Step 1:

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The functions are $\"\"$ and $\"\"$ .

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Region bounded is : $\"\"$ .

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Graph:

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Graph the functions are $\"\"$ and $\"\"$ .

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Shade the region bounded by the curves between $\"image\"$ and $\"\"$ .

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$\"\"$

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Note: The region shaded in blue color is the required area of region.

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Step 2:

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Definite integral as area of the region:

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If $\"image\"$ and $\"image\"$ are continuous and non-negative on the closed interval $\"image\"$ ,

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then the area of the region bounded by the graphs of $\"image\"$ and $\"image\"$ and the vertical lines $\"image\"$ and $\"image\"$ is given by

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$\"image\"$ .

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Observe the graph:

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Upper curve is $\"\"$ .

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Lower Curve is $\"\"$ .

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The vertical lines are $\"image\"$ and $\"\"$ .

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

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$\"\"$

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Let $\"\"$ .

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Apply derivative on each side.

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$\"\"$

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Substitute $\"\"$ and $\"\"$ in the integral.

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$\"\"$

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Substitute $\"\"$ .

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$\"\"$

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Area of the region is 0.316 sq-units.

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Solution:

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Graph the region of graph of $\"\"$ and $\"\"$ between $\"image\"$ and $\"\"$ is

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$\"\"$

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Area of the region is 0.316 sq-units.