# Sketch the region bounded by the curves y=sec^2x and y=4  on the interval [-pi/2, pi/2].  Find the area of the sketched region.

justaguide | College Teacher | (Level 2) Distinguished Educator

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The region bounded by the curves y = sec^2 x and y = 4 is shown in the graph below:

The curves `y = sec^2x` and y = 4 intersect at the point where `sec^2x = 4`

=> `cos^2x = 1/4`

=> `cos x = (1/2)` and `cos x = -1/2`

`x = pi/3` and `x = -pi/3`

The area of the bounded area is `int_(-pi/3)^(pi/3) 4 - sec^2x dx`

=> `4x - tan x|_(-pi/3)^(pi/3)`

=> `4*2*pi/3 - 2*sqrt 3`

=> `8*pi/3 - 2*sqrt 3`

The required area is `8*pi/3 - 2*sqrt 3`

quantatanu | Student, Undergraduate | (Level 1) Valedictorian

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I will continue again:

so the curves are:

y = Sec^2 (x)   [our f(x)]

y = 4               [our g(x)]

but this is a tricky problem, as sec^2 (x) becomes infinite at the points -pi/2 & pi/2 so we must avoid taking the integration limits to be these two. But no problem what we need to integrate is the bounded region and that region is within the interval [-pi/2,pi/2], to see this just plot y = Sec^2 (x) and draw y = 4 line and see where this line cuts the Sec^2 (x) curves, these 2 points will be enough for us to take as integration limits. To find out these two points we must solve,

Sec^2 (x) = 4

=> Sec(x) = +2            [as Sec(x) > 0, in  - Pi/2 to Pi/2]

=> x = +/- Pi/3            [in the interval -Pi/2 to Pi/2]

So the area bounded by the curves will be:

Area = Integral [ (4 - Sec^2 (x) ) dx ] from - Pi/3 to Pi/3

= 8 Pi/3  - 2 Sqrt [3 ]

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For understanding the theory I am pasting my last answer below:

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Suppose two curves are given to you,

1) y = f(x),

2) y = g(x),

Plot the two curves, then draw lines parallel to y-axis, one at x=-Pi/2 and the other at x=Pi/2,  blacken the regin bounded between these two curves and the lines. Suppose y=f(x) curve is above the y=g(x) curve, then area bounded by the curves or the area of the blackened region will be:

area = Integral[ {f(x) - g(x)} dx] from -Pi/2 to Pi/2

For example, suppose you have,

y = f(x) = 2 x^2 and

y = g(x) = x^2

These are two parabolas, now draw them you will see f(x) will be steeper than g(x), now drop lines at x = -Pi/2 and x = Pi/2 parallel to y-axis, now blacken the region, so area bounded will be:

area = Integral [  (2 x^2 - x^2) dx ] from -Pi/2 to Pi/2

= [ x^3 /3 ] from -Pi/2 to Pi/2

= Pi^3 /12

schoolsucks112 | Student, Undergraduate | eNotes Newbie

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Sorry the question cut out the section sketch the region bounded by the curves y=sec^2x and y=4 on the interval [-pi/2, pi/2]

quantatanu | Student, Undergraduate | (Level 1) Valedictorian

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Suppose two curves are given to you,

1) y = f(x),

2) y = g(x),

Plot the two curves, then draw lines parallel to y-axis, one at x=-Pi/2 and the other at x=Pi/2,  blacken the regin bounded between these two curves and the lines. Suppose y=f(x) curve is above the y=g(x) curve, then area bounded by the curves or the area of the blackened region will be:

area = Integral[ {f(x) - g(x)} dx] from -Pi/2 to Pi/2

For example, suppose you have,

y = f(x) = 2 x^2 and

y = g(x) = x^2

These are two parabolas, now draw them you will see f(x) will be steeper than g(x), now drop lines at x = -Pi/2 and x = Pi/2 parallel to y-axis, now blacken the region, so area bounded will be:

area = Integral [  (2 x^2 - x^2) dx ] from -Pi/2 to Pi/2

= [ x^3 /3 ] from -Pi/2 to Pi/2

= Pi^3 /12