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Math 220-2: Kursobjekt

Aaron Greicius

Section 1.5 Fundamental theorem of calculus

We have now met the two main operations in calculus:
\begin{align*} \text{Derivative: } f \amp \mapsto f'\\ \text{Integral: } f \amp \mapsto \int_a^b f(x)\, dx \text{.} \end{align*}
The fundamental theorem of calculus (FTC) how these two operations (derivative and integral) interact. The theorem has two parts (FTC I and FTC II). We will focus on FTC II in this section, and FTC I in the next.

Proof of FTC II.

Since \(f\) is continuous on \([a,b]\) it is integrable, and thus we can compute its integral as a limit of Riemann sums corresponding to any sequence of pointed partitions \(P_n\) satisfying \(\lVert P_n\rVert\to\infty\text{.}\) The strategy of the proof that follows is to pick these pointed partitions in a very particular way that ensures the value of of the integral we seek.
Let \(F\) be an antiderivative of \(f\) on \(I=[a,b]\text{.}\) Since \(f\) is continuous on \(I\text{,}\) we see that \(F\) is differentiable on \(I\) and its derivative \(F'=f\) is continuous. The mean value theorem thus applies to \(F\text{:}\) given any interval \([c,d]\) on which \(F'\) is defined and continuous, there is an element \(x^*\in[c,d]\) such that
\begin{equation*} F'(x^*)=\frac{F(d)-F(c)}{d-c}\text{.} \end{equation*}
We use this fact to produce a sequence of pointed partitions \(P_n\) of a very particular sort. Namely for each \(n\text{,}\) we divide \([a,b]\) into \(n\) equal subintervals of length \(\Delta x=(b-a)/n\text{,}\) and for each subinterval \(I_k=[x_{k_1},x_k]\) we choose our sample input \(c_k\) to be a point where we have
\begin{equation*} F'(c_k)=\frac{F(x_k)-F(x_{k-1})}{x_k-x_{k-1}}=\frac{F(x_k)-F(x_{k-1})}{\Delta x}\text{.} \end{equation*}
As discussed above, such a \(c_k\) is guaranteed to exist by the mean value theorem. Since \(F'=f\) by assumption, these sample points satisfy
\begin{equation} f(c_k)=\frac{F(x_k)-F(x_{k-1})}{\Delta x}\text{.}\tag{1.5.4} \end{equation}
Note that we have \(\lVert P_n\rVert=\Delta x=(b-a)/n\text{,}\) and thus \(\lVert P_n\rVert\to\infty\text{.}\) It follows that if \(S_n\) is the Riemann sum corresponding to the pointed partition \(P_n\text{,}\) then
\begin{equation*} \int_a^b f(x)\, dx=\lim_{n\to\infty}S_n\text{.} \end{equation*}
We now compute \(S_n\) for any fixed \(n\text{:}\)
\begin{align*} S_n \amp =\sum_{k=1}^nf(c_k)\Delta x\\ \amp = \sum_{k=1}\left( \frac{F(x_k)-F(x_{k-1})}{\Delta x}\right)\Delta x \amp \knowl{./knowl/xref/eq_FTC_II_proof.html}{\text{(1.5.4)}}\\ \amp =\sum_{k=1}^n F(x_k)-F(x_{k-1}) \\ \amp =(\cancel{F(x_1)}-F(x_0))+(\cancel{F(x_2)}-\cancel{F(x_1)}+\cdots (F(x_n)-\cancel{F(x_{n-1}})\\ \amp =F(x_n)-F(x_0)\\ \amp =F(b)-F(a)\text{.} \end{align*}
We’ve shown that \(S_n=F(b)-F(a)\) for all \(n\text{.}\) It follows that
\begin{equation*} \int_a^b f(x)\, dx=\lim_{n\to\infty}S_n=\lim_{n\to\infty}F(b)-F(a)=F(b)-F(a)\text{,} \end{equation*}
as desired.
As mentioned above, we will focus here on FTC II. It main computational import of this theorem is that it allows us to compute definite integrals indirectly, by first finding an antiderivative of the integrand, and then using (1.5.3).
FTC II allows us to reduce a definite integral computation essentially to finding an antiderivative of the integrand: i.e., we integrate basically by “computing derivatives in reverse”. The Augustus De Morgan quote below captures this quite poetically.
Henceforth our integral computations will involve many expressions of the form \(F(b)-F(a)\text{,}\) which we call a difference evaluation. It will be useful to have some compact notation for this.

Definition 1.5.4. Difference evaluation notation.

Let \(g\) be a function containing \(a\) and \(b\) in its domain. We define \(g]_a^b\) to be the difference evaluation expression
\begin{equation*} g(x)\Bigr\vert_a^b=g(b)-g(a)\text{.} \end{equation*}

Remark 1.5.5. Difference evaluation notation.

  1. It is easy to see that the following difference evaluation identities hold:
    \begin{align*} (f(x)+g(x))\Bigr\vert_a^b \amp = f(x)\Bigr\vert_a^b+g(x)\Bigr\vert_a^b \\ cf(x)\Bigr\vert_a^b \amp = c\left(f(x)\Bigr\vert_a^b\right) \end{align*}
  2. Observe that (1.5.3) can be expressed using our new notation as
    \begin{equation*} \int_a^b f(x)\, dx= F(x)\Bigr\vert_a^b\text{.} \end{equation*}

Example 1.5.6. Computing integrals with antiderivatives.

Use the FTC II to compute the following definite integrals.
  1. \(\displaystyle \displaystyle\int_a^b 1-x^3\, dx\)
  2. \(\displaystyle \displaystyle\int_0^{10} \frac{1}{\sqrt{2t+1}}\, dt\)
  3. \(\displaystyle\int_{3\pi/4}^{\pi}\sec^2 u\, du\text{.}\)
Solution.
Following Procedure 1.5.2, for each integral we provide an antiderivative \(F\) of the integrand \(f\) and then compute the integral as a difference evaluation using FTC II. The integrands in question are familiar and/or simple enough that we do not justify the fact that \(F\) is indeed an antiderivative.
  1. The function \(F(x)=x-\frac{1}{4}x^4\) is an antiderivative of \(f(x)=1-x^3\text{.}\) It follows from FTC II that
    \begin{align*} \int_a^b 1-x^3\, dx\amp = F(b)-F(a)\\ \amp = b-\frac{1}{4}b^4-(a-\frac{1}{4}a^4\\ \amp =(b-a)-\frac{1}{4}(b^4-a^4)\text{.} \end{align*}
  2. The function \(F(x)=\sqrt{2t+1}\) is an antiderivative of \(f(t)=1/\sqrt{2t+1}\text{.}\) It follows from FTC II that
    \begin{align*} \int_0^10\frac{1}{\sqrt{2t+1}}\, dt \amp = F(10)-F(0)\\ \amp = \sqrt{21}-\sqrt{1} \\ \amp =\sqrt{21}-1\text{.} \end{align*}
  3. The function \(F(x)=\tan u\) is an antiderivative of \(f(u)=\sec^2 u\text{.}\) It follows from FTC II that
    \begin{align*} \int_{3\pi/4}^\pi \sec^2 u\, du \amp = F(\pi)-F(3\pi/4)\\ \amp =\tan \pi -\tan(3\pi/4)\\ \amp =0-(-1)\\ \amp = 1\text{.} \end{align*}

Example 1.5.7. Signed area using FTC II.

Let \(f(x)=\sin(x/3)\text{,}\) and let \(\mathcal{C}\) be the graph of \(f\text{.}\) For each region \(\mathcal{R}\) compute the area of \(\mathcal{R}\) and the signed area of \(\mathcal{R}\text{.}\) Include a diagram of \(\mathcal{C}\) and \(\mathcal{R}\text{.}\) Make sure your answer is consistent with your graph. If your answer happens to be 0, use the diagram to explain why.
  1. \(\mathcal{R}\) is the region between \(\mathcal{C}\) and the \(x\)-axis, from \(x=-\pi\) to \(x=\pi\text{.}\)
  2. \(\mathcal{R}\) is the region between \(\mathcal{C}\) and the \(x\)-axis, from \(x=-\pi\) to \(x=\pi/2\text{.}\)
  3. \(\mathcal{R}\) is the region between \(\mathcal{C}\) and the \(x\)-axis, from \(x=0\) to \(x=6\pi\text{.}\)
Solution.
In each of the cases below we use our definitions of the area and signed area of the region \(\mathcal{R}\) associated to \(f\text{:}\)
\begin{align*} \operatorname{sgn\, area}\mathcal{R} \amp = \int_a^b f(x)\, dx\\ \operatorname{area}\mathcal{R} \amp = \int_a^b\abs{f(x)}\, dx\text{.} \end{align*}
Throughout we will use the fact that \(F(x)=-3\cos(x/3)\) is an antiderivative of \(f(x)=\sin(x/3)\text{.}\)
  1. We have
    \begin{align*} \operatorname{sgn\, area}\mathcal{R} \amp = \int_{-\pi}^{\pi}f(x)\, dx\\ \amp =F(\pi)-F(-\pi)\\ \amp =-3(\cos(\pi/3)-\cos(-\pi/3))=0\\ \operatorname{area}\mathcal{R} \amp = \int_a^b\abs{f(x)}\, dx\\ \amp = \int_{-\pi}^0\abs{f}\, dx+\int_0^{\pi}\abs{f} \amp (\knowl{./knowl/xref/th_interval_partition.html}{\text{Interval partition}})\\ \amp = \int_{-\pi}^0-f(x)\, dx+\int_0^\pi f(x)\, dx \amp (\text{def. of } \abs{f})\\ \amp = -\int_{-\pi}^0f(x)\, dx+\int_0^\pi f(x)\, dx\\ \amp =-(F(0)-F(-\pi))+(F(\pi)-F(0))\\ \amp = -(-3\cos 0-(-3)\cos(-\pi/3))+(-3\cos(\pi/3)-(-3)\cos 0)\\ \amp = 6-3=3\text{.} \end{align*}
    Below you find a diagram of \(\mathcal{R}\) described as a union \(\mathcal{R}=\mathcal{R}_1\cup \mathcal{R}_2\text{.}\) By the symmetry in the diagram it should come as no surprise that
    \begin{equation*} \operatorname{sgn\, area}\mathcal{R}=-\operatorname{area}\mathcal{R}_1+\mathcal{R}_2=0\text{.} \end{equation*}
    Part (a) region
  2. The reasoning and computations here are very similar. We get
    \begin{align*} \operatorname{sgn\, area}\mathcal{R} \amp =\int_{-\pi}^{\pi/2}f(x)\, dx \\ \amp = \frac{3}{2}-\frac{3\sqrt{3}}{2}\\ \operatorname{area}\mathcal{R} \amp \int_{-\pi}^{\pi/2}\abs{f}\, dx \\ \amp = \frac{9}{2}-\frac{3\sqrt{3}}{2}\text{.} \end{align*}
  3. Again, the reasoning and computations are similar. We get
    \begin{align*} \operatorname{sgn\, area}\mathcal{R} \amp =\int_{0}^{6\pi}f(x)\, dx \\ \amp =0 \\ \operatorname{area}\mathcal{R} \amp \int_{0}^{6\pi}\abs{f}\, dx \\ \amp = \int_0^{3\pi}f(x)\, dx-\int_{3\pi}^{6\pi}f(x)\, dx\\ \amp = 6+6=12\text{.} \end{align*}
    Both the fact that signed area here is zero and the symmetry appearing in the two terms of the area computation arises from the fact that \(\mathcal{R}=\mathcal{R}_1\cup \mathcal{R}_2\) is a union of two symmetric regions.
    Signed area over entire period

Example 1.5.9. Leaking water tank revisited.

Recall the set up from Example 1.1.9, where we determined that at time \(t\text{,}\) the amount of water \(A=f(t)\) has rate of change \(f'(t)=-1/\sqrt{2t+1}\text{.}\)
According to Theorem 1.5.8, for any \(t=b\) the integral
\begin{equation*} \int_0^b f'(t)\, dt=\int_0^b-1/\sqrt{2t+1}\, dt \end{equation*}
computes the net change in the amount of water in the tank between time \(t=0\) and \(t=b\) minutes: i.e., it computes \(\Delta A=f(b)-f(0)\text{.}\) We can use FTC II to compute this integral. The function \(F(x)=-\sqrt{2t+1}\) is an antiderivative of \(-1/\sqrt{2t+1}\) and thus
\begin{equation*} \int_0^b -1/\sqrt{2t+1}\, dt=F(b)-F(0)=1-\sqrt{2b+1}\text{.} \end{equation*}
This tells us that \(\Delta A=1-\sqrt{2b+1}\) at any time \(b\text{.}\) Since \(\Delta A=f(b)-f(0)\text{,}\) we conclude that \(f(b)-f(0)=1-\sqrt{2b+1}\text{,}\) or equivalently, \(f(b)=f(0)+1-\sqrt{2b+1}=101-\sqrt{2t+1}\) for all \(b\text{.}\) (Here we use the fact that \(f(0)=100\text{.}\)) We have just given another derivation of the formula
\begin{equation*} f(t)=101-\sqrt{2t+1} \end{equation*}
for the function \(A=f(t)\text{.}\)
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