`y = sec^2 (x), 0<=x<=(pi/3)` Use a graph to give a rough estimate of the area of the region that lies beneath the given curve. Then find the exact area.

Textbook Question

Chapter 5, 5.3 - Problem 52 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`y=sec^2(x)`

Refer the graph in the attached image.

From the graph,

Area of the region beneath the curve `~~` 4/10(Area of the Rectangle)

Area of the region=`~~(4/10)(pi/3*4)`

Area of the region`~~1.676`

Exact Area of the region=`int_0^(pi/3)sec^2(x)dx`

`=[tan(x)]_0^(pi/3)`

`=tan(pi/3)-tan(0)`

`=sqrt(3)-0`

=1.73205

 

 

 

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