Definite Integrals, Areas, And Parametric Equations In Calculus

This article delves into the solutions of three integral calculus problems, focusing on evaluating a definite integral, calculating the area between curves, and analyzing a parametric equation. Each problem will be addressed with detailed steps and explanations, providing a comprehensive understanding of the underlying concepts and techniques.

Q3: Evaluating the Definite Integral and Interpreting the Result

Definite integrals are a cornerstone of calculus, providing a powerful tool for calculating the area under a curve. This section focuses on evaluating the definite integral ${ \int_0^2 \left(2x^2 - 6x + \frac{1}{x^2 + 1}\right) dx }$ and interpreting the result in terms of areas. Understanding the geometric interpretation of definite integrals is crucial for grasping their significance in various applications, from physics to economics.

To evaluate the definite integral, we need to find the antiderivative of the integrand, which is a function whose derivative is the integrand. The integrand in this case is a polynomial plus a rational function, making it a good example for demonstrating integral calculus techniques. By applying the power rule and recognizing the arctangent form, we can systematically find the antiderivative and then evaluate it at the limits of integration. This process not only yields the numerical value of the integral but also provides insights into the area bounded by the function and the x-axis over the given interval.

Let's begin by finding the antiderivative of each term in the integrand:

• The antiderivative of ${2x^2}$ is ${\frac{2}{3}x^3}$.
• The antiderivative of ${-6x}$ is ${-3x^2}$.
• The antiderivative of ${\frac{1}{x^2 + 1}}$ is ${\arctan(x)}$.

Therefore, the antiderivative of the entire integrand is: ${ F(x) = \frac{2}{3}x^3 - 3x^2 + \arctan(x) }$ Now, we evaluate this antiderivative at the limits of integration, 0 and 2: ${ F(2) = \frac{2}{3}(2)^3 - 3(2)^2 + \arctan(2) = \frac{16}{3} - 12 + \arctan(2) }$ ${ F(0) = \frac{2}{3}(0)^3 - 3(0)^2 + \arctan(0) = 0 }$ By the Fundamental Theorem of Calculus, the definite integral is the difference between the antiderivative evaluated at the upper and lower limits: ${ \int_0^2 \left(2x^2 - 6x + \frac{1}{x^2 + 1}\right) dx = F(2) - F(0) = \frac{16}{3} - 12 + \arctan(2) - 0 }$ ${ = \frac{16}{3} - \frac{36}{3} + \arctan(2) = -\frac{20}{3} + \arctan(2) }$ Thus, the value of the definite integral is: ${ -\frac{20}{3} + \arctan(2) \approx -5.264 }$

Now, let's interpret this result in terms of areas. The definite integral represents the signed area between the curve defined by the integrand and the x-axis, from ${x = 0}$ to ${x = 2}$. The "signed" aspect means that areas above the x-axis are counted as positive, while areas below the x-axis are counted as negative. In this case, the negative value of the integral indicates that the area below the x-axis is larger than the area above the x-axis within the interval ${0, 2}$.

To visualize this, consider the graph of the function ${f(x) = 2x^2 - 6x + \frac{1}{x^2 + 1}}$. From ${x = 0}$ to ${x = 2}$, the function dips below the x-axis, creating a region where the function values are negative. The integral effectively calculates the net area, subtracting the area below the x-axis from the area above it. The negative result confirms that the area below the x-axis dominates in this interval.

In summary, evaluating the definite integral allows us to quantify the net area between a curve and the x-axis, providing a geometric interpretation of the integral's value. This concept is fundamental in calculus and has wide-ranging applications in various fields.

Q4: Finding the Area Enclosed by Parabolas

Determining the area enclosed by curves is a classic application of integral calculus. This section addresses the problem of finding the area enclosed by the parabolas ${y = x^2}$ and ${y = 2x - x^2}$. This involves identifying the points of intersection of the curves and setting up the appropriate integral to calculate the area between them. Understanding how to find the area between curves is essential for numerous applications, including optimization problems and calculations in physics and engineering.

To find the area enclosed by two curves, we first need to determine where they intersect. These intersection points will define the limits of integration. Then, we integrate the difference between the functions over this interval. The key is to correctly identify which function is above the other within the interval, as the order of subtraction affects the sign of the area.

First, let's find the points of intersection by setting the equations of the parabolas equal to each other: ${ x^2 = 2x - x^2 }$ ${ 2x^2 - 2x = 0 }$ ${ 2x(x - 1) = 0 }$ This gives us two solutions: ${x = 0}$ and ${x = 1}$. These are the x-coordinates of the intersection points. To find the corresponding y-coordinates, we can plug these x-values into either equation. Let's use ${y = x^2}$:

• When ${x = 0}$, ${y = 0^2 = 0}$.
• When ${x = 1}$, ${y = 1^2 = 1}$.

So, the parabolas intersect at the points (0, 0) and (1, 1).

Now, we need to determine which function is above the other in the interval ${0, 1}$. We can do this by picking a test point within the interval, say ${x = 0.5}$, and evaluating both functions:

• For ${y = x^2}$, when ${x = 0.5}$, ${y = (0.5)^2 = 0.25}$.
• For ${y = 2x - x^2}$, when ${x = 0.5}$, ${y = 2(0.5) - (0.5)^2 = 1 - 0.25 = 0.75}$.

Since ${0.75 > 0.25}$, the parabola ${y = 2x - x^2}$ is above the parabola ${y = x^2}$ in the interval ${0, 1}$.

Now we can set up the integral to calculate the area. The area A between the curves is given by: ${ A = \int_0^1 \left[(2x - x^2) - x^2\right] dx }$ ${ = \int_0^1 (2x - 2x^2) dx }$ Now, we find the antiderivative of the integrand: ${ \int (2x - 2x^2) dx = x^2 - \frac{2}{3}x^3 + C }$ We evaluate the antiderivative at the limits of integration: ${ \left[x^2 - \frac{2}{3}x^3\right]_0^1 = \left[(1)^2 - \frac{2}{3}(1)^3\right] - \left[(0)^2 - \frac{2}{3}(0)^3\right] }$ ${ = 1 - \frac{2}{3} - 0 = \frac{1}{3} }$ Therefore, the area enclosed by the parabolas ${y = x^2}$ and ${y = 2x - x^2}$ is ${\frac{1}{3}}$ square units. This example demonstrates the systematic approach to finding areas between curves, which is a fundamental skill in calculus with applications in geometry, physics, and engineering.

Q5: Sketching and Identifying a Curve Defined by Parametric Equations

Parametric equations provide a powerful way to describe curves in the plane by expressing the x and y coordinates as functions of a third variable, often denoted as ${t}$. This section focuses on sketching and identifying the curve defined by a given set of parametric equations. Understanding parametric equations is crucial for describing complex curves that cannot be easily represented in Cartesian form, and they have significant applications in computer graphics, physics, and engineering.

To sketch a curve defined by parametric equations, we typically analyze the behavior of the x and y coordinates as the parameter ${t}$ varies. We can create a table of values, plotting points for different values of ${t}$, and then connect these points to visualize the curve. Identifying the curve often involves eliminating the parameter ${t}$ to obtain a Cartesian equation, which can reveal the familiar form of the curve, such as a circle, ellipse, parabola, or hyperbola.

Let's consider a set of parametric equations as an example: ${ x = \cos(t) }$ ${ y = \sin(t) }$ where ${0 \leq t \leq 2\pi}$.

To sketch the curve, we can create a table of values for different values of ${t}$:

Plotting these points on the Cartesian plane, we see that they form a circle centered at the origin with a radius of 1. To confirm this, we can eliminate the parameter ${t}$ by using the trigonometric identity: ${ \cos^2(t) + \sin^2(t) = 1 }$ Substituting the parametric equations into this identity, we get: ${ x^2 + y^2 = 1 }$ This is the equation of a circle with radius 1 centered at the origin. Therefore, the parametric equations ${x = \cos(t)}$ and ${y = \sin(t)}$ for ${0 \leq t \leq 2\pi}$ define a unit circle traced counterclockwise.

This example illustrates the process of sketching and identifying curves defined by parametric equations. By analyzing the behavior of x and y as functions of ${t}$ and eliminating the parameter, we can gain a comprehensive understanding of the curve's shape and properties. Parametric equations are an essential tool in calculus and have wide-ranging applications in mathematics, science, and engineering.

In conclusion, this article has explored the solutions to three integral calculus problems, covering definite integrals, areas between curves, and parametric equations. Each problem has been addressed with detailed explanations and steps, providing a thorough understanding of the concepts and techniques involved. These fundamental concepts are essential for further studies in mathematics and their applications in various fields.