MATH 145 – Calculus I
November 6, 2025
Suppose \(v(t)\) is the velocity of an object of an object at time \(t\) and consider the distance traveled over the time interval \(\Delta t\).
\[ \begin{aligned} \int_a^b v(t) \, dt &= \lim\limits_{n \to \infty} \sum_{i = 1}^n v(t_i) \Delta t \\ &= \textrm{distance traveled (displacement)} \\ &= \textrm{area under the graph of } v(t) \\ &= f(b) - f(a) \end{aligned} \]
Suppose \(f(t)\) is a continuous function on the interval \([a, b]\) and that \(F(t)\) is an antiderivative of \(f(t)\). Then
\[ \int_a^b f(t) \, dt = F(b) - F(a) \]
Notes:
This makes evaluating definite integrals easy, provided you can find an antiderivative.
Alternatively, \(\displaystyle\int_a^b f'(t) \, dt = f(b) - f(a)\).
Also, \(F(b) = F(a) + \displaystyle\int_a^b f(t) \, dt\).
Example: At 10 a.m. A cistern at a rate of \(R(t) \, \frac{gallons}{hour}\), where \(R(t)\) has the graph given below.
Estimate the amount of water in the tank at noon.
Examples:
Theorem: \(\displaystyle\int_a^b c \cdot f(x) + d \cdot g(x) \, dx = c \displaystyle\int_a^b f(x) \, dx + d \displaystyle\int_a^b g(x) \, dx\).
Theorem: If \(a < c < b\), \(\displaystyle\int_a^b f(x) \, dx = \displaystyle\int_a^c f(x) \, dx + \displaystyle\int_c^b f(x) \, dx\).
Note: There is no product or quotient rule for definite integrals, at least not directly. (Take Calc II to learn more!)
Every continuous function \(f(x)\) has an antiderivative.
If \(f(x)\) is a continuous function on an interval and \(a\) is any point, then the function:
\[ F(x) = \int_a^x f(t) \, dt, \]
is an antiderivative of \(f(x)\).
If \(f(x)\) is a continuous function on an interval and \(a\) is any point, then the function:
\[ F(x) = \int_a^x f(t) \, dt, \]
is an antiderivative of \(f(x)\).
In other words:
\[ \frac{d}{dx} \left[ \int_a^x f(t) \, dt \right] = f(x). \]