Definite Integrals

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What is integration good for?

As we have seen, definite integrals can be used to compute area, just as derivatives can be used to compute slopes of tangent lines. But there is much more to derivatives than slopes of tangent lines, and there is much more to integrals than area. Integrals can be used to compute any bulk quantity.

To compute any bulk quantity:

Break it up into small pieces (of size $\Delta x$) that are easier to understand.
Estimate the contribution $f(x)\, \Delta x$ of each piece.
Add up the pieces to get the Riemann Sum $\displaystyle\sum_{i=1}^n f(x_i^*) \,\Delta x$.
Take a limit to slice finer and finer, infinitely fine, to get the exact answer: $$\int_a^b f(x) \,dx = \lim_{n\to\infty} \,\sum_{i=1}^n f(x_i^*) \,\Delta x.$$

Example: Distance

Suppose that we know a particle's velocity as a function of time. If the velocity is constant, then it's easy to find the distance traveled: $$ \hbox{Rate} \times \hbox{Time} = \hbox{Constant Velocity} \times \hbox{Time} = \hbox{Distance}.$$ If the velocity is almost constant, then we can get a good approximation by pretending that it is constant. If the velocity varies a lot, then we break our time interval into little sub-intervals on which the velocity doesn't change much, estimate each one, and add up the pieces. This is exactly the same strategy that we used for area under a curve.

Example 1: Find the distance traveled by a person walking at 2 MPH from 1:00 PM to 4:00 PM.

Solution: 3 hours at 2 MPH makes $3 \times 2 = 6$ miles.

Example 2: Find the approximate distance traveled by a person walking at velocity $v(t)=2 + 0.02 t$ MPH between $t=1$ (o'clock) and $t=4$ (o'clock).

Solution: The velocity doesn't change much between 1 and 4 PM, so we approximate our velocity as $v(1)=2.02$ and compute $(4-1)v(1) = 3(2.02)=6.06$ miles traveled.

The answer in Example 2 is an underestimate, a little bit too small, since the velocity goes up from $v=2.02$ at $t=1$ to $v=2.08$ at $t=4$. If we wanted an overestimate of the area, we would use the value of $v(t)$ at the right endpoint instead: $(4-1)v(4)=3(2.08)=6.24$. And if we wanted a more accurate guess, we might use the midpoint $f(2.5)=2.05$ and estimate the area as $(4-1)v(2.5)=3(2.05)=6.015$. But whether we use the left endpoint, the right endpoint, or the midpoint, we get roughly the same answer, since the velocity just isn't changing very much between $t=1$ and $t=4$.

If the paragraph and examples above look familiar, it is because this is exactly like finding the area under $y= 2 + 0.02x$ between $x=1$ and $x=4$. Only, in this case, the function is $v$ instead of $f$, and the variable $t$ instead of $x$, but it's the same calculation.

To compute the distance traveled between time $t=a$ and time $t=b$ from the function $v(t)$, we

Break our time interval into $n$ pieces, so that the velocity is practically constant in each piece.
The distance traveled in the $i$-th time interval is approximately $v(t_i^*)\Delta t$.
Add these up and take a limit to get a total distance of $$\int_a^b v(t) dt = \lim_{n\to\infty} \sum_{i=1}^n v(t_i^*) \Delta t,$$ where we may as well take $t_i^*$ to be $t_i$.

It follows that distance is the area between the velocity curve and the $t$-axis.

DO: What does it mean if the velocity is negative? How would that affect the distance traveled? Sketch a graph of a velocity curve that is positive for some times and negative for others to help you see what is happening.

The Six Pillars of Calculus

Trigonometry Review

Exponential Functions

Logarithms and Inverse functions

Intro to Limits

Limit Laws and Computations

Continuity and the Intermediate Value Theorem

Limits at Infinity

Rates of Change

The Derivative Function

Basic Differentiation Rules

Product and Quotient Rules

Derivatives of Trig Functions

The Chain Rule

Implicit Differentiation

Derivatives of Logs

Derivatives in Science

Related Rates

Linear Approximation and Differentials

Differentiation Review

Absolute and Local Extrema

The Mean Value and other Theorems

$f$ vs. $f'$

Indeterminate Forms and L'Hospital's Rule

Optimization

Newton's Method

Anti-derivatives

The Area under a curve

Definite Integrals

The Fundamental Theorem of Calculus

The Indefinite Integral and the Net Change

Substitution

Area Between Curves

Volumes

What is integration good for?

Example: Distance