Problem 53 - Biocalculus Calculus for the Life Sciences

Question

Evaluate \(\int_{-2}^{2}(x+3) \sqrt{4-x^{2}} d x\) 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: Break Down the Integral into Sum of Two Integrals

Start by rewriting the integral \( \int_{-2}^{2}(x+3) \sqrt{4-x^{2}} \, dx \) as the sum of two separate integrals. These would be \( \int_{-2}^{2}x \sqrt{4-x^{2}} \, dx + \int_{-2}^{2}3 \sqrt{4-x^{2}} \, dx \).

Step 2: Interpret the First Integral

Focus on the first integral \( \int_{-2}^{2}x \sqrt{4-x^{2}} \, dx \). Notice that the integrand, \( x \sqrt{4-x^{2}} \), is an odd function over a symmetric interval about the origin \([-2, 2]\). The integral of an odd function over a symmetric interval is zero. Therefore, this integral is zero.

Step 3: Evaluate the Second Integral as an Area

Now consider the second integral \( \int_{-2}^{2}3 \sqrt{4-x^{2}} \, dx \). Recognize that \( \sqrt{4-x^{2}} \) describes the top half of a circle with radius 2 centered at the origin. Thus, the integral \( 3 \cdot \int_{-2}^{2} \sqrt{4-x^{2}} \, dx \) represents the area of the semicircle multiplied by 3. The area of a circle with radius 2 is \( \pi \cdot 2^2 = 4\pi \), and thus, the area of the upper half is \( 2\pi \). Multiplying by 3 gives \( 6\pi \).

Step 4: Sum the Two Integral Results

Since the first integral was zero, the sum of the two integrals is just the result of the second integral, which is \( 6\pi \).

Step 5: Final Result

Put it all together to state the final result of evaluating the integral. The integral \( \int_{-2}^{2}(x+3) \sqrt{4-x^{2}} \, dx \) evaluates to \( 6\pi \).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus that calculate the accumulation of quantities, such as areas under a curve, along an interval on the x-axis. Given a function \( f(x) \), its definite integral from \( a \) to \( b \) is represented as \( \int_{a}^{b} f(x) \, dx \). In this notation, \( a \) and \( b \) are the limits of integration, defining the interval over which the function is evaluated.

The result of a definite integral is typically a real number rather than a function, which differentiates them from indefinite integrals that yield a family of functions. To solve a definite integral, it's common to use the Fundamental Theorem of Calculus, which connects differentiation and integration. This theorem states that if \( F \) is the antiderivative of \( f \), then:

• \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \)

This principle allows us to compute the total accumulation of \( f(x) \) over the interval \([a, b]\). In the exercise above, the integral was split into two parts, each calculated separately, to simplify the finding of their sum.

Even and Odd Functions

In mathematics, functions can be classified as even or odd based on their symmetry properties. Identifying whether a function is even or odd can be particularly helpful in simplifying definite integrals when the interval is symmetric about the y-axis, such as \([-a, a]\).

- **Even Functions**: A function \( f(x) \) is even if \( f(-x) = f(x) \) for all \( x \) in its domain. Graphically, even functions are symmetrical with respect to the y-axis. When integrating an even function over a symmetric interval \([-a, a]\), the result is double the integral from \( 0 \) to \( a \), i.e., \( 2 \int_{0}^{a} f(x) \, dx \).- **Odd Functions**: A function \( f(x) \) is odd if \( f(-x) = -f(x) \) for all \( x \). These functions exhibit symmetry around the origin. When an odd function is integrated over a symmetric interval, the result is zero.

In the given exercise, the function \( x \sqrt{4-x^2} \) was identified as an odd function. Therefore, the integral of this function from \(-2\) to \(2\) was zero, simplifying the calculation significantly.

Geometric Interpretation of Integrals

Integrating functions doesn't just yield numerical results; it can often give us a geometric picture, which is particularly intuitive in understanding calculus. The integral of a function over an interval represents the net area under its curve, relative to the x-axis.

In the exercise, this idea is clearly illustrated by the second integral, \( \int_{-2}^{2} 3 \sqrt{4-x^2} \, dx \). The term \( \sqrt{4-x^2} \) corresponds to the equation of a circle's upper half with radius 2, centered at the origin. The integral over the interval \([-2, 2]\) computes the area of this semicircle.

- The area of a full circle is \( \pi r^2 \) where \( r \) is the radius.- The complete area of a circle with radius 2 is \( 4\pi \).- Therefore, the area of the upper half is \(2\pi\).

When multiplied by the constant 3, the integral represents three times this semicircle's area, which results in \(6\pi\). Such geometric interpretations provide valuable insights into these calculations, ensuring a deeper understanding of their physical meaning.