Problem 7 - Calculus: Early Transcendentals

Question

Find the area bounded by the curves. \(y=\sqrt{x}\) and \(y=x^{2}\)

Step-by-Step Solution

Answer

The area bounded by the curves is \( \frac{1}{3} \).

Step 1: Set the Equations Equal to Find Intersection Points

To find the points where the curves intersect, set the equations equal to each other: \( \sqrt{x} = x^2 \). Squaring both sides gives \( x = x^4 \). Rearrange to get \( x^4 - x = 0 \). Factor this to \( x(x^3 - 1) = 0 \), giving solutions \( x = 0 \) and \( x^3 = 1 \), so \( x = 1 \). Thus, the curves intersect at \( x = 0 \) and \( x = 1 \).

Step 2: Set Up the Integral to Find the Area Between Curves

The area between curves can be found by integrating the difference between upper and lower functions from the intersection points. Here, from \( x = 0 \) to \( x = 1 \), the upper function is \( y = \sqrt{x} \) and the lower function is \( y = x^2 \). Set up the integral as follows: \[ \int_{0}^{1} (\sqrt{x} - x^2) \, dx \]

Step 3: Evaluate the Integral

Evaluate the integral \( \int_{0}^{1} (\sqrt{x} - x^2) \, dx \). Break it into two separate integrals: 1. \( \int_{0}^{1} \sqrt{x} \, dx \)2. \( -\int_{0}^{1} x^2 \, dx \)First, the integral of \( \sqrt{x} \) is: \[ \int \sqrt{x} \, dx = \frac{2}{3}x^{3/2} \] Evaluating from 0 to 1 gives \( \frac{2}{3}(1^{3/2}) - \frac{2}{3}(0^{3/2}) = \frac{2}{3} \).The integral of \( x^2 \) is: \[ \int x^2 \, dx = \frac{1}{3}x^3 \] Evaluating from 0 to 1 gives \( \frac{1}{3}(1^3) - \frac{1}{3}(0^3) = \frac{1}{3} \).

Step 4: Compute the Final Area

Subtract the area under \( y = x^2 \) from the area under \( y = \sqrt{x} \). \( \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \).Thus, the area between the curves \( y = \sqrt{x} \) and \( y = x^2 \) from \( x = 0 \) to \( x = 1 \) is \( \frac{1}{3} \).

Key Concepts

Integral Calculus

Integral calculus is a powerful mathematical tool used to calculate areas under curves, among other things. It revolves around the concept of a limit, which helps us sum up infinitesimally small quantities. In this particular problem, we use integral calculus to find the area between two curves, namely, \( y = \sqrt{x} \) and \( y = x^2 \). Given that these functions define different shapes on the Cartesian plane, the area between them can be thought of as the space trapped by these boundaries.

Here's how we use integrals:

• We first determine the curves that form the boundary of the area we're interested in. For this exercise, they are \( y = \sqrt{x} \) and \( y = x^2 \).
• Next, we calculate where these curves intersect. These points will serve as the limits or boundaries of our integration.
• Once we have the functions and their intersection points, we identify which function is on top (the upper function) and which is on the bottom (the lower function) within this region.

The integral of the difference between the upper and lower functions, evaluated within the limits of their intersection points, gives us the area of the region of interest.

Intersection Points

Intersection points are where two curves meet or cross each other on a graph. To find these points, we set the equations defining the curves equal and solve for \( x \). In the case at hand, we have:

• Set \( \sqrt{x} = x^2 \), which can then be manipulated to \( x = x^4 \). By rearranging, we have \( x^4 - x = 0 \).
• Factor the expression to \( x(x^3 - 1) = 0 \), resulting in the solutions \( x = 0 \) and \( x = 1 \).
• Thus, the intersection points are at \( x = 0 \) and \( x = 1 \).

These points are crucial as they provide the boundaries for calculating the area between the curves using integrals. They signify where the curves transition between being the upper and lower functions relative to each other.

Definite Integral

The definite integral is a specific type of integral that calculates the accumulated area under a curve, between two defined points on the x-axis. For the problem at hand, we are working with the definite integral:

• The bounds, 0 and 1, come from the intersection points of the curves \(y = \sqrt{x}\) and \(y = x^2\).
• The function we integrate is the difference between the upper function \( \sqrt{x} \) and the lower function \( x^2 \).

The definite integral is set up as \( \int_{0}^{1} (\sqrt{x} - x^2) \, dx \). This expression represents the total area "sandwiched" between the curves from \(x = 0\) to \(x = 1\).

Evaluating this integral involves finding the antiderivative of each function separately:

• For \( \sqrt{x} \), the antiderivative is \( \frac{2}{3}x^{3/2} \).
• For \( x^2 \), this is \( \frac{1}{3}x^3 \).

Finally, we compute the set values from the integral limits to arrive at the total bounded area, which in this case is \( \frac{1}{3} \). This technique allows integral calculus to provide precise numerical measurements for areas defined by complex, curved boundaries.