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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 6, Problema 6.2.35a

For the given regions R₁ and R₂, complete the following steps.


a. Find the area of region R₁.


R₁is the region in the first quadrant bounded by the line x=1 and the curve y=6x(2−x^2)^2; R₂ is the region in the first quadrant bounded the curve y=6x(2−x^2)^2and the line y=6x.

Guida verificata passo dopo passo
1
Identify the boundaries of region R₁. It is bounded by the line \(x=1\), the curve \(y=6x(2 - x^2)^2\), and the coordinate axes in the first quadrant. Since it is in the first quadrant, \(x\) ranges from 0 to 1.
Set up the integral for the area of region R₁. Because the region is bounded vertically by the curve \(y=6x(2 - x^2)^2\) and horizontally by \(x=1\), the area can be found by integrating the function \(y=6x(2 - x^2)^2\) with respect to \(x\) from 0 to 1:
\[\text{Area}_{R_1} = \int_0^1 6x(2 - x^2)^2 \, dx.\]
Before integrating, consider expanding the integrand \(6x(2 - x^2)^2\) to simplify the integral. First, expand \((2 - x^2)^2\) using the binomial formula:
\[(2 - x^2)^2 = 4 - 4x^2 + x^4.\]
Multiply this expansion by \$6x$ to get the integrand in polynomial form, then integrate term-by-term over the interval $[0,1]$ to find the area of region R₁.

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Definite Integrals for Area Calculation

Definite integrals are used to find the area under a curve between two points on the x-axis. By integrating the function that defines the curve over the given interval, we obtain the exact area of the region bounded by the curve and the x-axis or vertical lines.
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Determining the correct limits of integration is essential and depends on the boundaries of the region. For regions bounded by vertical lines and curves, the x-values where these boundaries intersect define the integration limits, ensuring the integral covers the entire region.
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When finding the area between two curves, subtract the lower function from the upper function and integrate the difference over the interval where they intersect. This method calculates the net area enclosed between the two curves accurately.
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