Problem 34 - Applied Calculus

Question

Compute the definite integral \(\int_{0}^{4} \cos \sqrt{x} d x\) and interpret the result in terms of areas.

Step-by-Step Solution

Answer

The signed area is \( 2 \sin(2) + 2\cos(2) - 2 \).

Step 1: Interpret the Integral

The integral \( \int_{0}^{4} \cos \sqrt{x} \, dx \) represents the signed area between the curve \( y = \cos \sqrt{x} \) and the \( x \)-axis, from \( x = 0 \) to \( x = 4 \).

Step 2: Substitution Method

To solve \( \int_{0}^{4} \cos \sqrt{x} \, dx \), use the substitution \( u = \sqrt{x} \). Then \( du = \frac{1}{2\sqrt{x}} \, dx \) or equivalently, \( dx = 2u \, du \).

Step 3: Change Limits of Integration

With \( u = \sqrt{x} \), the limits change from \( x = 0 \) to \( u = 0 \) and from \( x = 4 \) to \( u = 2 \).

Step 4: Substitute and Integrate

Substitute \( x = u^2 \) and \( dx = 2u \, du \) into the integral. This gives: \[\int_{0}^{2} \cos(u) \, 2u \, du = 2 \int_{0}^{2} u \cos(u) \, du\]

Step 5: Integration by Parts

To integrate \( 2 \int_{0}^{2} u \cos(u) \, du \), use integration by parts. Let \( v = u \) and \( dw = \cos(u) \, du \). Then \( dv = du \) and \( w = \sin(u) \).Substituting, we get:\[2 \left[ u \sin(u) \right]_{0}^{2} - 2 \int_{0}^{2} \sin(u) \, du\]

Step 6: Evaluate Definite Integrals

Calculate the boundary values:\[[u \sin(u)]_{0}^{2} = 2\sin(2) - 0 = 2\sin(2)\]Now, solve \( 2 \int_{0}^{2} \sin(u) \, du \):\[\int \sin(u) \, du = -\cos(u)\]\[2 \left[ -\cos(u) \right]_{0}^{2} = -2 \left( -\cos(2) + 1 \right) = 2 - 2\cos(2)\]

Step 7: Combine Results

Combine the evaluated parts:\[2 \sin(2) - (2 - 2 \cos(2)) = 2 \sin(2) + 2\cos(2) - 2\]

Step 8: Interpret Result

The result \( 2 \sin(2) + 2\cos(2) - 2 \) gives the signed area between the \( x \)-axis and the curve \( \cos \sqrt{x} \) from \( x = 0 \) to \( x = 4 \).

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used to solve integrals involving products of functions. It is based on the product rule for differentiation and is particularly helpful when one part of the product becomes simpler upon differentiation. The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]In the exercise, we applied integration by parts to solve the integral \( \int_{0}^{2} u \cos(u) \, du \). We chose \( u = u \) and \( dv = \cos(u) \, du \), then calculated \( v = \sin(u) \) and \( du = du \). This substitution simplified the integration process, reducing it to simpler terms. After solving, we obtained the expression as part of the overall solution.When using integration by parts, always aim to simplify the integral. Often, you will differentiate the function that simplifies upon differentiation and integrate the other. This strategic choice often leads to a solvable equation.

Substitution Method

The substitution method is a clever technique used to simplify integration by transforming the variable of integration. It is particularly valuable when a function and its derivative both appear in the integral. By substituting a part of the integral expression with a new variable, we can transform a complicated integral into a simple one.In the given exercise, we used the substitution \( u = \sqrt{x} \). This choice transformed the original integral \( \int_{0}^{4} \cos \sqrt{x} \, dx \) into \( 2 \int_{0}^{2} u \cos(u) \, du \). By changing the variables, the integral became easier to manage. We also adjusted our initial integration limits to reflect this substitution, converting \( x=0 \) to \( u=0 \) and \( x=4 \) to \( u=2 \).Choosing the right substitution often simplifies the integral immensely. As you practice, identifying the right substitution will become more intuitive, especially when you see repeated structures or compositions that can be altered by new variables.

Signed Area

The concept of signed area is crucial when interpreting the result of a definite integral geometrically. It represents the algebraic sum of areas above and below the x-axis for a given range of integration. If the curve lies above the x-axis, the area contributes positively and vice versa.In the original exercise, we computed the signed area of \( \int_{0}^{4} \cos \sqrt{x} \, dx \). Here, the integral's result, \( 2 \sin(2) + 2\cos(2) - 2 \), is a representation of this area between the curve \( y = \cos \sqrt{x} \) and the x-axis, from \( x = 0 \) to \( x = 4 \).Understanding signed area helps relate the mathematical computation of integrals to real-world concepts. It provides insights into how much "positive area" is gained versus lost due to negative contributions under the axis. This approach is essential in calculus when interpreting physical scenarios, from physics to finance.