Question

Evaluate the iterated integrals. $$ \int_{1 / 2}^{1} \int_{0}^{2 x} \cos \left(\pi x^{2}\right) d y d x $$

Short answer

The value of the integral is \(-\frac{\sqrt{2}}{2\pi}\).

Step-by-step solution

Identify the Order of Integration

The given integral is an iterated integral with limits \(\int_{1/2}^{1} \int_{0}^{2x} \cos(\pi x^2) \ dy \ dx\). Here, the integration with respect to \(y\) is performed first followed by integration with respect to \(x\).

Perform the Inner Integral

For the inner integral \(\int_0^{2x} \cos(\pi x^2) \ dy\), where \(\cos(\pi x^2)\) is a constant with respect to \(y\), simply multiply the cosine term by the difference of the limits of \(y\):\[\int_0^{2x} \cos(\pi x^2) \ dy = \cos(\pi x^2) [y]_0^{2x} = \cos(\pi x^2)(2x - 0) = 2x \cos(\pi x^2)\]

Perform the Outer Integral

Substitute the result from Step 2 into the outer integral:\[\int_{1/2}^{1} 2x \cos(\pi x^2) \ dx\]Let \(u = \pi x^2\), then \(du = 2\pi x \, dx\) or \(dx = \frac{du}{2\pi x}\). Substitute and simplify:\[\int \cos(u) \cdot \frac{1}{\pi} \ du = \frac{1}{\pi} \sin(u) + C\]Now back-substitute \(u = \pi x^2\):\[\frac{1}{\pi} \sin(\pi x^2)\]Evaluate this from \(x = 1/2\) to \(x = 1\):\[\frac{1}{\pi}[\sin(\pi \times 1^2) - \sin(\pi \times (1/2)^2)] = \frac{1}{\pi}[\sin(\pi) - \sin(\pi/4)] = \frac{1}{\pi}(0 - \frac{\sqrt{2}}{2}) = -\frac{\sqrt{2}}{2\pi}\]

Combine Results

The solution of the original iterated integral is the result from Step 3: \[-\frac{\sqrt{2}}{2\pi}\]

Key concepts

Integral Calculus

Integral Calculus is a branch of mathematics that focuses on the concept of integration. It allows us to compute areas under curves, solve differential equations, and find volumes. With integral calculus, we explore the reverse process of differentiation, which is called integration.

When dealing with iterated integrals like in our exercise, we perform integration repeatedly over multiple variables in a specified order. Here, we first integrate with respect to one variable, holding the other constant, and then integrate the resulting expression with respect to the other variable. In our example, we integrated first with respect to the variable \(y\) and then \(x\).

Iterated integrals can look daunting, but by focusing on one integral at a time, they become manageable. The main goal is to simplify the problem step by step until you reach the final result. Each step involves straightforward arithmetic when boundaries and constants in the integral are clear.

Trigonometric Functions

Trigonometric functions are fundamental to calculus, essential for modeling periodic phenomena such as sound and light waves. In this exercise, we encounter \(\cos(\pi x^2)\), a trigonometric function influencing how the integral behaves.

These functions, including sine and cosine, emerged from the study of angles in triangles but evolved to represent much more in calculus. Trigonometric functions are periodic, which means they repeat their values in regular intervals, and this property often simplifies calculation, especially when computing integrals.

The cosine function, specifically, is known for its unique wave-like symmetry. When integrated, it yields another trigonometric function. This symmetry can help reduce complexity when finding antiderivatives during integration.

• Understand the periodic nature: solutions can often simplify.
• Symmetry properties: leverage them to make calculations easier.
• Applications: help in solving real-world problems involving waves.

Integration Techniques

The complexity of integration problems varies, and mastering different integration techniques is crucial. For our iterated integrals, we applied both substitution and the fundamental theorem of calculus.

The substitution technique, also known as \(u\)-substitution, plays a key role in simplifying integrands. By changing the variable, substitution can transform a difficult integral into an easier problem.

In this exercise, we substitute \(u = \pi x^2\) to simplify the integration of \(\cos(\pi x^2)\). This transformation allows the complex trigonometric expression to be integrated more easily. Substitution helps not only in finding antiderivatives but also in evaluating definite integrals by altering the limits accordingly.

• Substitution: Change of variables to simplify integration.
• Antiderivatives: Identify functions whose derivative gives back the integrand.
• Boundaries: Adjust limits according to substitution changes.

Familiarity with these techniques will enhance your problem-solving skills in calculus, making challenging integrals approachable and manageable.