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5.3 The Fundamental Theorem of Calculus

last time: \(\int_{a}^{b} f(x) dx\) represents area between \(f(x)\) and x-axis from \(x=a\) to \(x=b\)

Coordinate graph showing a curve with shaded area under it between vertical lines at a and b.

now let's look at an area function

\[g(x) = \int_{a}^{x} f(t) dt\]
Graph of f(t) with shaded area from t=a to a variable point t=x on the horizontal axis.

\(x\) can move around

area as function of \(x\)

\(t\) is called a "dummy variable" we don't want to reuse \(x\)

\[\int_{a}^{x} f(t) dt = g(x)\]
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\[g(x) = \int_{a}^{x} f(t) dt\]

what is \(g'(x)\)? what is the rate of change of the accumulated area?

\[g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h}\]

\(g(x) = \int_{a}^{x} f(t) dt\)    â‘ 

\(g(x+h) = \int_{a}^{x+h} f(t) dt\)

Graph of a curve over t-axis with regions labeled 1 (from a to x) and 2 (from x to x+h).
\[g(x+h) - g(x) = \underbrace{\int_{a}^{x+h} f(t) dt}_{total} - \underbrace{\int_{a}^{x} f(t) dt}_{â‘ } = â‘¡\]

when \(h \to 0\), â‘¡ becomes approximately a rectangle

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The Fundamental Theorem of Calculus

Consider a function curve where we analyze a small vertical slice to understand the rate of change of the accumulated area.

  • height is \( f(x) \)
  • width is \( x + h - x = h \)
  • Area \( \approx f(x) \cdot h \)

which is the numerator of the derivative definition:

\[ g'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Graph of a curve f(x) on t-y axes with a shaded vertical rectangular slice of width h between x and x+h.

So,

\[ g'(x) = \lim_{h \to 0} \frac{f(x) \cdot h}{h} = \lim_{h \to 0} f(x) = f(x) \]

this gives us

\[ \frac{d}{dx} \int_{a}^{x} f(t) dt = f(x) \]

This is called the Fundamental Theorem of Calculus (part 1) (FTC 1)

It says the rate of change of accumulated area is equal to the function bounding above.

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Example

Given the function:

\[ g(x) = \int_{0}^{x} \cos(\pi t) dt \]

find \( g'(x) \).

Graph of cos(pi t) showing the area shaded from t=0 to t=x under the curve.

Applying FTC 1

\[ \text{FTC 1: } \frac{d}{dx} \int_{a}^{x} f(t) dt = f(x) \]

By substituting our specific function into the theorem:

\[ \text{so, } \frac{d}{dx} \int_{0}^{x} \cos(\pi t) dt = \cos(\pi x) \]

Note: The variable of integration \( t \) is replaced by the upper limit variable \( x \).

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Example: Rate of Change of Area

Find the rate of change of the function representing the area under \( f(t) = t + 5 \) from \( t = -5 \) to \( t = x \).

Coordinate graph showing a linear function f(t)=t+5 and a shaded triangular area from t=-5 to t=x.

Area is defined as:

\[ A(x) = \int_{-5}^{x} (t + 5) \, dt \]
\[ A'(x) = \frac{d}{dx} \int_{-5}^{x} (t + 5) \, dt = \boxed{x + 5} \]

Note the region is a triangle with base \( x + 5 \) and height \( f(x) = x + 5 \).

\[ A(x) = \frac{1}{2} (x + 5)(x + 5) = \frac{1}{2} (x + 5)^2 \]
\[ A'(x) = \frac{1}{2} \cdot 2 \cdot (x + 5) = x + 5 \]

Which matches the result from FTC 1.

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Fundamental Theorem of Calculus Part 1 (FTC 1)

\[ \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \]

Only true if the variable \( x \) is the upper limit of integration.

If it's not, we can use the property:

\[ \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \]

Example

\[ \frac{d}{dx} \int_{x}^{1} \sqrt{t^5 + 1} \, dt \]

Note: \( x \) is not the upper limit.

\[ = \frac{d}{dx} \left( -\int_{1}^{x} \sqrt{t^5 + 1} \, dt \right) \]
\[ = -\frac{d}{dx} \int_{1}^{x} \sqrt{t^5 + 1} \, dt = \boxed{-\sqrt{x^5 + 1}} \]

(Use FTC 1 for the final step)

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What if \( x \) is not simply \( x \)?

Example

\[ y = \int_{2}^{e^{3x}} \sin^2(5t) \, dt \quad y' = ? \]

FTC 1 needs the upper limit to be just a variable.

Let's use the Chain Rule.

Let \( u = e^{3x} \)

then \( y \) becomes \( y = \int_{2}^{u} \sin^2(5t) \, dt = y(u) \text{ w/ } u = e^{3x} \)

\[ \text{Chain Rule: } \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} \]
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\[ y = \int_{2}^{u} \sin^2(5t) \, dt \quad u = e^{3x} \]
\[ \begin{aligned} \frac{dy}{dx} &= \frac{dy}{du} \frac{du}{dx} \\ &= \left( \frac{d}{du} \int_{2}^{u} \sin^2(5t) \, dt \right) \frac{d}{dx} e^{3x} \end{aligned} \]

use FTC 1:

\[ \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \]\[ \frac{d}{du} \int_{a}^{u} f(t) \, dt = f(u) \]
\[ = \sin^2(5u) \cdot 3e^{3x} \]

\( \nwarrow u = e^{3x} \)

\[ = \sin^2(5e^{3x}) \cdot 3e^{3x} = \boxed{3e^{3x} \sin^2(5e^{3x})} \]
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The Fundamental Theorem of Calculus (Part 2)

The second part of the Fundamental Theorem of Calculus allows us to calculate \[ \int_{a}^{b} f(x) dx \] exactly (no more Riemann sums).

Derivation from FTC 1

\[ \text{FTC 1 : } \frac{d}{dx} \int_{a}^{x} f(t) dt = f(x) \]

So the antiderivative of \( \frac{d}{dx} \int_{a}^{x} f(t) dt \) is the antiderivative of \( f(x) \).

\[ \rightarrow \int_{a}^{x} f(t) dt = F(x) + C \quad \text{where } F'(x) = f(x) \]

We know:

\[ \int_{a}^{a} f(t) dt = 0 \]

So,

\[ \int_{a}^{a} f(t) dt = F(a) + C = 0 \rightarrow \]
\[ C = -F(a) \]
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From:

\[ \int_{a}^{x} f(t) dt = F(x) + C \]

We get, by letting \( x = b \):

\[ \int_{a}^{b} f(t) dt = F(b) + C \quad \text{but } C = -F(a) \]\[ = F(b) - F(a) \]

Therefore,

\[ \int_{a}^{b} f(x) dx = F(b) - F(a) \]\[ \text{where } F' = f \]

Fundamental Theorem of Calculus (part 2)

FTC 2

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Fundamental Theorem of Calculus (FTC) 2 Example

Example Problem

\[ \int_{0}^{3} 2x \, dx \]

Where \( f(x) = 2x \).

FTC 2:

\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]

Find \( F(x) \) such that \( F'(x) = f(x) \) (in other words, \( F \) is the antiderivative of \( f(x) \)).

\[ f(x) = 2x \]\[ F(x) = 2 \cdot \frac{x^2}{2} + C = x^2 + C \]

Now use FTC 2:

\[ \int_{0}^{3} 2x \, dx = F(3) - F(0) \]\[ = (9 + C) - (0 + C) = 9 \]

Notice \( C \) disappears.

Graph of y=2x from x=0 to 3. Shaded triangle area under the line equals 9.

Check: \( \frac{1}{2}(3)(6) = 9 \)

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FTC 2 Notation and Further Examples

Notation

FTC 2 is written as:

\[ \int_{a}^{b} f(x) \, dx = F(x) \Big|_a^b = F(b) - F(a) \]

Example

\[ \int_{-1}^{2} (x^2 - 2x + 7) \, dx \]

Here, \( f(x) = x^2 - 2x + 7 \) and \( F(x) = \frac{1}{3}x^3 - x^2 + 7x + C \).

\[ = \left[ \frac{1}{3}x^3 - x^2 + 7x \right]_{-1}^{2} \]
Don't write \( +C \); it goes away anyway.
\[ = \left[ \frac{1}{3}(2)^3 - (2)^2 + 7(2) \right] - \left[ \frac{1}{3}(-1)^3 - (-1)^2 + 7(-1) \right] \]

\( F(b) - F(a) \)

= 21