Definite integrals are a cornerstone in calculus, representing the accumulation of quantities. They play a pivotal role in determining the area under a curve, the length of curves, and have a myriad of applications in physics, engineering, and economics. This set of notes will delve deep into the fundamental theorem of calculus and the concept of the area under a curve.
Fundamental Theorem of Calculus
The fundamental theorem of calculus is a bridge between the two primary operations in calculus: differentiation and integration.
Statement:
If a function f is continuous over a closed interval [a, b] and F is its antiderivative, then the integral of f from a to b is equal to F(b) minus F(a).
Historical Context:
This theorem emerged in the late 17th century with significant inputs from mathematicians like James Gregory, Isaac Barrow, Gottfried Leibniz, and Isaac Newton. Their collective endeavours laid the groundwork for what we now understand as modern calculus.
Implication:
This theorem offers a method to evaluate definite integrals without directly calculating the area under the curve. Instead, one can find an antiderivative and evaluate it at the interval's endpoints.
Applications:
The fundamental theorem of calculus is versatile. In physics, it's used to compute quantities like displacement, work, and electric charge. In economics, it helps determine consumer and producer surplus. It's also a tool in probability theory to ascertain expected values.
Area Under a Curve
The concept of the area under a curve is visualised as the region enclosed by the curve, the x-axis, and two vertical lines indicating the limits of integration.
Explanation:
The area under a curve pertains to the definite integral of a function over a set interval. This area is the "net area", meaning areas above the x-axis are positive, while those below are negative. The definite integral provides the net accumulation of the function's values across the interval.
Example:
Take the function f(x) = sin(x). The area under its curve from 0 to π is 2. This is the space between the sine curve, the x-axis, and the lines x = 0 and x = π.
Properties of Definite Integrals:
- Additivity Over Intervals: The integral from a to b added to the integral from b to c equals the integral from a to c.
- Reversal of Limits: Switching the limits of integration changes the sign of the integral's value.
- Constant Multiple: The integral of a constant multiplied by a function equals the constant multiplied by the integral of the function.
- Additivity Over Functions: The integral of a sum equals the sum of the integrals.
Practice Questions:
1. Evaluate the integral of x squared from 0 to 2.
2. Determine the area under the curve y = x cubed from -1 to 1.
3. Calculate the net area enclosed by the function f(x) = cos(x) from 0 to 2π.
FAQ
Yes, the value of a definite integral can be negative. This occurs when the function lies below the x-axis for the interval of integration. In geometric terms, it represents the area below the x-axis and above the function. In such cases, the negative value signifies the "direction" or "orientation" of the quantity, rather than its magnitude.
In physics, definite integrals are used to compute accumulated quantities over a specific interval. For instance, when determining the total displacement of an object given its velocity function, the definite integral of the velocity function over a time interval gives the net displacement. Similarly, in the context of work done by a variable force, the definite integral of the force function over a distance interval provides the total work done. Essentially, whenever there's a need to accumulate a quantity over an interval, definite integrals come into play.
The definite integral of a constant function over an interval is simply the product of the constant value and the width of the interval. Geometrically, this represents the area of a rectangle with height equal to the constant value and width equal to the interval's length. It showcases the simplicity and directness of integrating constant functions.
The width of the interval directly impacts the value of the definite integral. A wider interval generally results in a larger absolute value for the integral, as it covers more area under the curve. However, the function's behaviour over the interval is crucial. If the function has both positive and negative values within the interval, the areas above and below the x-axis might cancel each other out, leading to a smaller net area.
The geometric interpretation of a definite integral is the net area between the function and the x-axis over a specified interval. If the function lies above the x-axis for the interval, the definite integral represents the area under the curve. If the function is below the x-axis, the area is considered negative. For functions that cross the x-axis, the definite integral calculates the net area, taking into account the portions above and below the x-axis.
Practice Questions
To evaluate the definite integral of x3 from 0 to 2, we integrate x3 with respect to x over the interval [0, 2]. The antiderivative of x3 is (x4)/4. Evaluating this antiderivative at the upper limit 2 and subtracting its value at the lower limit 0, we get: (24)/4 - (04)/4 = 4. Thus, the value of the definite integral is 4.
To find the value of the definite integral of sqrt(x) from 1 to 4, we integrate sqrt(x) with respect to x over the interval [1, 4]. The antiderivative of sqrt(x) is (2x(3/2))/3. Evaluating this antiderivative at the upper limit 4 and subtracting its value at the lower limit 1, we get: (2(4(3/2)))/3 - (2(1(3/2)))/3 = 14/3 or approximately 4.67. Therefore, the value of the definite integral is 14/3 or approximately 4.67.
