Welcome to a major turning point in calculus. So far, our entire journey has been about the "forward" problem: given a function $f(x)$, find its derivative $f'(x)$. We've become experts at finding the rate of change of a function.
Now, we're going to ask the backward problem: given a derivative $f'(x)$, can we find the original function $f(x)$?
This "backward" operation is called antidifferentiation, and it's the bridge to the second half of calculus. Why do we care? Imagine you know the velocity of a car (its speedometer reading). How can you find the total distance it has traveled? Velocity is the derivative of position, so to find the position, we must "un-differentiate" the velocity. This is the core idea.
Let's formalize this "backward" idea.
A function $F$ is called an antiderivative of $f$ on an interval $I$ if $F'(x) = f(x)$ for all $x$ in $I$.
In simple terms, an antiderivative is the function you get when you "un-do" a derivative.
Verify that $F(x) = x^3 + 2x$ is an antiderivative of $f(x) = 3x^2 + 2$.
We just need to check if $F'(x) = f(x)$. We take the derivative of $F(x)$:
$F'(x) = \frac{d}{dx}(x^3 + 2x) = 3x^2 + 2$.
Since $F'(x) = f(x)$, we have confirmed that $F(x)$ is indeed an antiderivative of $f(x)$.
Let $f(x) = 2x$. Find an antiderivative $F(x)$.
We need to think backward. What function, when we differentiate it, gives us $2x$? From the power rule, we know that $\frac{d}{dx}(x^2) = 2x$. So, one antiderivative is $F(x) = x^2$.
But wait... what about $G(x) = x^2 + 5$? Let's check its derivative: $G'(x) = \frac{d}{dx}(x^2 + 5) = 2x + 0 = 2x$.
And what about $H(x) = x^2 - 100$? $H'(x) = \frac{d}{dx}(x^2 - 100) = 2x + 0 = 2x$.
It seems that $x^2$, $x^2 + 5$, and $x^2 - 100$ are all antiderivatives of $f(x) = 2x$. This isn't a problem, but a crucial property.
Find an antiderivative for the function $f(x) = 5x^4$.
Example 2 showed us that if a function has one antiderivative, it has infinitely many. They all differ by a constant. This makes sense: the derivative of any constant is zero, so adding a constant doesn't change the derivative.
This leads to a key visual idea. The functions $y = x^2$, $y = x^2 + 1$, $y = x^2 - 1$, etc., form a "family" of parallel curves. They are all vertical shifts of each other. At any given $x$-value (say, $x=1$), the tangent lines on all these curves have the exact same slope (in this case, $m=2$).
If $F$ is one antiderivative of $f$, then the most general antiderivative of $f$ is given by:
$$ F(x) + C $$
where $C$ is an arbitrary constant, called the constant of integration.
The process of finding the most general antiderivative is so important that it gets its own name and symbol. We call it the indefinite integral.
$$ \int f(x) \, dx = F(x) + C $$
The expression $\int f(x) \, dx$ is read as "the indefinite integral of $f(x)$ with respect to $x$."
Find $\int 2x \, dx$.
From our work in Part 1, we know that an antiderivative of $2x$ is $x^2$. To find the most general antiderivative, we just add the constant of integration.
$$ \int 2x \, dx = x^2 + C $$
Find $\int \cos(x) \, dx$.
We ask ourselves: "What function, when differentiated, gives $\cos(x)$?" We know from our derivative rules that $\frac{d}{dx}(\sin(x)) = \cos(x)$.
Therefore, the most general antiderivative is:
$$ \int \cos(x) \, dx = \sin(x) + C $$
What is the most general antiderivative of $f(x) = e^x$?
Write your answer using indefinite integral notation.
Our strategy for finding antiderivatives is to reverse our known derivative rules. This gives us a table of basic antiderivatives that you should memorize. (This table is from page 358 of Stewart's 9th ed.)
| Function $f(x)$ | Antiderivative $\int f(x) \, dx$ |
|---|---|
| $k$ (a constant) | $kx + C$ |
| $x^n$ (where $n \neq -1$) | $\frac{x^{n+1}}{n+1} + C$ (The Power Rule for Integrals) |
| $x^{-1}$ or $\frac{1}{x}$ | $\ln|x| + C$ |
| $e^x$ | $e^x + C$ |
| $b^x$ | $\frac{b^x}{\ln(b)} + C$ |
| $\cos(x)$ | $\sin(x) + C$ |
| $\sin(x)$ | $-\cos(x) + C$ |
| $\sec^2(x)$ | $\tan(x) + C$ |
| $\csc^2(x)$ | $-\cot(x) + C$ |
| $\sec(x)\tan(x)$ | $\sec(x) + C$ |
| $\csc(x)\cot(x)$ | $-\csc(x) + C$ |
| $\frac{1}{1+x^2}$ | $\arctan(x) + C$ |
| $\frac{1}{\sqrt{1-x^2}}$ | $\arcsin(x) + C$ |
This rule is a major point of confusion, so let's clarify it.
Just like derivatives, integrals have sum, difference, and constant multiple rules:
Find $\int (4x^3 - \sqrt{x} + \frac{5}{x^2}) \, dx$.
Step 1: Rewrite all terms as powers of $x$. This is the most important step! Do not try to integrate until you do this.
$\sqrt{x} = x^{1/2}$
$\frac{5}{x^2} = 5x^{-2}$
So, we are finding: $\int (4x^3 - x^{1/2} + 5x^{-2}) \, dx$
Step 2: Integrate term-by-term using the Power Rule.
$$ = 4\left(\frac{x^{3+1}}{3+1}\right) - \left(\frac{x^{1/2 + 1}}{1/2 + 1}\right) + 5\left(\frac{x^{-2+1}}{-2+1}\right) + C $$
Step 3: Simplify.
$$ = 4\left(\frac{x^4}{4}\right) - \left(\frac{x^{3/2}}{3/2}\right) + 5\left(\frac{x^{-1}}{-1}\right) + C $$
$$ = x^4 - \frac{2}{3}x^{3/2} - 5x^{-1} + C $$
$$ = x^4 - \frac{2}{3}x^{3/2} - \frac{5}{x} + C $$
(Note: We only need to write one $+C$ at the end, as it represents the sum of all the individual constants.)
Find $\int (3\sec(x)\tan(x) - 7e^x + 2\sin(x)) \, dx$.
We integrate term-by-term by pulling out the constants and using our table:
$$ = 3\int \sec(x)\tan(x) \, dx - 7\int e^x \, dx + 2\int \sin(x) \, dx $$
Now, substitute the rules from the table:
$$ = 3(\sec(x)) - 7(e^x) + 2(-\cos(x)) + C $$
$$ = 3\sec(x) - 7e^x - 2\cos(x) + C $$
Find $\int (x^5 + \frac{3}{x} - 2) \, dx$.
The general antiderivative $F(x) + C$ represents an infinite family of functions. How do we find one specific function? We need more information. We need a single point that our function must pass through. This is called an initial condition.
Find $f(x)$ given that $f'(x) = 10x - 3$ and $f(1) = 4$.
Step 1: Find the general antiderivative.
$f(x) = \int f'(x) \, dx = \int (10x - 3) \, dx$
$f(x) = 10\left(\frac{x^2}{2}\right) - 3x + C$
$f(x) = 5x^2 - 3x + C$
Step 2: Use the initial condition to solve for $C$.
We are given $f(1) = 4$. This means "when $x=1$, $y=4$". We plug these values into our general solution:
$4 = 5(1)^2 - 3(1) + C$
$4 = 5 - 3 + C$
$4 = 2 + C$
$C = 2$
Step 3: Write the specific solution.
The specific function we're looking for is $f(x) = 5x^2 - 3x + 2$.
Find $f(x)$ given that $f''(x) = \sin(x)$, $f'(0) = 5$, and $f(0) = 1$.
We have to antidifferentiate twice, and we will have to solve for two constants ($C_1$ and $C_2$).
Step 1: Find $f'(x)$ from $f''(x)$.
$f'(x) = \int f''(x) \, dx = \int \sin(x) \, dx = -\cos(x) + C_1$
Step 2: Use the first initial condition to find $C_1$.
We use $f'(0) = 5$ on $f'(x) = -\cos(x) + C_1$:
$5 = -\cos(0) + C_1$
$5 = -1 + C_1$
$C_1 = 6$
So, we now have the specific first derivative: $f'(x) = -\cos(x) + 6$.
Step 3: Find $f(x)$ from $f'(x)$.
$f(x) = \int f'(x) \, dx = \int (-\cos(x) + 6) \, dx = -\sin(x) + 6x + C_2$
Step 4: Use the second initial condition to find $C_2$.
We use $f(0) = 1$ on $f(x) = -\sin(x) + 6x + C_2$:
$1 = -\sin(0) + 6(0) + C_2$
$1 = -0 + 0 + C_2$
$C_2 = 1$
Step 5: Write the final solution.
$f(x) = -\sin(x) + 6x + 1$.
Find $g(x)$ given $g'(x) = 4x^3 + e^x$ and $g(0) = 7$.
This is a classic physics application of initial-value problems. Recall the relationship:
This means we can move "up" the chain by antidifferentiating:
A particle moves with acceleration $a(t) = 3t + 8$. Its initial velocity is $v(0) = -6$ m/s and its initial position is $s(0) = 10$ m. Find the position function $s(t)$.
This is a second-order initial-value problem, just like Example 2 in Part 4.
Step 1: Find $v(t)$ from $a(t)$.
$v(t) = \int a(t) \, dt = \int (3t + 8) \, dt = 3\left(\frac{t^2}{2}\right) + 8t + C_1$
Step 2: Find $C_1$ using $v(0) = -6$.
$-6 = \frac{3}{2}(0)^2 + 8(0) + C_1 \implies C_1 = -6$.
So, $v(t) = \frac{3}{2}t^2 + 8t - 6$.
Step 3: Find $s(t)$ from $v(t)$.
$s(t) = \int v(t) \, dt = \int (\frac{3}{2}t^2 + 8t - 6) \, dt = \frac{3}{2}\left(\frac{t^3}{3}\right) + 8\left(\frac{t^2}{2}\right) - 6t + C_2$
$s(t) = \frac{1}{2}t^3 + 4t^2 - 6t + C_2$
Step 4: Find $C_2$ using $s(0) = 10$.
$10 = \frac{1}{2}(0)^3 + 4(0)^2 - 6(0) + C_2 \implies C_2 = 10$.
Step 5: Write the final solution.
$s(t) = \frac{1}{2}t^3 + 4t^2 - 6t + 10$.
A ball is thrown upward with a speed of 15 m/s from the edge of a cliff 80 m high. Find its height $s(t)$ above the ground $t$ seconds later. (Assume acceleration due to gravity is $a(t) = -9.8$ m/s²).
Step 1: Identify all initial conditions.
The acceleration is given: $a(t) = -9.8$.
The initial velocity is $v(0) = +15$ m/s (positive because it's thrown upward).
The initial position (height) is $s(0) = 80$ m.
Step 2: Find $v(t)$.
$v(t) = \int a(t) \, dt = \int -9.8 \, dt = -9.8t + C_1$
Using $v(0) = 15$: $15 = -9.8(0) + C_1 \implies C_1 = 15$.
So, $v(t) = -9.8t + 15$.
Step 3: Find $s(t)$.
$s(t) = \int v(t) \, dt = \int (-9.8t + 15) \, dt = -9.8\left(\frac{t^2}{2}\right) + 15t + C_2$
$s(t) = -4.9t^2 + 15t + C_2$
Step 4: Find $C_2$ using $s(0) = 80$.
$80 = -4.9(0)^2 + 15(0) + C_2 \implies C_2 = 80$.
Step 5: Write the final solution.
The height of the ball at time $t$ is: $s(t) = -4.9t^2 + 15t + 80$.
A car is traveling at 88 ft/s (which is 60 mph). The driver brakes, and the car experiences a constant deceleration of $a(t) = -16$ ft/s². How far does the car travel before it comes to a complete stop?
Antidifferentiation is the inverse operation of differentiation. It allows us to "go backward" from a rate of change to an original function. This concept is the foundation for the entire second half of calculus (Integral Calculus) and is essential for solving problems in physics, engineering, biology, and economics.
After this lecture, you should be able to: