The Fundamental Theorem of Calculus Part 1

We are now going to look at one of the most important theorems in all of mathematics known as the Fundamental Theorem of Calculus (often abbreviated as the F.T.C). Traditionally, the F.T.C. is broken up into two part. We will look at the first part of the F.T.C., while the second part can be found on The Fundamental Theorem of Calculus Part 2 page.

Theorem 1 (The Fundamental Theorem of Calculus Part 1): If a function $f$ is continuous on the interval $[a, b]$, such that we have a function $g(x) = \int_a^x f(t) \: dt$ where $a ≤ x ≤ b$, and $g$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then $g'(x) = f(x)$.

• Proof: Suppose that $x \in (a, b)$, and $(x + h) \in (a, b)$. Then:

(1)

\begin{align} g(x + h) - g(x) = \int_a^{x + h} f(t) \: dt - \int_a^x f(t) \: dt \end{align}

• We can take the first integral and split it up such that:

(2)

\begin{align} \quad g(x + h) - g(x) = \left ( \int_a^x f(t) \: dt + \int_x^{x + h} f(t) \: dt \right ) - \int_a^x f(t) \: dt \\ \quad g(x + h) - g(x) = \int_x^{x + h} f(t) \: dt \end{align}

• Now let's divide both sides by $h > 0$ to obtain:

(3)

\begin{align} \frac{g(x + h) - g(x)}{h} = \frac{1}{h} \cdot \int_x^{x + h} f(t) \: dt \end{align}

• By The Extreme Value Theorem, there exists values $x = u$, and $x =v$ such that $f(u) = m$, and $f(v) = M$, where $m$ is the absolute minimum on the interval $[x, x+h]$, and $M$ is the absolute maximum on the interval $[x, x+h]$. Hence it follows that

(4)

\begin{align} f(u) \: h ≤ \int_x^{x + h} f(t) \: dt ≤ f(v) \: h \end{align}

• Essentially, $\int_x^{x + h} f(t) \: dt$ has an area less or equal to that area contained within the box formed by $Mh - mh$. Now let's divide each term in our inequality by $h > 0$ (there is no problem here since $h ≠ 0$). We obtain the following inequality:

(5)

\begin{align} f(u) ≤ \frac{1}{h} \int_x^{x + h} f(t) \: dt ≤ f(v) \end{align}

• Substituting back, we obtain:

(6)

\begin{align} f(u) ≤ \frac{g(x+h) - g(x)}{h} ≤ f(v) \end{align}

• As $h \to 0$, $u \to x$ (since $u$ is contained on the interval $[x, x + h]$) and $v \to x$ (since $v$ is also contained on the interval $[x, x + h]$). Hence it follows that $\lim_{h \to 0} \frac{g(x + h) - g(x)}{h} = g'(x) = f(x)$

• But also by the Squeeze Theorem for our inequality:

(7)

\begin{align} \lim_{h \to 0} f(x) ≤ \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} ≤ \lim_{h \to 0} f(x) \\ \lim_{u \to x} f(u) ≤ \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} ≤ \lim_{v \to x} f(v) \\ f(x) ≤ g'(x) ≤ f(x) \\ f(x) = g'(x) \end{align}

• Hence it follows that $\frac{d}{dx} \int_a^x f(t) \: dt = f(x)$. $\blacksquare$

Note: The process works analogously for h < 0.

Example 1

Differentiate the function $g(x) = \int_{0}^{x} \sqrt{3 + t} \: dt$.

We note that $f(t) = \sqrt{3 + t}$ is a continuous function, and by the fundamental theorem of calculus part 1, it follows that:

(8)

\begin{align} \frac{d}{dx} g(x) = \sqrt{3 + x} \end{align}

Example 2

Differentiate the function $g(x) = \int_{2}^{x} 4t^2 + 1 \: dt$.

Once again, $f(t) = 4t^2 + 1$ is a continuous function, and by the fundamental theorem of calculus part, it follows that:

(9)

\begin{align} \frac{d}{dx} g(x) = 4x^2 + 1 \end{align}

Example 3

Differentiate the function $g(x) = \int_{1}^{x^3} 3t + \sin t \: dt$.

We know that $3t + \sin t$ is a continuous function. We should note that we must apply the chain rule however, since our function is a composition of two parts, that is $m(x) = \int_{1}^{x} 3t + \sin t \: dt$ and $n(x) = x^3$, then $g(x) = (m \circ n)(x)$. Thus, applying the chain rule we obtain that:

(10)

\begin{align} \frac{d}{dx} g(x) = [3x^4 + \sin (x^4)] \cdot 4x^3 \end{align}

Example 4

Differentiate the function $g(x) = \int_{x}^{x^3} 2t^2 + 3 \: dt$.

We will first begin by splitting the integral as follows, and then flipping the first one as shown:

(11)

\begin{align} g(x) = \int_{x}^{0} 2t^2 + 3 \: dt + \int_{0}^{x^3} 2t^2 + 3 \: dt \\ \: g(x) = -\int_{0}^{x} 2t^2 + 3 \: dt + \int_{0}^{x^3} 2t^2 + 3 \: dt \end{align}

Since $2t^2 + 3$ is a continuous function, we can apply the fundamental theorem of calculus while being mindful that we have to apply the chain rule to the second integral, and thus:

(12)

\begin{align} \frac{d}{dx} g(x) = -(2x^2 + 3) + (2(x^3)^2 + 3) \cdot 3x^2 \end{align}