Question

Evaluate the integral and subsequent iterated integral. (a) \(\int_{0}^{\pi}(2 x \cos y+\sin x) d x\) (b) \(\int_{0}^{\pi / 2} \int_{0}^{\pi}(2 x \cos y+\sin x) d x d y\)

Short answer

(a) Result is \( \pi^2 \cos y + 2 \). (b) Result is \( \pi(\pi + 1) \).

Step-by-step solution

Evaluate the inner integral for (a)

For the integral \( \int_{0}^{\pi}(2 x \cos y+\sin x) d x \), we treat \( y \) as a constant and integrate with respect to \( x \). Splitting the integral, we have \( \int_{0}^{\pi} 2x \cos y \, dx + \int_{0}^{\pi} \sin x \, dx \).

Compute the integral of the cosine term

Evaluate \( \int_{0}^{\pi} 2x \cos y \, dx \). This integral simplifies to \( 2\cos y \int_{0}^{\pi} x\, dx \). Using the power rule, \( \int x\, dx = \frac{x^2}{2} \), so \( \int_{0}^{\pi} x\, dx = \frac{x^2}{2} \Big|_0^{\pi} = \frac{\pi^2}{2} \). Thus, \( 2\cos y \cdot \frac{\pi^2}{2} = \pi^2 \cos y \).

Compute the integral of the sine term

For \( \int_{0}^{\pi} \sin x \, dx \), the antiderivative of \( \sin x \) is \( -\cos x \). Thus, \( -\cos x \Big|_0^{\pi} = (-\cos \pi) - (-\cos 0) = 1 + 1 = 2 \).

Combine results for integral (a)

Combine the results from Steps 2 and 3. \( \int_{0}^{\pi}(2 x \cos y+\sin x) d x = \pi^2 \cos y + 2 \). This is the evaluated integral for part (a).

Set up the iterated integral for (b)

For the iterated integral \( \int_{0}^{\pi / 2} \int_{0}^{\pi} (2x \cos y + \sin x) \, dx \, dy \), we use the result from part (a) as the inner integral. This becomes \( \int_{0}^{\pi / 2} (\pi^2 \cos y + 2) \, dy \).

Compute the iterated integral for (b)

The integral simplifies to two separate integrals: \( \pi^2 \int_{0}^{\pi / 2} \cos y \, dy + \int_{0}^{\pi / 2} 2 \, dy \). The antiderivative of \( \cos y \) is \( \sin y \), so \( \pi^2 \sin y \Big|_0^{\pi/2} = \pi^2(1 - 0) = \pi^2 \). For the second integral, \( 2y \Big|_0^{\pi/2} = \pi \).

Combine results for integral (b)

Add the results from Step 6: \( \pi^2 + \pi \). This is the final result of the iterated integral \( \int_{0}^{\pi / 2} \int_{0}^{\pi} (2x \cos y + \sin x) \, dx \, dy = \pi(\pi + 1) \).

Key concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus that allow us to calculate the exact accumulation of quantities. When we compute a definite integral, we're finding the net area under a curve between two points, often symbolized by \(a\) and \(b\). This process is essential in numerous fields such as physics, engineering, and economics.

Evaluating a definite integral involves calculating the difference between the antiderivative evaluated at the upper and lower bounds. The standard notation is: \[\int_{a}^{b} f(x)\, dx = F(b) - F(a)\]\ where \(F(x)\) is the antiderivative of \(f(x)\).

• The integration removes the uncertainty about the area due to approximation.
• The definite integral accounts for every bit of area precisely, including little regions under the curve.

In our exercise, integrating from 0 to \(\pi\) involves finding the components of a function like \(2x \cos y + \sin x\). Each part is integrated separately using appropriate techniques, ensuring an accurate total representation of area. The result gives valuable information about the behaviour and characteristics of the function over the specified interval.

Trigonometric Integration

Trigonometric integration refers to integrating functions that involve trigonometric functions such as sine and cosine. These integrations utilize specific identities and techniques to simplify and solve integrals, making them more manageable.

• Using the antiderivatives of basic trigonometric functions is common: \(\int \sin x\, dx = -\cos x + C\) and \(\int \cos x\, dx = \sin x + C\).
• Identities like \(\sin^2 x + \cos^2 x = 1\) can aid in transformation.

In our specific solution, we managed the integral involving \(\cos y\) multiplied by a polynomial in \(x\) by treating \(y\) as constant. This is key in iterated integrals—when one variable is constant while addressing the other.

The integration of terms like \(2x \cos y + \sin x\) is split into fundamental cosines and sines for efficiency. Recognizing when a trigonometric identity can be applied or separating integrals simplifies calculating an otherwise complex expression.

Calculus Problem Solving

Solving calculus problems, especially involving integrals, requires a structured approach. This structure involves identifying the type of integral and the appropriate techniques for solving it.

When facing complex integrals like the ones in the exercise, consider the following steps:

• Break down the integral into more manageable parts or simpler terms.
• Utilize known identities or rules (such as trigonometric identities or power rules) to simplify integrals.
• Compute each segment methodically, ensuring intermediate steps are correct before combining them for a final solution.
• For iterated integrals, solve the inner integral completely before addressing the outer integral.

Analyzing our original exercise, the problem was initially divided into easier components by separately evaluating \(2x \cos y\) and \(\sin x\). Each integral was handled using rules of integration and substitution where necessary. The idea is to simplify, strategize, and then synthesize—crafting all individual pieces into the complete solution.

This methodical approach is crucial in calculus problem solving, especially for complex, multi-variable integrals, ensuring accuracy and comprehension.