Section 5.3 The Fundamental Theorem of Calculus
References.
- Calculus, Early Transcendentals by Stewart, Section 5.3.
The Fundamental Theorem of Calculus relates derivatives to definite integrals, giving an easy way to evaluate many definite integrals using antiderivatives. One half is the formula
\begin{equation}
\int_a^b f(x) \, dx = F(b) - F(a),\tag{5.3.1}
\end{equation}
true when \(F\) is any antiderivative of \(f\) on the interval \([a,b]\text{:}\) \(F' = f\text{.}\)
This equation can for example be corroborated for some simple definite integrals whose values are clear from geometry:
For \(f(x)=c\text{,}\) any constant, all antiderivatives have the form \(cx+C\) (\(C\) another constant), so Equation (5.3.1) says
\begin{equation*}
\int_a^b c \, dx = F(b) - F(a) = (cb+C) - (ca+C) = c(b-a),
\end{equation*}
the expected area of the rectangle of width \(b-a\text{,}\) height \(c\) under this curve.
Example 5.3.2.
For \(f(x)=x\text{,}\) antiderivatives have the form \(x^2/2+C\) (\(C\) a constant),
\begin{equation*}
\int_a^b x \, dx
= F(b) - F(a)
= \left(\frac{b^2}{2}+C\right) - \left(\frac{a^2}{2}+C\right)
= \frac{b^2}{2} - \frac{a^2}{2}
= \frac{a+b}{2}(b-a)
\end{equation*}
which is the area of the trapezoid under this line \(y=x\text{:}\) the difference of the areas of two right triangular regions, or the width of the trapezoid times its average height. (Case \(a=0\) is the area \(1/2 \cdot b \cdot b\) of a right triangle of width \(b\text{,}\) height \(b\text{.}\))
Next, one that requires a far less obvious anti-derivative:
Checkpoint 5.3.3.
- Verify that \(f(x) = \sqrt{1 - x^2}\) has antiderivative \(\displaystyle F(x) = \frac{x \sqrt{1 - x^2} + \arcsin(x)}{2}\text{.}\)
- Sketch a graph of \(y = \sqrt{1 - x^2}\) on interval \([-1, 1]\text{.}\)
- Use this graph to explain why the value of \(2\int_{-1}^1 \sqrt{1 - x^2} dx\) should be \(\pi\text{.}\)
- Verify this by evaluating this integral, using FTC, Equation (5.3.1).
The moral here is that we would like to know how to find many more anti-derivatives, like the one seen above.
Getting Antiderivatives from Definite Integrals.
To understand why the above result is true, we do something a bit more ambitious: using the definite integral over intervals of variable width \([a, x]\) for \(x\) between \(a\) and \(b\text{,}\) so that the value of the definite integral depends on the choice of \(x\text{.}\) This gives a function of \(x\text{,}\) which turns out to be an antiderivative of \(f\text{.}\)
That is, we define a function \(g(x)\text{,}\) \(a \leq x \leq b\) by
\begin{equation*}
g(x) = \int_a^x f(t) \,dt
\end{equation*}
Let us compute the derivative of \(g\text{,}\) using the definition
\begin{equation*}
g'(x) = \lim_{h \to 0} \frac{g(x+h)-g(x)}{h}.
\end{equation*}
The numerator in the difference quotient is
\begin{align*}
\end{align*}
so
\begin{equation*}
\frac{g(x+h) - g(x)}{h} = \frac{1}{h} \int_x^{x+h} f(t) \,dt.
\end{equation*}
Intuitively, the value of \(f(t)\) over the small interval is close to \(f(x)\text{,}\) so the area given by the interval is close to that of a rectangle of height \(f(x)\text{,}\) width \(h\text{.}\) That is, the integral here is approximately \(f(x) h\text{,}\) so that the difference quotient is approximately \(f(x)\text{,}\) and then the limit gives
\begin{equation*}
g'(x)= \lim_{h \to 0} \frac{g(x+h)-g(x)}{h}
= \lim_{h \to 0} \frac{1}{h} \int_x^{x+h} f(t) \,dt = \lim_{h \to 0} \frac{1}{h} f(x) h = f(x).
\end{equation*}
This can be shown more carefully using the Extreme Value Theorem and the comparison properties of definite integrals.
So as claimed, \(\displaystyle g(x)= \int_a^x f(t) \, dt\) is an antiderivative of \(f\text{:}\)
\begin{equation*}
\frac{d}{dx} \int_a^x f(t) \,dt = f(x).
\end{equation*}
Getting Definite Integrals from Antiderivatives.
If \(F\) is any antiderivative of \(f\) on interval \([a,b]\text{,}\) it differs from the above antiderivative \(g\) only by an added constant, so \(\displaystyle F(x) = \int_a^x f(t) \,dt + C.\) Thus
\begin{align*}
\end{align*}
The last is true because the variable name used, \(t\) or \(x\text{,}\) has no effect on the value of a definite integral, which is a number, not a function of \(x\) or of \(t\text{.}\)
\begin{equation*}
\int_a^b f(x) \,dx = F(b)-F(a).
\end{equation*}
\begin{equation*}
\Big[ F(x) \Big]_a^b = F(b) - F(a), \quad = \begin{array}{r} F(b) \\ - F(a)\end{array}
\end{equation*}
I sometimes use the “vertical” form at right above to keep straight which term is added and which subtracted.
Integration and differentiation as inverse processes.
The two parts of the Fundamental Theorem of Calculus can be summarized by the idea that integration and differentiation are like inverses:- Computing the integral of a function \(f\) [to upper limit \(x\)] and then differentiating the result gets you back to where you started: function \(f\text{.}\)
- Differentiating a function \(F\) (getting \(f=F'\)) and then integrating over an interval \([a,x]\) the result gets you back to where you started: function \(F\) (up to adding a constant.)
Exercises Exercises
Study Calculus Volume 1, Section 5.3 2 ; in particular Theorems 4 and 5, Examples 17, 18, 20 and 21; Checkpoints 16, 17 and 19; and Exercises 149, 153, 155, 157, 161, 171, 177, 179, 183, 190, 191 and 195.
For further practice, look at several exercises from each of the following ranges: 148–159, 160–163, 170–189, 190–193, and 194–197.
openstax.org/books/calculus-volume-1/pages/5-3-the-fundamental-theorem-of-calculusopenstax.org/books/calculus-volume-1/pages/5-3-the-fundamental-theorem-of-calculus
