Question

Find the area under the graph of \(y=\sec ^{3} x\) from \(x=0\) to \(x=1\). (Round the answer to two significant digits.)

Short answer

The area is approximately 4.0.

Step-by-step solution

Understand the problem

We need to find the area under the curve of the function \( y = \sec^3 x \) from \( x = 0 \) to \( x = 1 \). This area can be found using definite integration.

Set up the integral

To find the area under the curve, set up the integral as follows: \[ \int_{0}^{1} \sec^3 x \, dx \]

Use Integration by Parts

Integration by parts is necessary to evaluate this integral. The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]. For this integral, let's choose:- \( u = \sec x \) implies \( du = \sec x \tan x \, dx \)- \( dv = \sec^2 x \, dx \) implies \( v = \tan x \)

Apply Integration by Parts

Apply the integration by parts formula:\[ \int \sec^3 x \, dx = \sec x \tan x - \int \tan x \cdot \sec x \tan x \, dx \]This simplifies to:\[ \int \sec^3 x \, dx = \sec x \tan x - \int \sec x \tan^2 x \, dx \]

Simplify the Integral Further

Use the identity \( \tan^2 x = \sec^2 x - 1 \) to simplify the integral:\[ \int \sec x \tan^2 x \, dx = \int \sec^3 x \, dx - \int \sec x \, dx \] Reorganize:\[ \int \sec^3 x \, dx = \sec x \tan x - (\int \sec^3 x \, dx - \int \sec x \, dx) \]

Solve for the Integral

Rearrange the equation to isolate \( \int \sec^3 x \, dx \):\[ 2 \int \sec^3 x \, dx = \sec x \tan x + \ln|\sec x + \tan x| \]\[ \int \sec^3 x \, dx = \frac{1}{2}(\sec x \tan x + \ln|\sec x + \tan x|) \]

Evaluate the Definite Integral

We now evaluate the integral from \(0\) to \(1\):\[ \int_{0}^{1} \sec^3 x \, dx = \frac{1}{2}\left( [\sec x \tan x + \ln|\sec x + \tan x|] \Big|_0^1 \right) \]Calculate separately for \( x = 1 \) and \( x = 0 \) and find the difference.

Calculate Each Boundary

For \( x = 1 \), find \( \sec 1 \cdot \tan 1 + \ln|\sec 1 + \tan 1| \).For \( x = 0 \), find \( \sec 0 \cdot \tan 0 + \ln|\sec 0 + \tan 0| \). Substitute and simplify to find:\[ \frac{1}{2}([\sec(1)\tan(1) + \ln|\sec(1) + \tan(1)|] - [0]) \]

Finish Calculations and Round

Finally, calculate the expression:\[ \frac{1}{2} \left( \sec(1)[\tan(1) + \ln|\sec(1) + \tan(1)|] \right) \approx 3.99 \] Round the calculated area to two significant digits, resulting in approximately 4.0.

Key concepts

Integration by Parts

Integration by parts is a fundamental technique in calculus used to solve integrals where the standard methods are not easily applicable. The idea is to transform the original integral into a simpler form using the formula:

• \( \int u \, dv = uv - \int v \, du \)

Here, \( u \) and \( dv \) are parts of the original integral that are chosen strategically to simplify the process. Typically, one chooses \( u \) to be a function that becomes simpler when differentiated, and \( dv \) is chosen to be easily integrable.
In the given problem, we chose \( u = \sec x \) and \( dv = \sec^2 x \, dx \). When applying integration by parts, this choice led us to break down the original complex integral into more manageable parts. This technique is particularly useful when dealing with functions involving products of polynomial and transcendental functions, as seen in many calculus problems.

Area Under the Curve

Finding the area under a curve represented by a function between two points is a common application of definite integration. In essence, computing the integral from a lower limit \( a \) to an upper limit \( b \) of a function \( f(x) \) gives the accumulated area under the curve of \( f(x) \) between those points.
For the function \( y = \sec^3 x \), we needed to find the area under this curve from \( x = 0 \) to \( x = 1 \). Setting up the integral as \( \int_{0}^{1} \sec^3 x \, dx \) allows us to evaluate the total area bounded by the curve, the x-axis, and the vertical lines at \( x = 0 \) and \( x = 1 \).
The result after evaluating this definite integral gives the area, which is rounded to two significant digits to provide a precise approximation. This process of determining the area under a curve is a powerful tool in mathematics, utilized in various fields such as physics, engineering, and economics.

Trigonometric Integrals

Trigonometric integrals involve the integration of functions containing trigonometric functions such as sine, cosine, secant, etc. These can often be challenging due to the periodic and oscillatory nature of trigonometric functions.
In our exercise, we dealt with \( \sec^3 x \), which is a higher power of the secant function. Such integrals often require special techniques or identities for simplification. For instance, using the identity \( \tan^2 x = \sec^2 x - 1 \) was crucial for further simplifying our integral.
The common strategy for handling these integrals, aside from integration by parts, includes recognizing trigonometric identities that can transform or simplify the functions involved. Understanding these identities is key to tackling a wide range of integrals involving trigonometric expressions. It opens the door to solving otherwise complex problems methodically.