Definite Integral - Upper And Lower Bound Of Area Under A Continuous Function

Definite Integral - Upper and lower bound of area under a continuous function

The definite integral of a function f(x) over a closed interval [a, b] gives us the area between the graph of the function and the x-axis.
The upper and lower bounds of this area can be determined by evaluating the definite integral using the lower limit a and upper limit b.
The integral is denoted as: ∫_a^b f(x) dx.
The upper bound represents the maximum value of the area, while the lower bound represents the minimum value.
The definite integral can also be used to calculate other quantities such as displacement, velocity, and acceleration. Example: Consider the function f(x) = x² over the interval [0, 2]. We can find the upper and lower bounds of the area between the graph and the x-axis by evaluating the definite integral ∫₀² x² dx.
The upper bound gives us the maximum possible area, which is obtained by assuming that the function is always positive.
The lower bound gives us the minimum possible area, which is obtained by assuming that the function is always negative.
By solving the definite integral, we find that the upper bound is 8/3 and the lower bound is 0. Equations:
Upper bound: A_upper = ∫_a^b |f(x)| dx
Lower bound: A_lower = -∫_a^b |f(x)| dx Applications of definite integral:
Calculating areas under curves
Finding displacement, velocity, and acceleration
Calculating work and energy
Solving differential equations
Calculating probability densities Next slide…

The upper and lower bounds of the definite integral can provide a range of possible values for the area under a function.
The bounds are influenced by the sign and shape of the function.
If the function is always positive, the upper bound will be positive, and if the function is always negative, the lower bound will be negative.
If the function changes sign, the area between the graph and the x-axis can cancel out, resulting in a smaller value for the definite integral.
The bounds can be used to estimate the area when the exact value is difficult to determine.

The definite integral can be evaluated using the Fundamental Theorem of Calculus.
The theorem states that if f(x) is continuous on the interval [a, b], and F(x) is an antiderivative of f(x), then ∫_a^b f(x) dx = F(b) - F(a).
This theorem allows us to find the area under a curve by finding an antiderivative of the function and evaluating it at the upper and lower limits.
The definite integral represents the difference between the values of the antiderivative at the upper and lower limits.
Example: The function f(x) = 3x² has antiderivative F(x) = x³. Evaluating F(2) - F(0) gives us the area between the graph and the x-axis, which is 8.

The definite integral can be extended to multiple dimensions.
In two dimensions, the definite integral can be used to calculate the area between a curve and the x-y plane.
The bounds for the integral will be specified for both the x and y axes, resulting in a rectangular region.
Example: The definite integral ∬ R (x² + y²) dA represents the volume under the surface z = x² + y² within the region R.
In three dimensions, the definite integral can be used to calculate the volume between a surface and the x-y-z space.
The bounds for the integral will be specified for all three axes, resulting in a cuboid or irregular region.

The definite integral can also be used to calculate the average value of a function over an interval.
The average value of a function f(x) on the interval [a, b] is given by: Average value = 1 / (b - a) * ∫_a^b f(x) dx.
This formula calculates the area under the curve and divides it by the length of the interval.
Example: Find the average value of the function f(x) = x³ on the interval [-1, 1]. The integral evaluates to 0, so the average value is also 0.

The definite integral can be used to calculate the length of a curve.
For a curve defined by a function y = f(x) on the interval [a, b], the length L is given by: L = ∫_a^b √(1 + (f’(x))²) dx.
This formula considers the infinitesimal length of each segment of the curve and adds them up.
The derivative f’(x) provides information about the slope of the curve.
Example: Find the length of the curve y = x² on the interval [0, 1]. By the formula, L = ∫₀¹ √(1 + (2x)²) dx, which evaluates to approximately 1.478.

The definite integral can also be used to calculate moments and centers of mass.
The moment of a function f(x) about the y-axis is given by: M_y = ∫_a^b xf(x) dx.
The center of mass of a function f(x) on the interval [a, b] is given by: x-bar = (1 / A) * ∫_a^b xf(x) dx, where A is the area between the curve and the x-axis.
These formulas consider the distribution and mass of the function, giving insights into its geometry.
Example: Find the moment about the y-axis and the center of mass of the function f(x) = x² on the interval [0, 2]. The moment is M_y = ∫₀² x³ dx = 8/4 = 2. The center of mass is x-bar = (1 / (8/3)) * ∫₀² x³ dx = 9/4.

The definite integral can be used to solve problems involving accumulation.
If a quantity is changing over time at a rate given by a function f(t), the total accumulated quantity can be found using the definite integral.
Example: If water is flowing into a tank at a rate of 5t liters per hour, where t represents time in hours, the total amount of water in the tank at time t can be found by evaluating the integral ∫₀^t 5t dt.

The definite integral can also be used to calculate average rate of change.
If a function f(t) represents the position of an object at time t, the average velocity over an interval [a, b] can be found using the definite integral.
Example: If the position of an object is given by f(t) = t² + 3t, the average velocity from t = 1 to t = 4 can be found by evaluating the integral ∫₁⁴ (2t + 3) dt.

The definite integral also has applications in probability theory.
If a probability density function f(x) represents the distribution of a random variable, the probability that the variable falls within a certain interval can be found using the definite integral.
Example: If the probability density function of the height of adults is given by f(x), the probability that an adult’s height falls within a range of 150 cm to 170 cm can be found by evaluating the integral ∫₁₅₀¹⁷⁰ f(x) dx.

The definite integral is a powerful tool in calculus that allows us to calculate areas, volumes, averages, lengths, moments, and solve various other mathematical problems.
It provides a rigorous and systematic approach to analyzing functions and their properties.
Mastering the concept of definite integrals is crucial for success in calculus and its applications in various fields.
Practice is essential to develop proficiency in evaluating definite integrals and applying them to real-world problems.
With sufficient practice and understanding, you can harness the power of definite integrals to solve complex mathematical problems.

The definite integral is a fundamental concept in calculus that has wide-ranging applications in various fields of study.
It allows us to calculate the area, volume, and other quantities between curves and the coordinate axes.
The bounds of the definite integral determine the range over which we calculate these quantities.
The upper and lower bounds can provide a range of possible values for the area or volume in question.
The specific values of the area or volume will depend on the function being integrated and the interval of integration.

The definite integral can be graphically represented as the area under the curve y = f(x) between the limits of integration.
This area is bounded by the x-axis and the vertical lines corresponding to the lower and upper limits of the interval.
The sign of the function also impacts the result of the integral.
If the function is positive over the interval, the definite integral will give the area above the x-axis.
Conversely, if the function is negative, the definite integral will give the area below the x-axis.

Evaluating the definite integral often involves finding an antiderivative or evaluating the limit of a summation.
The antiderivative of a function f(x) provides the indefinite integral, or the collection of all possible antiderivatives.
The definite integral, however, involves evaluating the antiderivative at the lower and upper bounds of the interval of integration.
This process helps determine the exact value of the area or volume under consideration.

The Fundamental Theorem of Calculus is a major result that connects differentiation and integration.
This theorem states that if F(x) is the antiderivative of the function f(x), then the definite integral of f(x) from a to b is equal to F(b) - F(a).
In other words, the definite integral can be calculated by subtracting the value of the antiderivative at the lower bound from the value at the upper bound.
This theorem provides a powerful tool for evaluating definite integrals and connecting them to other areas of calculus.

The definite integral can have both practical and theoretical applications.
In physics, it is used to calculate quantities such as work, energy, and probability.
In economics, it is used to analyze supply and demand curves to determine optimal production levels.
In engineering, it is used to calculate moments of inertia in structural design.
These are just a few examples of the broad applications of definite integrals in various disciplines.

The definite integral can be used to solve problems involving rates, such as finding the average value or average rate of change of a function.
It can also be used to solve optimization problems, where we seek to find the maximum or minimum value of a function subject to certain constraints.
Additionally, it plays a key role in solving differential equations, which are mathematical models that describe the relationship between a function and its derivatives.

The definite integral can be used to calculate areas between curves.
To find the area between two curves y = f(x) and y = g(x) over an interval [a, b], we can evaluate the integral of the difference between the two functions.
The upper and lower bounds are determined by the points where the two curves intersect.
Example: Find the area between the curves y = x² and y = 2x - 1 over the interval [0, 2]. The area is given by ∫₀² (x² - (2x - 1)) dx.

The definite integral can also be used to calculate volumes of solid shapes.
For shapes with known cross-sectional areas, the volume can be found by integrating the area function over a specified interval.
Example: To find the volume of a cone with radius r and height h, we can slice the cone into infinitesimally thin discs perpendicular to its central axis.
The area of each disc is given by A = πr², and integrating this function from 0 to h gives us the volume of the cone: ∫₀^h πr² dx.

The definite integral can be used to calculate the average value of a function.
This value represents the y-coordinate of the horizontal line that divides the area under the curve into two equal parts.
To calculate the average value, we divide the definite integral of the function by the length of the interval [a, b].
Example: Find the average value of the function f(x) = 3x² over the interval [-1, 1]. The average value is given by (1 / (2 - (-1))) * ∫_-1¹ 3x² dx.

In conclusion, the definite integral is a powerful mathematical tool with various applications in calculus and related fields.
It allows us to calculate areas, volumes, averages, and other quantities by integrating over a specified interval.
The upper and lower bounds of the integral play a significant role in determining the range of values for the area or volume under consideration.
The Fundamental Theorem of Calculus provides a key relationship between differentiation and integration, enabling us to evaluate definite integrals more efficiently.
Understanding and practicing the concepts of definite integration are essential for success in advanced mathematics and its practical applications.