Lecture 26: The Fundamental Theorem of Calculus✅
Connection Between Displacement and Velocity
Given a displacement function, the velocity can be found by taking the derivative of the displacement function.
v(t) = \frac{d}{dt} x(t)
Displacement with Constant Velocity
If velocity (v) is constant from time t1 to t2, the displacement is:
v \cdot (t2 - t1)
This can be visualized as the area of a rectangle with height v and base (t2 - t1).
Displacement with Non-Constant Velocity
When velocity is not constant but varies with time v(t), displacement is found via approximation using equally spaced intervals \Delta t.
The displacement is approximated by a sum of rectangular areas.
The integral is the area under the curve y = v(t) between t1 and t2.
\int{t1}^{t2} v(t) dt = \lim{ \Delta t \to 0} A
Where A is the approximation of displacement.
Fundamental Theorem of Calculus
The integral of the velocity function v(t) from t1 to t2 is the displacement between those times.
\int{t1}^{t2} v(t) dt = x(t2) - x(t_1)
If v(t) = \frac{d}{dt}x(t), then:
\int{t1}^{t2} \frac{d}{dt} x(t) dt = x(t2) - x(t_1)
Calculating the integral using the limit of the sum of rectangular areas is tedious.
With the fundamental theorem of calculus, if we know the displacement at t2 and t1, we can easily calculate the integral.
General Context of the Fundamental Theorem of Calculus
Let f(x) be a continuous function on the interval [a, b].
Let F(x) be a function such that F'(x) = f(x).
Then, the fundamental theorem of calculus states:
\int_{a}^{b} f(x) dx = F(b) - F(a)
Example 1
Find the area under the curve f(x) = 1 between 0 and 1.
We need to find F(x) such that F'(x) = 1.
F(x) = x
Therefore, \int_{0}^{1} 1 dx = F(1) - F(0) = 1 - 0 = 1
Example 2
Find the area under the curve f(x) = x between -1 and 1.
We need to find F(x) such that F'(x) = x.
F(x) = \frac{x^2}{2}
Therefore, \int_{-1}^{1} x dx = F(1) - F(-1) = \frac{1^2}{2} - \frac{(-1)^2}{2} = 0
The positive area cancels the negative area.
Example 3
Find the area under the curve f(x) = sin(x) from 0 to \pi.
We need to find F(x) such that F'(x) = sin(x).
F(x) = -cos(x)
Therefore, \int_{0}^{\pi} sin(x) dx = F(\pi) - F(0) = -cos(\pi) - (-cos(0)) = -(-1) + 1 = 2
Antiderivatives
Capital F is called an antiderivative of lowercase f.
An antiderivative is not unique because F(x) + C also works, where C is any constant, because constant will be cancelled out by subtraction.
F(b) - F(a) = [F(b) + c] - [F(a) + c].
Short notation for representing F(b) - F(a) is F(x)]_a^b
Corollary of the Fundamental Theorem of Calculus
If F(x) is an antiderivative of f(x), then:
\frac{d}{dx} \int_{a}^{x} f(t) dt = f(x)
Indefinite Integrals
Given f(x) = 3x^2, then g(x) = x^3 is an antiderivative of f(x).
Also, h(x) = x^3 + 1 and k(x) = x^3 - 65 are antiderivatives of f(x).
If F(x) is an antiderivative of f(x), then the indefinite integral of f(x) is F(x) + C, where C is any arbitrary constant.
\int f(x) dx = F(x) + C
A definite integral is a number (area under the curve), while an indefinite integral is a family of functions.
For example, \int 3x^2 dx = x^3 + C where C represents any constant.
Indefinite Integrals List
\int f(x) dx = F(x) + C
\int 1 dx = x + C
\int x^\alpha dx = \frac{x^{\alpha + 1}}{\alpha + 1} + C if \alpha \neq -1
\int \frac{1}{x} dx = ln|x| + C
\int cos(x) dx = sin(x) + C
\int sin(x) dx = -cos(x) + C
\int e^x dx = e^x + C
Properties for the Indefinite Integral
\int [f(x) + g(x)] dx = \int f(x) dx + \int g(x) dx
\int [f(x) - g(x)] dx = \int f(x) dx - \int g(x) dx
\int k \cdot f(x) dx = k \cdot \int f(x) dx
Example: \int \pi x^\alpha dx = \pi \int x^\alpha dx = \pi [\frac{x^{\alpha + 1}}{\alpha + 1} + c] = \frac{\pi x^{\alpha + 1}}{\alpha + 1} + C'
Important Comments
Memorize the table of indefinite integrals via practice and checking the derivatives.
The table provided in the lecture won't be in the final exam.
There are methods for finding the derivative of any function, but not always an antiderivative.
Use Mathematica and Maple to check antiderivatives.