Math Insight

We have defined the definite integral as a limit of Riemann sums. Specifically, we said that $$\int_a^b f(t) \, dt$$ is the limit as $n$ goes to infinity of the Riemann sum $$\sum_{i=1}^n f(t_i) \Delta t$$ where the interval $[a,b]$ is divided into $n$ intervals, $\Delta t$ is the width of these intervals, and $t_i$ denotes either the left or right endpoints of these $n$ intervals.

We will now look at how to evaluate the definite integral.

1. Recall that we came up with the Riemann sum as a way to estimate the total change from the rate of change. As the number of intervals increases, the Riemann sum approaches the total change. This means that an alternative meaning for the definite integral $\displaystyle \int_a^b f(t) \, dt$ is the total change from $a$ to $b$ of some state variable. Let's call this state variable $F$. Then $\displaystyle \int_a^b f(t) \, dt$ represents the total change in $F$ from $a$ to $b$ if $f(t)$ represents the rate of change of $F$. In other words, $\displaystyle \int_a^b f(t) \, dt$ represents the total change in $F$ from $a$ to $b$ if $F$ satisfies the differential equation $\frac{dF}{dt} =$
   .

   Suppose we knew some $F(t)$ that satisfied this differential equation 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:
   $\displaystyle \int 3 t^{2} - 1 \, dt =$
   
   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? No effect. None whatsoever! None.
   

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

   If $f(t)$ is a continuous function and $F(t)$ is any differentiable function such that $F'(t)=f(t)$, 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. There are many functions for which we cannot find an antiderivative, however, and in those case we must evaluate the definite integral through Riemann sums. If we have to use Riemann sums, we typically cannot find an exact value, but we can get a good estimate by taking small intervals.

   It may be helpful to think about the relationship between Forward Euler and Riemann sums. With Forward Euler, we are estimating the value $F(b)$ by taking $F(a)$ and adding $\displaystyle \sum_{i=1}^{n} f(t_i)\Delta t$, where $t_i$ are the left endpoints of the $n$ intervals into which we split $[a,b]$. With left-hand Riemann sums, we are estimating the difference $F(b)-F(a)$ by $\displaystyle \sum_{i=1}^{n} f(t_i)\Delta t$. How can we derive the Riemann sum estimate from Forward Euler? Multiply by 2. Add f(a). Subtract f(a).
   

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.

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

2. $\displaystyle \int_{0}^{1} 2 t + e^{4 t}\, dt$

   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.

   $\displaystyle \int_{0}^{1} 2 t + e^{4 t}\, dt =$
   $\Big|_0^1=$
   $-$
   $=$
   

   Need help?Hide help

   Hint

   To enter $e^{x}$, type e^(x) or exp(x).

   Hide help

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

   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. We can also ignore the absolute values that would be necessary in this problem for an indefinite integral, because we know that $t$ is always positive in this definite integral.

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

   Need help?Hide help

   Hint

   To enter $\log (x)$, type log(x).

   Hide help