Fundamental Theorem of Calculus Made Easy

Fundamental Theorem of Calculus — or short for FTC. Wow, it sounds scary and new, but it's all about the relationship between derivatives and integrals. FTC shows that derivatives and integrals are indeed inverse of each other.

Real-world applications of derivatives

velocity, acceleration, optimization, etc.

Real-world applications of integrals

finding areas, accumulations, net changes, etc.

Part 1

Part 1 is about the formula that relates the derivative to the integral. If $r(x)$ is continuous on $[a, b]$ then:

$$f(x) = \int_{a}^{x} r(t) \, dt$$

$f(x)$ is continuous on $[a, b]$, it is differentiable on $(a, b)$, and $f'(x) = r(x)$.

Type	Given integral	Rules to follow
# as endpoints	$f(x)= \int_{a}^{x} r(t) \,dt$	Plug in $x$ for $t$.
# as endpoints	$f(x)= \int_{x}^{a} r(t) \,dt$	Reverse limits of integration, multiply by $-1$, then plug in $x$ for $t$.
function as endpoint	$f(x)= \int_{a}^{g(x)} r(t) \,dt$	Plug in $g(x)$ for $t$, then multiply by $\frac{dg}{dx}$.
function as endpoint	$f(x)= \int_{g(x)}^{a} r(t) \,dt$	Reverse limits, multiply by $-1$, plug in $g(x)$ for $t$, then multiply by $\frac{dg}{dx}$.
functions as endpoints	$f(x)= \int_{g(x)}^{h(x)} r(t) \,dt$	Step 1: Split: $\int_{g(x)}^{0} r(t)\,dt + \int_{0}^{h(x)} r(t)\,dt$ Step 2: Reverse the first: $\int_{0}^{g(x)} r(t)\,dt$ Step 3: Multiply by $-1$: $-\int_{0}^{g(x)} r(t)\,dt$ Step 4: Plug in $g(x)$ and $h(x)$ for $t$ Step 5: Multiply by $\frac{dg}{dx}$ and $\frac{dh}{dx}$

Basically, if $f(x)$ is an integral of $r(x)$ then $r(x)$ is a derivative of $f(x)$.

It is like saying the square root of 4 is 2, then 2 squared is 4.

Before we move on to Part 2, let me give you an example. Try it yourself first and then move on to the explanation for better results.

In the next example, we are being asked to find the derivative of an antiderivative. The result should be the original function. Step 4 is in every case — plugging in the endpoints gives you the initial and final accumulated quantities, allowing you to find the net change.

When the result of an integral is a function, the slope of that function represents the net change of the quantity being measured, and its positive or negative nature can be determined based on the slope's sign.

Net change — signifies how much a quantity has increased or decreased during that interval, taking into account all positive and negative influences (overall trend of net accumulation or net reduction of the quantity).

Example

Given:

$$f(x) = \int_{x^2}^{\cos(x)} t^2 + 2 \, dt$$

Find $f'(x)$.

$$f'(x) = (\cos(x))^2 + 2 - 2x(x^4 + 2)$$

The resulting function has a negative slope. What does that mean in the context of net change?

The function is decreasing as you move along the x-axis. When the rate of change is negative, it indicates a decrease in the quantity — an overall reduction.

graph of $f'(x) = (\cos x)^2 + 2 - 2x(x^4+2)$

In real-world scenarios, a negative slope and negative net change could represent: decreasing temperature, declining population, decreasing stock prices, or decreasing levels of a substance.

General Formula

$$f'(x) = r[h(x)] \cdot h'(x) - r[g(x)] \cdot g'(x)$$

$h(x)$ — upper limit
$g(x)$ — lower limit

Exercise

$$f(x) = \int_{2x^2}^{6} 6t^3 - 6t \, dt \quad \text{Find } f'(x).$$

Check your answer

$$f'(x) = -192x^7 + 48x^3$$

Part 2

The second part of the theorem provides a way to compute definite integrals using antiderivatives. If we have a continuous function $f(x)$ and its antiderivative $F(x)$ on $[a, b]$, then:

$$\int_{a}^{b} f(x) \, dx = F(x) \Big|_{a}^{b} = F(b) - F(a)$$

Both limits of integration must be constant. Also, the interval must be closed.

Example

$$\int_{\pi}^{2\pi} \cos\theta \, d\theta \quad \text{where } F(x) = \sin\theta$$

Solution:

$$F(b) = \sin(2\pi), \quad F(a) = \sin(\pi)$$ $$F(b) - F(a) = \sin(2\pi) - \sin(\pi) = 0 - 0 = 0$$

We just found the net area — the area above the x-axis is the same as the area below it. The cosine curve is symmetric over $[\pi, 2\pi]$.

graph of $\cos(\theta)$ — first period $[0, 2\pi]$

Net Change Theorem (advanced)

The Net Change Theorem is closely related to the FTC and provides a practical way to compute net changes over a specific interval using definite integrals.

If we integrate a rate of change of a function, we find the net amount of change of the function during the period of integration.

If $f'(x)$ is the rate of change of $f(x)$, then:

$$\int_{a}^{b} f'(x) \, dx = f(b) - f(a)$$

Example

Water is being pumped into a tank at a rate (gallons) given by $w(t) = 60 - 10\sqrt{t}$, with $0 \leq t \leq 120$ where $t$ is time in minutes. The tank had 1200 gallons when pumping began. How much water is in the tank after 49 minutes?

$$1200 + \int_{0}^{49} 60 - 10\sqrt{t} \, dt$$ $$= 1200 + \int_{0}^{49} 60 - 10t^{\frac{1}{2}} \, dt$$ $$= 1200 + \left[60t - \frac{20}{3} t^{\frac{3}{2}}\right]_{0}^{49}$$ $$= 1200 + \left[60 \cdot 49 - \frac{20}{3} \cdot 49^{\frac{3}{2}}\right] - [0] = \frac{5560}{3}$$

About 1853.3 gallons of water will be in the tank after 49 minutes of pumping.