Proving the fundamental theorem

The area function A(x), why its derivative is f(x), and how this proves the fundamental theorem of calculus; differentiating an integral with a variable upper limit.

The area function

Fix a starting point $a$ and let the right-hand edge move. The area function $$A(x) = \int_a^x f(t)\,dt$$ is the area under $y = f(t)$ from $t = a$ to $t = x$. The letter $t$ is used inside so that it does not clash with the moving limit $x$.

The area grows at the rate f(x)

As the edge moves right, the area grows. It grows fastest where the curve is highest.

Try this

Move $X$ slowly. Where is the area growing fastest?

Why $A'(x) = f(x)$

The proof in three steps●●

Let $f$ be continuous.

$A(x + h) - A(x)$ is the area of a thin strip of width $h$ from $x$ to $x + h$. Its height is very close to $f(x)$, so $A(x + h) - A(x) \approx h\,f(x)$.

A thin strip

So $\dfrac{A(x + h) - A(x)}{h} \approx f(x)$, and the approximation becomes exact as $h \to 0$. Hence $A'(x) = f(x)$.

Take the limit

$A$ is therefore a primitive of $f$, so $A(x) = F(x) + C$. Since $A(a) = 0$, $C = -F(a)$. Then $\displaystyle\int_a^b f(t)\,dt = A(b) = F(b) - F(a)$.

Use the primitive

Differentiating an integral

Example 1●

Find $\dfrac{d}{dx}\displaystyle\int_1^x \sqrt{1 + t^4}\,dt$.

Replace $t$ with $x$: the answer is $\sqrt{1 + x^4}$. No primitive is needed.

Example 2●●

Let $A(x) = \displaystyle\int_2^x 3t^2\,dt$. Find $A(x)$ and check that $A'(x) = 3x^2$.

$A(x) = \Big[t^3\Big]_2^x = x^3 - 8$.

$A'(x) = 3x^2$, as expected, and $A(2) = 0$.

Flashcards

Area function

$A(x) = \displaystyle\int_a^x f(t)\,dt$

$A'(x)$

$f(x)$

$A(a)$

$0$, because the strip has zero width.

Practice

$\dfrac{d}{dx}\displaystyle\int_0^x t^2\,dt = $
$\dfrac{d}{dx}\displaystyle\int_1^x \sqrt{1 + t^4}\,dt = $
$A(x) = \displaystyle\int_2^x 3t^2\,dt$. Find $A(2)$.
For the same $A$, find $A(3)$.
$A(x) = \displaystyle\int_0^x (2t + 1)\,dt$. Find $A(2)$.
For the same $A$, find $A'(4)$.
In the proof, $A(x + h) - A(x)$ is approximately:
Where $f(x)$ is negative, the area function $A(x)$ is:
Which fact fixes the constant in $A(x) = F(x) + C$?