Question

Let A denote the area between the graph of \(\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}}\) and the interval \([0,1]\), and let \(\mathrm{B}\) denote the area between the graph of \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}\) and the interval \([0,1]\). Explain geometrically why \(\mathrm{A}+\mathrm{B}=1\)

Short answer

Question: Prove that the total area under the curves f(x) = √x and f(x) = x^2 on the interval [0, 1] equals the area of a unit square. Answer: We can demonstrate this by calculating the area under each curve on the given interval and showing that their sum equals 1. The area under f(x) = √x (denoted as A) is 2/3, and the area under f(x) = x^2 (denoted as B) is 1/3. Adding A and B together (2/3 + 1/3), we find that the total area under both curves equals 1, which is the area of the unit square.

Step-by-step solution

Understand the graphs of the functions

First, we need to understand the graphs of the functions f(x) = √x and f(x) = x^2. The function f(x) = √x is a monotonically increasing function that starts from origin and has the shape of the sideways parabola, and the function f(x) = x^2 is a parabolic function that has a minimum value of 0 at x = 0 and grows towards infinity as x increases.

Sketch the graphs

Next, sketch the graphs of the two functions on the interval [0, 1]. We observe that f(x) = √x lies above the curve f(x) = x^2 since √x ≥ x^2 for all x in the interval [0, 1]. Furthermore, both functions start at the origin (0, 0) and end at the point (1, 1).

Calculate A - the area under the curve f(x) = √x

To calculate the area A under the curve f(x) = √x, we need to integrate the function within the interval [0, 1]: A = \int_{0}^{1} \sqrt{x}\,dx = \frac{2}{3}(x^{\frac{3}{2}})|_0^1 = \frac{2}{3}(1)^{\frac{3}{2}} - \frac{2}{3}(0)^{\frac{3}{2}} = \frac{2}{3}

Calculate B - the area under the curve f(x) = x^2

Similarly, calculate the area B under the curve f(x) = x^2 within the same interval: B = \int_{0}^{1} x^2\, dx = \frac{1}{3}(x^3)|_0^1 = \frac{1}{3}(1)^3 - \frac{1}{3}(0)^3 = \frac{1}{3}

Add the areas A and B together

Now that we have the areas A and B, we can add them together to see if their sum equals 1: A + B = \frac{2}{3} + \frac{1}{3} = 1 Since A + B = 1, the total area under both curves equals the area of the unit square, proving the statement geometrically.

Key concepts

Area Under Curve

Understanding the concept of the area under a curve is a fundamental part of integral calculus, particularly for students preparing for IIT JEE. The area under the curve of a function can be represented visually on a graph. It is bounded by the function's graph, the x-axis, and vertical lines at the endpoints of the interval.

In our exercise, we're considering two functions over the interval [0,1], and by calculating the area under each curve, we are essentially finding the space that lies between the graph of the function and the x-axis. It's important to visualize these areas because they can inform us about quantities that are changing, such as distance covered over time when analyzing a velocity-time graph.

Now, let's dive into why A + B equals 1 geometrically. The sum of the areas under the curves of \(\sqrt{x}\) and \(x^2\) from 0 to 1 is equal to the entire area of a unit square bound by [0,1] on the x-axis and [0,1] on the y-axis, since these functions are inverses of each other in this interval and partition the square into two complementary regions.

Integration

Integration serves as the mathematical tool for finding the area under the curve. In broader terms, integration is the inverse operation of differentiation, and it can be thought of as 'summing up' infinitesimally small quantities. For the IIT JEE syllabus, mastering integration techniques is essential, as it applies to a wide range of problems beyond finding areas.

For the given exercise, integration is used to calculate the exact areas A and B. The process involves taking the integral of the function over the specified interval. In our step-by-step solution, \(A = \int_0^1 \sqrt{x} dx\) and \(B = \int_0^1 x^2 dx\), both integrals are evaluated using antiderivatives, showcasing a practical application of integration in computing areas under curves.

Graph of Functions

The graph of a function is a visual representation that shows how values of the function relate to each other. For IIT JEE aspirants, interpreting the graphs of functions is crucial for solving problems related to calculus, as it often provides insights into the behavior of functions.

The function \(\sqrt{x}\), depicted in our exercise, is a half-parabola opening to the right, and the graph of \(x^2\) is a parabola opening upwards. By graphing these functions on an interval from [0,1], we observe their geometrical properties, such as points of intersection, slopes, and symmetry, which are pivotal in understanding why their combined area equals 1. This symmetry between the graphs of \(\sqrt{x}\) and \(x^2\) within this interval is key to explaining the result geometrically.

Definite Integral

The definite integral of a function over an interval provides the net area under the curve within that interval. This core concept is essential when dealing with exercises related to integral calculus in IIT JEE. When we talk about net area, we consider the orientation of the area with respect to the x-axis; areas above the axis are positive, and those below are negative.

In the exercise at hand, we calculated the definite integrals \(A = \int_0^1 \sqrt{x} dx\) and \(B = \int_0^1 x^2 dx\). Each one represents the definite integral, hence the net area under each curve from 0to 1. Since the functions do not cross the x-axis in this interval, both areas are positive and add up to exactly 1, the total area of the square bounded by the interval [0,1] on both axes.