Math Insight

Integrals and the Fundamental Theorem of Calculus

Math 1241, Fall 2017
Name:
ID #:
Due date: Nov. 13, 2017, 11:59 p.m.
Table/group #:
Group members:
Total points: 3
  1. We have defined the definite integral as a limit of Riemann sums. The integral $\int_a^b f(t) \, dt$ is the limit as $n$ goes to infinity of the Riemann sum $\sum_{i=1}^n f(t_i) \Delta t$. To define the Riemann sum, we divide the interval $[a,b]$ into $n$ intervals of width $\Delta t$. We evaluate the function $f$ at the points $t_i$, which could be either the left or right endpoints of these $n$ intervals.

    Here we show how to use the the Fundamental Theorem of Calculus to evaluate the definite integral without calculating a Riemann sum, which works as long as we can calculate the indefinite integral, or antiderivative, $\int f(t) \, dt$.

    1. If $f(t)$ is the rate at which a function $F(t)$ is changing, then the Riemann sum estimates the total change in $F(t)$ from $a$ to $b$. As we increase $n$, this estimate improves. The limit as $n$ goes to infinity, i.e., the definite integral $\displaystyle \int_a^b f(t) \, dt$, is the actual value of the total change in $F(t)$ from $a$ to $b$.

      Now, what does it mean for $f(t)$ to be the rate at which $F(t)$ is changing? It means that $F$ satisfies the differential equation $\frac{dF}{dt} =$
      .

      Suppose we knew what the function $F(t)$ was, and we wanted to know the total change in $F(t)$ from $t=a$ to $t=b$. We could do this just by subtracting two values of $F$:
      .

    2. Let's look at a specific example. Suppose $f(t)=3 t^{2} - 1$ and we want to know the change in $F$ from $t=0$ to $t=2$. We can solve the differential equation $\frac{dF}{dt}=3 t^{2} - 1$ using an indefinite integral:
      $\displaystyle F(t) = \int 3 t^{2} - 1 \, dt =$

      To finish solving the differential equation, we need an initial condition. Let's see what happens when we pick three different initial conditions.

      First, let's try $F(0)=0$. In this case, what is $C$?
      So $F(t)=$
      . To find the total change, we need to subtract $F(0)$ from $F(2)$. The total change is
      -
      =
      .

      Next, let's try $F(0)=2$. In this case, what is $C$?
      So $F(t)=$
      . To find the total change, we need to subtract $F(0)$ from $F(2)$. The total change is
      -
      =
      .

      Now, you pick an initial condition to try: $F(0)=$
      . In this case, what is $C$?
      So $F(t)=$
      . To find the total change, we need to subtract $F(0)$ from $F(2)$. The total change is
      -
      =
      .

      What effect does the initial condition have on the total change?

    3. We can evaluate a definite integral either by taking the limit of Riemann sums or by solving a pure time differential equation. This is the idea of the Fundamental Theorem of Calculus:

      If $f(t)$ is a continuous function and $F(t)$ any differentiable function with $F'(t)=f(t)$, then $$\int_a^b f(t)\, dt=F(b)-F(a)=F(t)\Big |_a^b$$ The notation $F(t)\Big |_a^b$ is just short-hand notation for $F(b)-F(a)$.

      In other words, if we can find an antiderivative of $f(t)$, we can evaluate the definite integral just by plugging in the limits of integration and subtracting. (We can't always find an antiderivative, but we won't worry about such cases here.)

  2. Evaluate the following definite integrals using the Fundamental Theorem of Calculus.
    1. $\displaystyle \int_{0}^{2} 4 t^{3} + 3 t - 2\, dt$

      The Fundamental Theorem of Calculus says we just need to find an antiderivative, evaluate it at the limits of integration, and subtract. For the antiderivative, we just use the same rules we found for evaluating indefinite integrals. We can forget about the constant for definite integrals, because any antiderivative will work.

      $\displaystyle \int_{0}^{2} 4 t^{3} + 3 t - 2\, dt =$
      $\Bigg|_0^2=$
      $-$
      $=$

    2. $\displaystyle \int_{0}^{1} 2 x + e^{4 x}\, dx =$
      $\Bigg|_0^1=$
      $-$
      $=$

    3. $\displaystyle \int_{1}^{3} t^{2} - \frac{1}{3 t}\, dt =$
      $\Bigg|_1^3=$
      $-$
      $=$

      (In this case, since $t$ is always positive, we can ignore the absolute value. However, it never hurts to keep the absolute value.)

  3. Evaluate the following integrals. (Don't round your answers.)
    1. $\displaystyle \int_{-2}^{0} 2 e^{3 x}\, dx$ =
    2. $\displaystyle \int_{-3}^{-2} 1 - \frac{1}{t}\, dt$ =