Welcome to the most important day of the semester. Today, we cover the Fundamental Theorem of Calculus (FTC).
Until now, we have treated Differential Calculus (tangents, rates of change) and Integral Calculus (areas) as separate topics. It was Isaac Newton and Gottfried Leibniz who discovered that these two seemingly different concepts are actually inverses of each other.
This discovery fueled the scientific revolution. It is the reason we can compute areas without taking limits of Riemann Sums. Let's dive in.
Before we state the theorem, we need to define a new type of function called an Accumulation Function (or Area Function).
Let $f(t)$ be a continuous function. We define $g(x)$ as the area under $f$ from a fixed point $a$ to a moving point $x$.
$$ g(x) = \int_a^x f(t) \,dt $$
Notice carefully:
Instruction: Move the slider n to see the area accumulate for each value of x.
Let $g(x) = \int_0^x t \,dt$ for $x \ge 0$. Find a formula for $g(x)$.
This integral represents the area under the line $y=t$ from $0$ to $x$.
Geometrically, this forms a triangle with base $x$ and height $x$ (since $y=x$).
$$ \text{Area} = \frac{1}{2}(\text{base})(\text{height}) = \frac{1}{2}(x)(x) = \frac{1}{2}x^2 $$
So, $g(x) = \frac{1}{2}x^2$.
Notice something amazing: The derivative of the area function $g(x) = \frac{1}{2}x^2$ is $g'(x) = x$. This is exactly the function $f(t)=t$ that we started with!
The observation from Example 1 is that the derivative of the accumulation function is the integrand. It turns out this is true for all continuous functions $f$!
If $f$ is continuous on $[a, b]$, then the function $g$ defined by
$$ g(x) = \int_a^x f(t) \,dt $$
is continuous on $[a, b]$ and differentiable on $(a, b)$, and:
$$ g'(x) = f(x) $$
It can be confusing to look at, so let's break down the pieces of this equation:
Let's build the intuition for this geometrically. We want to find $g'(x)$. By the limit definition of the derivative:
$$ g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} $$
Let's analyze the numerator, $g(x+h) - g(x)$.
Therefore, the difference $g(x+h) - g(x)$ is the area of the small "slice" under the curve from $x$ to $x+h$.
$$ g(x+h) - g(x) = \int_x^{x+h} f(t) \,dt $$
When $h$ is very small, this thin slice is almost a perfect rectangle.
So, we can approximate the area of this slice:
$$ \text{Area of slice} \approx f(x) \cdot h $$
Now, let's substitute this approximation back into the limit definition:
$$ g'(x) = \lim_{h \to 0} \frac{f(x) \cdot h}{h} = \lim_{h \to 0} f(x) = f(x) $$
This shows that the rate of change of the area (the derivative $g'(x)$) is precisely the height of the original function (the integrand $f(x)$) at that exact point.
Instruction: slide a until it matches u
Find the derivative of the function $g(x) = \int_1^x \sqrt{1+t^3} \,dt$.
We can apply FTC Part 1 directly. Here, $f(t) = \sqrt{1+t^3}$.
Since the lower limit is a constant (1) and the upper limit is $x$, the derivative is simply the integrand with $x$ plugged in.
$$ g'(x) = \sqrt{1+x^3} $$
It really is that simple!
Find $\frac{d}{dx} \int_1^{x^4} \sec(t) \,dt$.
Be careful! The upper limit is not just $x$, it is $x^4$. We must use the Chain Rule.
Let $u = x^4$. Then we are taking $\frac{d}{dx} \int_1^u \sec(t) \,dt$.
$$ \frac{d}{dx} \text{Integral} = \frac{d}{du} \text{Integral} \cdot \frac{du}{dx} $$
$$ = \sec(u) \cdot \frac{du}{dx} $$
$$ = \sec(x^4) \cdot (4x^3) $$
General Rule: $\frac{d}{dx} \int_a^{u(x)} f(t) \,dt = f(u(x)) \cdot u'(x)$
Find $h'(x)$ if $h(x) = \int_x^5 \sin(t^2) \,dt$. (Hint: Look at the limits of integration!)
It is standard to start the accumulation function $g(x) = \int_{a=0}^x f(t) \,dt$ at $0$, but what if we started somewhere else, like $a = -1$? How would that change the function?
Let's compare two functions using the same curve $f(t)=t$:
Geometrically, $B(x)$ includes the exact same area as $A(x)$, plus the extra chunk of area between $-1$ and $0$. Since that extra chunk is just a fixed number (a constant $C$), the two functions differ only by that constant.
This means changing the lower limit acts as a vertical shift of the accumulation function. It moves the graph up or down, but it does not change its shape or its slope.
This is why FTC1 ($g'(x)=f(x)$) works no matter what the starting point $a$ is. The derivative of the constant $C$ is just zero!
Instruction: Change the slider a to see how the accumulation function (the curve) shifts vertically, but its shape (and thus its slope) remains the same.
Part 1 showed us that differentiation and integration are inverses. Part 2 gives us the powerful computational method we've been waiting for. We no longer need to use Riemann sums to calculate areas!
If $f$ is continuous on $[a, b]$, then:
$$ \int_a^b f(x) \,dx = F(b) - F(a) $$
where $F$ is any antiderivative of $f$, that is, a function such that $F'(x) = f(x)$.
Evaluation Notation: We often use the bracket notation to represent the subtraction:
$$ F(x) \Big]_a^b = F(b) - F(a) $$
Evaluate $\int_1^3 e^x \,dx$.
Step 1: Find an antiderivative $F(x)$.
We know that the derivative of $e^x$ is $e^x$. So, $F(x) = e^x$.
Step 2: Evaluate at the endpoints.
$$ \int_1^3 e^x \,dx = e^x \Big]_1^3 = e^3 - e^1 $$
The exact area is $e^3 - e$.
Evaluate $\int_3^6 (4x + 3) \,dx$.
Step 1: Find the antiderivative.
The antiderivative of $4x$ is $2x^2$. The antiderivative of $3$ is $3x$.
So, $F(x) = 2x^2 + 3x$.
Step 2: Apply FTC2.
$$ \int_3^6 (4x + 3) \,dx = [2x^2 + 3x]_3^6 $$
$$ = (2(6)^2 + 3(6)) - (2(3)^2 + 3(3)) $$
$$ = (2(36) + 18) - (2(9) + 9) $$
$$ = (72 + 18) - (18 + 9) $$
$$ = 90 - 27 = 63 $$
Evaluate $\int_0^{\pi} \sin(x) \,dx$.
Evaluate $\int_0^{2\pi} \sin(x) \,dx$.
The antiderivative is the same as before: $F(x) = -\cos(x)$.
$$ \int_0^{2\pi} \sin(x) \,dx = [-\cos(x)]_0^{2\pi} $$
$$ = (-\cos(2\pi)) - (-\cos(0)) $$
$$ = (-1) - (-1) = 0 $$
Wait, the area is 0? Of course not. This shows a critical concept.
The result $0$ is not "area," it is the net area. The integral from $0$ to $\pi$ was $2$ (as we saw in CYU #3). The integral from $\pi$ to $2\pi$ is $-2$ (an equal "area" below the x-axis, which the integral treats as negative).
The definite integral $\int_a^b f(x) \,dx$ sums up the positive and negative "signed areas."
If a question asks for "the area," it almost always means the total area, which may require you to split the integral.
FTC2 can be rewritten to relate the integral of a rate of change to the total change in the quantity. Remember our discussion in class with the velocity function: The area under the curve of the velocity function is the net change in position (displacement).
Let's see how this applies to the Fundamental Theorem of Calculus (Part 2). If we let $f(t) = v(t)$ (velocity) and its antiderivative be $F(t) = s(t)$ (position), then the theorem shows the parallel:
$$ \int_a^b v(t) \,dt = s(b) - s(a) = \text{Net Change in Position (Displacement)} $$
$$ \int_a^b f(t) \,dt = F(b) - F(a) = \text{Net Change in } F $$
This leads to the general idea. Since $F'(x)$ represents the rate of change of $F$, the theorem $\int_a^b F'(x) \,dx = F(b) - F(a)$ says:
The integral of a rate of change is the net change:
$$ \int_a^b r(t) \,dt = \text{Net Change in quantity from } t=a \text{ to } t=b $$
A particle moves along a line with velocity $v(t) = 3t^2 - 12$ m/s from $t=1$ to $t=3$.
(a) Find the displacement (net change in position).
(b) Find the total distance traveled.
Displacement is the integral of velocity (the net area).
$$ s(3) - s(1) = \int_1^3 (3t^2 - 12) \,dt = [t^3 - 12t]_1^3 $$
$$ = ((3)^3 - 12(3)) - ((1)^3 - 12(1)) $$
$$ = (27 - 36) - (1 - 12) = (-9) - (-11) = 2 \text{ meters} $$
The particle is 2 meters to the right of where it started.
Total distance requires us to find the total area, not the net area. We must account for where the velocity is negative (moving backward).
First, find roots: $v(t) = 3t^2 - 12 = 0 \implies t^2 = 4 \implies t = 2$ (since $t=-2$ is not in our interval).
We must split the integral at $t=2$ and take the absolute value of the negative part.
$$ \text{Total Distance} = \int_1^3 |v(t)| \,dt = \int_1^2 -(3t^2 - 12) \,dt + \int_2^3 (3t^2 - 12) \,dt $$
$$ = \int_1^2 (12 - 3t^2) \,dt + \int_2^3 (3t^2 - 12) \,dt $$
$$ = [12t - t^3]_1^2 + [t^3 - 12t]_2^3 $$
$$ = ((12(2) - 2^3) - (12(1) - 1^3)) + ((3^3 - 12(3)) - (2^3 - 12(2))) $$
$$ = ((24 - 8) - (12 - 1)) + ((27 - 36) - (8 - 24)) $$
$$ = (16 - 11) + (-9 - (-16)) $$
$$ = 5 + 7 = 12 \text{ meters} $$
(The particle moved 5m left, then 7m right, for a total distance of 12m).
If $C'(x)$ is the marginal cost (in dollars per unit) to produce $x$ units, what does $\int_{100}^{200} C'(x) \,dx$ represent?
We have now connected the two halves of calculus. The derivative of an integral is the original function (FTC1), and the integral of a derivative is the net change in the function (FTC2).
After this lecture, you should be able to:
Test your knowledge by trying to solve these problems, which cover all parts of the Fundamental Theorem.
Find the derivative $h'(x)$ for the function:
$$ h(x) = \int_{\cos(x)}^{x^2} \sqrt{t^2 + 1} \,dt $$
Evaluate the definite integral:
$$ \int_1^4 \frac{x^2 + \sqrt{x}}{x} \,dx $$
Let $g(x) = \int_0^x f(t) \,dt$. The graph of $f(t)$ is a line segment from $(0, 4)$ to $(5, -1)$. On what interval is $g(x)$ increasing?
Water is pumped into a tank at a rate of $r(t) = 200 - 4t$ liters per minute, where $t$ is in minutes and $0 \le t \le 30$. How much water is in the tank after 30 minutes, assuming it was empty at $t=0$?