$ -axis and the vertical lines $\"image\"$ and $\"image\"$ is given by

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$\"\"$ .

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$\"\"$

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$\"\"$

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$\"\"$

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Apply formula: $\"\"$ .

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$\"\"$

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$\"\"$

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$\"\"$

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$\"\"$

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$\"\"$ .

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$\"\"$

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Consider $\"\"$ .

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Apply parts of integration formula: $\"\"$ .

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

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$\"\"$

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$\"\"$ .

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$\"\"$

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Integrate on each side.

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$\"\"$

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$\"\"$ .

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Substitute the corresponding values in $\"\"$ .

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$\"\"$

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$\"\"$

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$\"\"$

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$\"\"$

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$\"\"$

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$\"\"$

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$\"\"$ .

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$\"\"$

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

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$\"\"$

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$\"\"$

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$\"\"$

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$\"\"$

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$\"\"$

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$\"\"$ .

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Area of the region is $\"\"$ square units.

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$\"\"$

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Area of the region is $\"\"$ square units.