Chapter 5: Problem 25
Let \(A\) denote the area between the graph of \(f(x)=\sqrt{x}\) and the interval \([0,1],\) and let \(B\) denote the area between the graph of \(f(x)=x^{2}\) and the interval \([0,1] .\) Explain geometrically why \(A+B=1\).
Short Answer
Expert verified
The sum of areas under the curves \( f(x) = \sqrt{x} \) and \( f(x) = x^2 \) over \([0, 1]\) equals 1 due to geometric division of a unit square.
Step by step solution
01
Understand the Problem
We need to find the areas under two different functions, \( f(x) = \sqrt{x} \) and \( f(x) = x^2 \), over the interval \([0, 1]\). We will express these areas as \( A \) and \( B \) respectively, and then show that \( A + B = 1 \) using geometric reasoning.
02
Visualize the Graphs
Plot the functions \( f(x) = \sqrt{x} \) and \( f(x) = x^2 \) on the same set of axes over the interval \([0, 1]\). Observe that \( f(x) = \sqrt{x} \) starts from 0 and reaches 1 at \( x = 1 \), while \( f(x) = x^2 \) also starts from 0 but curves upwards more sharply, reaching 1 at \( x = 1 \). The curve \( \sqrt{x} \) is always above \( x^2 \) in this interval.
03
Calculate Area A Under \( f(x) = \sqrt{x} \)
The area \( A \) is between the graph of \( f(x) = \sqrt{x} \) and the \( x \)-axis from \( x = 0 \) to \( x = 1 \). Mathematically, the area \( A \) can be found by the integral: \[ A = \int_0^1 \sqrt{x} \, dx. \] Compute this integral to find \( A = \frac{2}{3} \).
04
Calculate Area B Under \( f(x) = x^2 \)
The area \( B \) is between the graph of \( f(x) = x^2 \) and the \( x \)-axis from \( x = 0 \) to \( x = 1 \). Mathematically, the area \( B \) can be found by the integral: \[ B = \int_0^1 x^2 \, dx. \] Compute this integral to find \( B = \frac{1}{3} \).
05
Verify that \( A + B = 1 \)
From the calculated areas, \( A = \frac{2}{3} \) and \( B = \frac{1}{3} \). Add these areas together: \[ A + B = \frac{2}{3} + \frac{1}{3} = 1. \] This completes the proof that \( A + B = 1 \) for these particular curves over the given interval. Geometrically, this makes sense because the area under \( y = 1 \) from \( x = 0 \) to \( x = 1 \) forms a complete unit square, which is divided by the two curves.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Area Under a Curve
The area under a curve is a fundamental concept in integral calculus, representing the space between the curve and the x-axis over a given interval. This area is quantified using integrals, which effectively sum an infinite number of infinitesimally thin vertical rectangles under the curve.
For functions that lie above the x-axis in a given interval, the area is positive. In the given exercise, the goal is to find the areas under the curves of two functions: \( f(x) = \sqrt{x} \) and \( f(x) = x^2 \).
For functions that lie above the x-axis in a given interval, the area is positive. In the given exercise, the goal is to find the areas under the curves of two functions: \( f(x) = \sqrt{x} \) and \( f(x) = x^2 \).
- The area beneath \( f(x) = \sqrt{x} \) from \( x = 0 \) to \( x = 1 \) represents \( A \).
- Similarly, the area under \( f(x) = x^2 \) from \( x = 0 \) to \( x = 1 \) represents \( B \).
Definite Integral
The definite integral is a powerful tool in calculus, used specifically to calculate the precise area under a curve. Unlike an indefinite integral, which indicates a family of functions, a definite integral gives a specific numerical result.
To solve the original exercise and find \( A \) and \( B \), we evaluated the definite integrals:
To solve the original exercise and find \( A \) and \( B \), we evaluated the definite integrals:
- \( A = \int_0^1 \sqrt{x} \, dx = \frac{2}{3} \)
- \( B = \int_0^1 x^2 \, dx = \frac{1}{3} \)
Graphical Interpretation of Integrals
The graphical interpretation of integrals helps us visually understand the area concept. By plotting the functions \( f(x) = \sqrt{x} \) and \( f(x) = x^2 \) over \([0, 1]\) on the graph, we can see how these curves create distinct regions above the x-axis.
- The curve \( f(x) = \sqrt{x} \) traces a concave path from the origin, gently rising to reach (1,1), whereas \( f(x) = x^2 \) forms a quadratic or parabolic arc, which is much steeper.
- The area between each curve and the x-axis represents the respective definite integral over the interval \([0, 1]\). \( A \) lies under the curve \( \sqrt{x} \), while \( B \) is under \( x^2 \).