Step 1:

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The curve equation is $\"\"$ and a line equation $\"\"$ . $\"\"$

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Sketch the graphs :

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$\"\"$

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Shade the region above the curve $\"\"$ and region below the line $\"\"$ .

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Now find the intersection of the curve and a line.

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Find the intersection points from the graph $\"\"$ and $\"\"$ .

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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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$\"\"$

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$\"\"$

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Apply power rule of integration: $\"\"$ .

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$\"\"$

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Therefore area above the curve $\"\"$ and region below the line $\"\"$ is 36 sq units.

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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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$\"\"$

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$\"\"$

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$\"\"$

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$\"\"$

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$\"\"$

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$\"\"$ .

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Solution:

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Area of the region is $\"\"$ sq-units.

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(1)

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Step 1:

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

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Let $\"image\"$ and $\"image\"$ .

\

Definite integral as area of the region:

\

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

\

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

\

$\"image\"$ .

\

\

$\"image\"$

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$\"image\"$

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$\"image\"$

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Apply power rule of integration: $\"image\"$ .

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$\"image\"$

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$\"image\"$

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$\"image\"$

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$\"image\"$

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$\"image\"$

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$\"image\"$

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$\"image\"$

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$\"image\"$

\

$\"image\"$

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$\"image\"$

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$\"image\"$

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$\"image\"$

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$\"image\"$

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$\"image\"$

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$\"image\"$ sq-units.

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Solution:

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Area of the region is $\"image\"$ sq-units.