Problem 77 - Calculus: Early Transcendentals

Question

Evaluate \( \displaystyle \int^2_{-2} (x + 3) \sqrt{4 - x^2} \,dx \) by writing it as a sum of two integrals and interpreting one of those integrals in terms of an area.

Step-by-Step Solution

Answer

The integral evaluates to \(6\pi\).

Step 1: Breakdown the Integral into Two Parts

We start by rewriting the original integral as the sum of two separate integrals: \[ \int_{-2}^{2} (x + 3) \sqrt{4 - x^2} \, dx = \int_{-2}^{2} x \sqrt{4 - x^2} \, dx + \int_{-2}^{2} 3 \sqrt{4 - x^2} \, dx \]

Step 2: Interpret the Second Integral in Terms of Areas

The integral \( \int_{-2}^{2} 3 \sqrt{4 - x^2} \, dx \) can be related to the area of a semicircle. The function \( f(x) = \sqrt{4 - x^2} \) represents the upper half of a circle with radius 2, and the interval from \(-2\) to \(2\) covers the entire semicircle. The second integral can thus be expressed as 3 times the area of a semicircle with radius 2:\[ 3 \times \frac{1}{2} \pi r^2 = 3 \times \frac{1}{2} \pi (2)^2 = 6\pi \]

Step 3: Evaluate the First Integral

The integral \( \int_{-2}^{2} x \sqrt{4 - x^2} \, dx \) is an odd function over a symmetric interval around zero. Thus, due to symmetry, its value is zero:\[ \int_{-2}^{2} x \sqrt{4 - x^2} \, dx = 0 \]

Step 4: Sum the Integrals

Combine the results of the two integrals: The first integral evaluates to zero, and the second integral evaluates to \(6\pi\), giving:\[ \int_{-2}^{2} (x + 3) \sqrt{4 - x^2} \, dx = 0 + 6\pi = 6\pi \]

Key Concepts

Definite Integrals

Definite integrals are a cornerstone of integral calculus. They provide a way to calculate the accumulation of a quantity, often representing the area under a curve on a graph. When we compute a definite integral from a lower bound to an upper bound, we are essentially summing up infinitely small parts of a function over that interval. For example, the original exercise, \[ \int_{-2}^{2} (x + 3) \sqrt{4 - x^2} \, dx \] involves finding the total "area" or accumulated value created by the function over the range from \(-2\) to \(2\).

• The bounds indicate the interval over which the function is integrated.
• The result of evaluating this definite integral is a specific number, often representing area in practical problems.

Integral calculus often involves splitting complicated integrals into more manageable pieces, just as done by dividing into two separate integrals. This simplification helps calculate areas or understand behavior analytically over each portion of the interval.

Symmetry in Integrals

Symmetry can significantly simplify the evaluation of definite integrals. When a function exhibits symmetry over the integration interval, it can simplify calculations, sometimes even reaching results without lengthy computations. Considering the integral step from the exercise,\[ \int_{-2}^{2} x \sqrt{4 - x^2} \, dx \]is identified as an odd function. An odd function, denoted numerically by \( f(-x) = -f(x) \), reflects across the origin. If the integration limits are symmetric about zero, such as from \(-a\) to \(a\),

• The integral of an odd function over a symmetric interval results in zero.

The symmetry theorem uses the peculiar property of these odd functions and helps in deriving simplifications as seen where the odd part of the split integral results in 0.

Geometric Interpretation of Integrals

Geometric interpretation provides a visual understanding of what integrals represent. In the context of the exercise, understanding the integral \[ \int_{-2}^{2} 3 \sqrt{4 - x^2} \, dx \]relies on its geometric visualization. The function \(f(x) = \sqrt{4 - x^2} \)represents the upper half of a circle with a radius of 2. Therefore, when evaluating the integral over the interval from \(-2\) to \(2\), we are calculating the area of a semicircle.

• The integral returns 3 times the area of the semicircle, which can be computed as \( 3 \times \frac{1}{2} \pi \times (2)^2 \).
• The geometric interpretation links the algebraic process of integration to tangible areas and forms.

This is a powerful approach as it allows students to visualize and internalize integration, aiding in understanding and solving various problems using geometrical insights.