The Fundamental Theorem of Calculus connects the concepts of differentiation and integration, showing that these two processes are essentially inverse operations. It states that if a function is continuous on a closed interval
[ a, b ] and F is an antiderivative of f on that interval, then:
\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]
This powerful theorem not only helps us evaluate definite integrals but also provides insight into the relationship between a function and its rate of change. Let’s explore three practical examples to see how we can apply this theorem in real-world scenarios.
One common application of the Fundamental Theorem of Calculus is finding the area under a curve represented by a function. This is particularly useful in physics or economics to understand total quantities.
To find the area under the curve of the function f(x) = 2x from x = 1 to x = 3, we can apply the theorem.
We start by finding an antiderivative of f(x):
F(x) = x² + C
Now we evaluate the definite integral:
\[\int_{1}^{3} 2x \, dx = F(3) - F(1)\]
Calculating F(3) and F(1):
F(3) = 3² = 9
F(1) = 1² = 1
So, the area is:
\[9 - 1 = 8\]
This area represents the total accumulation of the quantity represented by the function between the two points on the x-axis. You could change the function or the limits to see how the area changes accordingly.
In economics, we often need to find the accumulated change over time, such as total revenue generated from selling a product over a specific period. Let’s say the rate of revenue growth is given by f(t) = 5t, where t is in hours.
To find the total revenue generated from time t = 0 to t = 4 hours, we again apply the Fundamental Theorem of Calculus.
First, we find an antiderivative of f(t):
F(t) = \(\frac{5}{2}t^2 + C\)
Now we evaluate the definite integral:
\[\int_{0}^{4} 5t \, dt = F(4) - F(0)\]
Calculating F(4) and F(0):
F(4) = \(\frac{5}{2}(4^2) = \frac{5}{2}(16) = 40\)
F(0) = 0
The total revenue generated is:
\[40 - 0 = 40\]
This example shows how we can use the theorem to determine accumulated revenue over time. Changing the function to reflect different growth rates or adjusting the time period can provide further insights into revenue generation.
In physics, we can use the Fundamental Theorem of Calculus to analyze motion. Suppose the velocity of an object is given by v(t) = 3t² - 2t, where t is in seconds. To find the total displacement from t = 1 to t = 3, we need to find the definite integral of the velocity function.
First, we find an antiderivative of v(t):
F(t) = t³ - t² + C
Now, we evaluate the definite integral:
\[\int_{1}^{3} (3t^2 - 2t) \, dt = F(3) - F(1)\]
Calculating F(3) and F(1):
F(3) = 3³ - 3² = 27 - 9 = 18
F(1) = 1³ - 1² = 1 - 1 = 0
Thus, the total displacement is:
\[18 - 0 = 18\]
This example illustrates how the theorem can be applied to solve problems involving motion. You can modify the velocity function or the time intervals to see how they affect displacement, enhancing your understanding of motion dynamics.
By using these examples, we can see how the Fundamental Theorem of Calculus is not just a theoretical concept but a practical tool that helps us solve real-world problems across various fields. Whether it’s calculating areas, understanding economic changes, or analyzing motion, this theorem is a cornerstone of calculus that enables us to make sense of continuous change.