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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
6장, 문제 6.2.61

Find the area of the region described in the following exercises.


The region in the first quadrant bounded by y=5/2−1/x and y=x

검증된 단계별 안내
1
First, identify the curves that bound the region: the line \(y = x\) and the curve \(y = \frac{5}{2} - \frac{1}{x}\), both in the first quadrant.
Next, find the points of intersection between the two curves by setting \(x = \frac{5}{2} - \frac{1}{x}\) and solving for \(x\). Multiply both sides by \(x\) to clear the denominator, resulting in a quadratic equation.
Solve the quadratic equation to find the \(x\)-coordinates of the intersection points. Since the region is in the first quadrant, only consider positive solutions for \(x\).
Set up the integral for the area between the curves from the smaller intersection point \(x = a\) to the larger intersection point \(x = b\). The area is given by \(\int_a^b \left( \left( \frac{5}{2} - \frac{1}{x} \right) - x \right) \, dx\).
Finally, evaluate the integral by integrating each term separately and then subtracting the values at the bounds \(a\) and \(b\) to find the area of the region.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
5m

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Finding Points of Intersection

To determine the area between two curves, first find where they intersect by setting the functions equal to each other. Solving y = 5/2 - 1/x and y = x gives the limits of integration, which define the region in the first quadrant.
추천 영상:
04:50
Critical Points

Setting Up the Definite Integral for Area

The area between two curves y = f(x) and y = g(x) over an interval [a, b] is found by integrating the difference f(x) - g(x). Here, identify which function is on top in the first quadrant and integrate their difference between the intersection points.
추천 영상:
가이드 코스
05:43
Definition of the Definite Integral

Properties of Functions in the First Quadrant

Since the region is in the first quadrant, both x and y are positive. This restricts the domain and range for the problem, ensuring the limits of integration and the functions are considered only where x > 0 and y > 0.
추천 영상:
가이드 코스
06:21
Properties of Functions