Question

Let \(\mathrm{f}(\mathrm{x})=\left\\{\begin{array}{ll}-\mathrm{x}-1 & \text { if }-3 \leq \mathrm{x}<0 \\ -\sqrt{1-\mathrm{x}^{2}} & \text { if } \quad 0 \leq \mathrm{x} \leq 1\end{array}\right.\) Evaluate \(\int_{-3}^{1} f(x) d x\) by interpreting the integral as a difference of areas.

Short answer

Answer: The integral of the piecewise function from -3 to 1 is -15/2 - (1/2)π.

Step-by-step solution

Split the integral into two parts

Since f(x) is defined as a piecewise function, we need to split the integral into two parts, corresponding to the two conditions on x: - For -3 ≤ x < 0, f(x) = -x - 1 - For 0 ≤ x ≤ 1, f(x) = -√(1-x^2). Therefore, the integral becomes the sum of two separate integrals: \(\int_{-3}^1 f(x) dx = \int_{-3}^0 (-x - 1) dx + \int_{0}^1 -\sqrt{1-x^2} dx\).

Evaluate the first integral (for -3 ≤ x < 0)

We will first evaluate the integral for the first condition: \(\int_{-3}^0 (-x - 1) dx\). To do this, we will find the antiderivative of the function inside, F(x): F(x) = \(-\frac{x^2}{2} - x + C\). Then we will apply the limits of integration according to the fundamental theorem of calculus: \(-\left[\left(-\frac{1}{2} \cdot 0^2 - 0\right) - \left(-\frac{1}{2} \cdot (-3)^2 - (-3)\right)\right] = -\left[0 + \left(\frac{9}{2} + 3\right)\right] = -\left[\frac{15}{2}\right]\).

Evaluate the second integral (for 0 ≤ x ≤ 1)

Now we will evaluate the integral for the second condition: \(\int_{0}^1 -\sqrt{1-x^2} dx\). Due to the complexity of this integral, it is easiest to evaluate it geometrically. This integral corresponds to the area of the lower semicircle of radius 1 centered at the origin. Thus, the area of this lower semicircle can be computed as half of a full circle's area: \(-\frac{1}{2}\pi(1)^2 = -\frac{1}{2}\pi\).

Sum the two separate integral results

Now we will sum the two areas computed in steps 2 and 3: \(\int_{-3}^1 f(x) dx = -\left[\frac{15}{2}\right] - \frac{1}{2}\pi\). Thus, the definite integral of f(x) from -3 to 1, interpreted as a difference of areas, is equal to: \(\int_{-3}^1 f(x) dx = -\frac{15}{2} - \frac{1}{2}\pi\).

Key concepts

Piecewise Functions

Piecewise functions, like the one given in the exercise (\f\( \text{f}(\text{x}) \f\)), are mathematical expressions defined by multiple sub-functions, each applying to a certain interval of the main function's domain. This particular structure allows one to create complex functions that can model different behaviors over different intervals.

In order to evaluate the definite integral of a piecewise function from \f\( \text{a} \f\) to \f\( \text{b} \f\), as with \f\( \text{f}(\text{x}) \f\) from \f\( -3 \f\) to \f\( 1 \f\) in our problem, we need to consider the interval boundaries of the piecewise sub-functions. We must split the integral at these boundaries and treat each piece according to its own rule. This technique ensures that we correctly accommodate the function's changing behavior across its domain.

When faced with a real-world problem involving piecewise functions, it's crucial to understand which part of the function applies to the given situation. As seen in the solution, for the intervals \f\( [-3, 0) \f\) and \f\( [0, 1] \f\), we deal with two different expressions requiring separate integration.

Antiderivative

An antiderivative of a function is another function that 'undoes' the effect of taking a derivative. For instance, when we say that the antiderivative of \f\( \text{f}(\text{x}) \f\) is \f\( \text{F}(\text{x}) \f\), we mean that the derivative of \f\( \text{F}(\text{x}) \f\), denoted as \f\( \text{F}'(\text{x}) \f\), yields \f\( \text{f}(\text{x}) \f\).

Calculating the antiderivative is the core step in finding the exact area under the curve described by \f\( \text{f}(\text{x}) \f\) between two points on a graph. This procedure, showcased in our step-by-step solution, involves finding the antiderivative \f\( \text{F}(\text{x}) \f\) and then using it in the Fundamental Theorem of Calculus to determine the integral's value. An important thing to remember is that the antiderivative is not unique; a constant of integration, \f\( \text{C} \f\), is always included since derivatives of constants are zero.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, two principal operations in calculus. It asserts that the definite integral of a function can be computed using its antiderivative. Essentially, if \f\( \text{F} \f\) is the antiderivative of \f\( \text{f} \f\), then the definite integral of \f\( \text{f} \f\) from \f\( \text{a} \f\) to \f\( \text{b} \f\) is given by \f\( \text{F}(\text{b}) - \text{F}(\text{a})\f\).

This theorem was applied in the given exercise to evaluate the definite integral of our piecewise function. This process entails computing the antiderivative of each sub-function over its corresponding interval, followed by applying the relevant limits of integration to calculate the difference in the antiderivative's values at those limits. The two results are then added to obtain the total area under the piecewise function across its entire interval.

Geometric Interpretation of Integrals

The geometric interpretation of integrals allows us to perceive definite integrals as net areas under a curve. When we integrate a function over an interval, we are effectively summing infinitely many infinitesimally thin rectangles under the curve. The net area can be positive, negative, or even zero, depending on whether the curve is above or below the horizontal axis.

In the context of our exercise, evaluating the integral geometrically is especially helpful for the segment involving the semicircle. By recognizing the geometric shape described by the function \f\( -\text{√}(1-\text{x}^2) \f\), we can directly use the formula for the area of a circle rather than working through a more complex integral calculus. This approach illustrates a powerful method to simplify certain types of definite integrals, where straightforward geometry can provide the answer in place of traditional antiderivatives and Fundamental Theorem calculations.