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.

y = sec^2 x, 0 ≤ x ≤ π/4

To get the area, we use integration. From the graph, we can draw vertical strips that run from 0 to π/4.

∫ y dx
= ∫ sec^2 (x) dx
= tan x | from 0 to π/4
= tan (π/4) - tan (0)
= 1 - 0
= 1

Thank you Jai

To find the rough estimate of the area, we can use a graph.

First, let's graph the curve y = sec^2 x in the given interval 0 ≤ x ≤ π/4.

The graph of y = sec^2 x is a continuous curve that starts at x = 0 and approaches infinity as x approaches π/4. However, we only need to consider the interval 0 ≤ x ≤ π/4.

Next, we approximate the area by dividing the region into rectangles and summing up their areas. The height of each rectangle is the function value at the midpoint of the corresponding interval, and the width of each rectangle is the size of the interval.

By using this approximation method, we can estimate the area under the curve.

To find the exact area, we need to integrate the function over the given interval.

∫[0,π/4] sec^2 x dx

To integrate, we can use the identity:

∫ sec^2 x dx = tan x + C

where C is the constant of integration. Applying this identity, we have:

∫[0,π/4] sec^2 x dx = [ tan x ]_[0,π/4] = tan(π/4) - tan(0)

Since tan(π/4) = 1 and tan(0) = 0, we have:

∫[0,π/4] sec^2 x dx = 1 - 0 = 1

Therefore, the exact area under the curve y = sec^2 x, on the interval 0 ≤ x ≤ π/4, is 1 square unit.

To estimate the area below the curve y = sec^2 x, 0 ≤ x ≤ π/4, we can create a rough graph of the function and approximate the area of the region using rectangles.

First, let's plot the graph of y = sec^2 x over the given interval.

To graph y = sec^2 x, we need to understand the characteristics of the sec^2 x function. The graph of sec^2 x is similar to the graph of y = 1/x^2, with the only difference being that the sec^2 x function is periodic with period 2π.

Since we are interested in the interval 0 ≤ x ≤ π/4, we only need to focus on one period of the sec^2 x function. Let's graph the function over the interval 0 ≤ x ≤ 2π.

First, we can identify critical points and vertical asymptotes of the function. The critical points occur when sec^2 x = 0, which happens when cos^2 x = 1/sec^2 x = 0. This occurs when cos x = 0. We know that cos x = 0 at x = π/2 and x = 3π/2. These values represent vertical asymptotes of the function.

Next, we can plot some key points on the graph. We know that sec^2 x = 1/cos^2 x, so we can choose some values of x and calculate their corresponding values of sec^2 x.

For example:
- x = π/4: sec^2 (π/4) = 1/(cos^2 (π/4)) = 1/(1/2) = 2
- x = π/3: sec^2 (π/3) = 1/(cos^2 (π/3)) = 1/(1/4) = 4

Using these key points and the knowledge that the graph of sec^2 x is similar to y = 1/x^2, we can sketch a rough graph of y = sec^2 x over the interval 0 ≤ x ≤ 2π.

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Now, let's focus on the region beneath the curve for the interval 0 ≤ x ≤ π/4.

We can estimate the area of the region by dividing it into several rectangles. The more rectangles we use, the more accurate our estimate will be.

Let's divide the region into four equal rectangles:

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| 2 | 3
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| | 4
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To estimate the area of each rectangle, we need to calculate the width and height.

The width of each rectangle is Δx = (π/4 - 0)/4 = π/16 since we divided the interval into 4 equal parts.

To calculate the height of each rectangle, we can choose a representative point in each rectangle and evaluate the function at that point.

For example:
- In rectangle 1, we can choose x = π/64 (the left endpoint) and evaluate f(x) = sec^2 x at that point.
- In rectangle 2, we can choose x = 3π/64 (the midpoint of the first subinterval) and evaluate f(x) = sec^2 x at that point.
- In rectangle 3, we can choose x = 5π/64 (the midpoint of the second subinterval) and evaluate f(x) = sec^2 x at that point.
- In rectangle 4, we can choose x = 7π/64 (the right endpoint) and evaluate f(x) = sec^2 x at that point.

Using these representative points and evaluating the function at each point, we can calculate the height of each rectangle.

Now we can estimate the area of each rectangle by multiplying the width by the height of each rectangle. Finally, we can sum up the areas of all the rectangles to get an estimate of the total area.

To find the exact area of the region beneath the curve y = sec^2 x, 0 ≤ x ≤ π/4, we can use definite integration.

Using the definite integral, the exact area A is given by:

A = ∫(0 to π/4) sec^2 x dx

By integrating this expression, we can find the exact area.

I hope this helps you understand how to estimate and find the exact area beneath the given curve.