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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 5 - IntegrationProblema 5.2.41
Capitolo 5, Problema 5.2.41

Definite integrals Use geometry (not Riemann sums) to evaluate the following definite integrals. Sketch a graph of the integrand, show the region in question, and interpret your result.                                                                                                                                      
                                                                                                                                                                                       
 ∫₋₁² ( ―|𝓍| ) d𝓍

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Identify the integrand function: here it is \(f(x) = -|x|\), which means the graph is the negative of the absolute value function. This creates a 'V' shape opening downward with its vertex at the origin (0,0).
Sketch the graph of \(f(x) = -|x|\) over the interval \([-1, 2]\). From \(x = -1\) to \(x = 0\), the function decreases linearly from \(-1\) to \(0\), and from \(x = 0\) to \(x = 2\), it decreases linearly from \(0\) to \(-2\).
Interpret the definite integral \(\int_{-1}^{2} -|x| \, dx\) as the net area between the curve and the x-axis from \(x = -1\) to \(x = 2\). Since the function is negative or zero in this interval, the integral corresponds to the negative of the total area of the region bounded by the curve and the x-axis.
Divide the region into two geometric shapes: a triangle from \(x = -1\) to \(x = 0\) with base length 1 and height 1, and another triangle from \(x = 0\) to \(x = 2\) with base length 2 and height 2. Calculate the area of each triangle using the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\).
Sum the areas of the two triangles to find the total area under the curve (above the function since it is negative). The value of the definite integral is then the negative of this total area, reflecting the fact that the function lies below the x-axis.

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Definite Integral as Area Under a Curve

A definite integral represents the net area between the graph of a function and the x-axis over a given interval. Areas above the x-axis contribute positively, while areas below contribute negatively. Understanding this geometric interpretation allows evaluation of integrals without limits or sums.
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Definition of the Definite Integral

Graphing the Integrand Function

Sketching the graph of the integrand, here f(x) = -|x|, helps visualize the region whose area is being calculated. The absolute value creates a V-shape, and the negative sign reflects it below the x-axis. This visualization is crucial for identifying shapes and calculating areas geometrically.
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Graph of Sine and Cosine Function

Using Geometric Shapes to Evaluate Integrals

When the integrand forms simple geometric shapes (triangles, rectangles), the definite integral can be found by calculating the area of these shapes. For f(x) = -|x| on [-1,2], the region forms triangles below the x-axis, so applying area formulas for triangles yields the integral value efficiently.
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Integration Using Partial Fractions
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