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4.2 Fundamental Theorem of Calculus

Theorem (Fundamental Theorem of Calculus)

Let \(f: [a, b] \rightarrow \mathbb{R}\) be an integrable function and define \(G: [a, b] \rightarrow \mathbb{R}\) by \(G(x) = \displaystyle \int_{a}^{x}f(t)dt \quad \forall a \leq x \leq b\)

If \(f\) is continuous at \(c \in [a, b]\), then \(G\) is differentiable at \(c\) and \(G'(c) = f(c)\).
If \(c = a\) or \(c = b\), then \(G'(c)\) is the left, respectively right, derivative of \(f\) at \(c\).

Proof

We know \(G\) is differentiable at \(c\) if \(\lim_{h\to 0}\frac{G(c+h)-G(c)}{h}=G'(c)\)

We want to prove that \(G'(c)=f(c)\), suppose first that \(h>0\), then \(a\leq c\leq c+h<b\)

Then \(G(c+h)={{{\displaystyle\int_{a}^{c+h}f\left(t\right)\mathrm{d}t=\int_{a}^{c}f\left(t\right)\mathrm{d}t+\int_{c}^{c+h}f\left(t\right)\mathrm{d}t}}}\)

And since \(G(c)=\displaystyle \int^{c}_{a} f(t)dt\), then \(G(c+h)-G(c)={\displaystyle\int_{c}^{c+h}f\left(t\right)\mathrm{d}t}\)

Thus \(\displaystyle\frac{G(c+h)-G(c)}{h}=\frac{\displaystyle\int_{c}^{c+h}f\left(t\right)\mathrm{d}t}{h}\)

Define \(m_{n}=\inf\{f(t):t\in[c,c+h]\}\), \(M_{n}=\sup\{f(t):t\in[c,c+h]\}\)

We know if \(f\) is integrable, then \(f\) is bounded. Thus \(m_n\leq f(t)\leq M_n,\forall t\in [c,c+h]\)

Thus \(\displaystyle\frac{m_n(c+h-c)}{h}\leq\frac{\displaystyle\int_{c}^{c+h}f\left(t\right)\mathrm{d}t}{h}\leq\frac{M_n(c+h-c)}{h}\), then \(m_{n}\leq\frac{G\left(c+h\right)-G\left(c\right)}{h}\leq M_{n}\)

Suppose \(h<0\), then \(a\leq c+h<c<b\), then \(G(c+h)=\displaystyle\int_{a}^{c+h}f\left(t\right)\mathrm{d}t=\int_{a}^{c+h}f\left (t\right)\mathrm{d}t+\int_{c+h}^{c}f\left(t\right)\mathrm{d}t-\int_{c+h}^{c}f\left (t\right)\mathrm{d}t=G(c)-\int_{c+h}^{c}f\left(t\right)\mathrm{d}t\)

Then \(\displaystyle\frac{G(c+h)-G(c)}{h}=\frac{{\displaystyle-\int_{c+h}^{c}f\left(t\right)\mathrm{d}t}}{h} =\frac{{\displaystyle\int_{c}^{c+h}f\left(t\right)\mathrm{d}t}}{h}\)

Again \(m_{n}^{\prime}\leq f(t)\leq M_{n}^{\prime},\forall t\in[c+h,c]\), then \({{\displaystyle\frac{m_{n}^{\prime}(c+h-c)}{h}\geq\frac{{{\displaystyle\int_{c}^{c+h}f\left(t\right)\mathrm{d}t}}}{h}\geq\frac{M_{n}^{\prime}(c+h-c)}{h}}}\) since \(h<0\)

Then \(m_{n}^{\prime}\leq\frac{G\left(c+h\right)-G\left(c\right)}{h}\leq M_{n}^{\prime}\)

Since \(f\) is continuous at \(c\), then \(\lim_{h\to 0}f(c+h)=f(c)\). This implies that \(M_n,m_n,M_n',m_n'\to f(c)\) when \(h\to 0\) since the definition of \(M_n\)...

Thus \(\lim_{h\to0}\frac{G\left(c+h\right)-G\left(c\right)}{h}=f\left(c\right)\)

Remark

We could defined \(H(x)=\displaystyle \int^{b}_{x}f(t)dt\), then \(\displaystyle\int^{b}_{a}f(t)dt=\int^{x}_{a}f(t)dt+\int^{b}_{x}f(t)dt=G(x)+H(x)\)

Then \(0=G'(c)+H'(c)\), then \(-G'(c)=H'(c)\) and \(H'(c)=-f(c)\)

Example

  1. \(f: \mathbb{R} \to \mathbb{R}\), \(f(x) = \begin{cases} 1 & x \geq 0 \\ 0 & x < 0 \end{cases}\)

    We know that \(f\) is integrable. Take the interval \([-1, 1]\) and the function \(\displaystyle G(x) = \int_{-1}^{x}f(t) \, dt\)

    If \(-1 \leq x < 0 \implies f(t) = 0 \quad \forall \, -1 \leq t \leq x\)​\(\implies \displaystyle G(x) = \int_{-1}^{x}f(t) \, dt = \int_{-1}^{x}0 \, dt = 0\)

    If \(0 \leq x \leq 1 \implies f(t) = 1\) if \(0 \leq t \leq x\)

    \(\displaystyle \implies G(x) = \int_{-1}^{x}f(t) \, dt = \int_{-1}^{0}f(t) \, dt + \int_{0}^{x} f(t) \, dt = 0 + \int_{0}^{x}1 \, dt = 1(x - 0) = x\)

    \(\implies G(x) = \begin{cases} 0 \quad \text{if} \quad -1 \leq x < 0 \\ x \quad \text{if} \quad 0 \leq x \leq 1 \end{cases}\)

    G is NOT differentiable at \(c=0\)

  2. \(f(x) = |x|\) in \([-1,1]\)image

    For \(-1 \leq x < 0\), \(f(t) = -t \quad \forall \ -1 \leq t \leq x < 0\)

    \(\displaystyle G(x) = \int_{-1}^{x}f(t) \, dt = \int_{-1}^{x}-t \, dt = - \int_{-1}^{x}t \, dt = - \left( \frac{x^{2}}{2}- \frac{(-1)^{2}}{2}\right) = \frac{1}{2}- \frac{x^{2}}{2}\)

    For \(0 \leq x \leq 1\), \(f(t) = t \quad \forall \ 0 \leq t \leq x \leq 1\)

    \(\displaystyle G(x) = \int_{-1}^{x}f(t) \, dt = \int_{-1}^{0}f(t) \, dt + \int_{0}^{x}f(t) \, d t = \int_{-1}^{0}-t \, dt + \int_{0}^{x}t \, dt = \frac{1}{2}+ \frac{x^{2}}{2}- \frac{(0)^{2}}{2} = \frac{1}{2}+ \frac{x^{2}}{2}\)

    image\(G(x) = \begin{cases} \frac{1}{2} - \frac{x^2}{2} &, -1 \leq x < 0 \\ \frac{1}{2} + \frac{x^2}{2} &, 0 \leq x \leq 1 \end{cases}\)

    Thus we get a continuous and differentiable function

Remark

If \(f\) is continuous on all \([a, b]\), then \(G\) is differentiable on all \([a, b]\) and \(G' = f\) on \([a, b]\). That is, \(G\) is a primitive of \(f\) on \([a, b]\).

Example

If \(G_1, G_2: [a,b] \to \mathbb{R}\) are primitives of \(f: [a,b] \to \mathbb{R}\)

\(\implies G_1'(x) = f(x) = G_2'(x) \implies G_1'(x) - G_2'(x) = 0 \quad \forall x \in [a,b]\)

\(\implies (G_1(x) - G_2(x))' = 0\)

\(\implies \exists c \in \mathbb{R}\) s.t. \(G_1(x) - G_2(x) = c \quad \forall x\)

Theorem

Let \(f: [a, b] \rightarrow \mathbb{R}\) be a continuous function and let \(F: [a, b] \rightarrow \mathbb{R}\) be a primitive, that is \(F'(x) = f(x) \quad \forall x \in [a, b]\)

Then \(\displaystyle\int_{a}^{b}f(x) dx = F(b) - F(a)\)

Proof

Let \(\displaystyle G(x) = \int_{a}^{x}f(t) dt \quad \forall x \in [a,b]\). By FTC, \(G'(x) = f(x) \quad \forall x \in [a,b]\)

By hypothesis, \(F'(x) = f(x) \quad \forall x \in [a,b]\)

\(\implies G(x)=F(x)+c,c\in\mathbb{R}\)(*)

So \({\displaystyle G(b)=\int_{a}^{b}f(t)dt=F(b)+c}\), then \({{\displaystyle G(x)=\int_{a}^{x}f(t)dt=F(x)+c}}\) (# #)

From (*) we may get \(G(a)=F(a)+c\). From (##) we get \(\displaystyle G(a)=\int_{a}^{a}f(t)dt=0\)

Thus \(0 = F(a) + c\)​\(\implies c = -F(a)\)

Remarks

  1. The theorem above tells us that we may compute the integral of a continuous function by evaluating in a primitive function (in case we know a primitive!!).

  2. The value of the integral does not depend on the primitive we choose:

    If \(F_{1}, F_{2}: [a, b] \rightarrow \mathbb{R}\) are two primitives of a continuous function \(f: [a, b] \rightarrow \mathbb{R}\), then there exists \(c \in \mathbb{R}\) such that \(F_{2}(x) = F_{1}(x) + c \quad \forall x \in [a, b]\)
    ​\(\displaystyle\int_{a}^{b}f(x) dx = F_{1}(b) - F_{1}(a)\) \(= F_{1}(b) + c - (F_{1}(a) + c)\) \(= F_{2}(b) + c - (F_{2}(a))\) \(= F_{2}(b) - F_{2}(a)\)

  3. \(f: \mathbb{R} \rightarrow \mathbb{R}, \quad f(x) = x^n, n \in \mathbb{N}\)

    \(\implies F(x) = \frac{x^{n+1}}{n+1}\) is a primitive

    Thus \(\displaystyle\int_{a}^{b}x^{n}dx = F(b) - F(a) = \frac{b^{n+1}}{n+1}- \frac{a^{n+1}}{n+1} = \left . \frac{x^{n+1}}{n+1}\right|_{a}^{b}\)

    For example \(\displaystyle\int_{0}^{2}x^{3} dx = \frac{2^{4}}{4}= \frac{16}{4}= 4\)

  4. Moreover, we can compute integrals of polynomials!
    ​\(p(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0\)

    \(\displaystyle\implies \int_{a}^{b}p(x) dx = \int_{a}^{b}(a_{n} x^{n} + a_{n-1}x^{n-1}+ \dots + a_{1} x + a_{0}) dx\)

    \(\displaystyle= \int_{a}^{b}a_{n} x^{n} dx + \int_{a}^{b}a_{n-1}x^{n-1}dx + \dots + \int_{a}^{b} a_{1} x dx + \int_{a}^{b}a_{0} dx\)

    \(\displaystyle= a_{n} \int_{a}^{b}x^{n} dx + a_{n-1}\int_{a}^{b}x^{n-1}dx + \dots + a_{1} \int _{a}^{b}x dx + a_{0} \int_{a}^{b}1 dx\)

    \(\displaystyle= \left(a_{n} \frac{x^{n+1}}{n+1}+ a_{n-1}\frac{x^{n}}{n}+ \dots + a_{1} \frac{x^{2}}{2} + a_{0} x \right) \Big|_{a}^{b}\)

    \(\displaystyle\int_{0}^{1}(x^{2} - 2x + 1) dx = \left(\frac{x^{3}}{3}- 2\frac{x^{2}}{2}+ x \right ) \Big|_{0}^{1}= \left( \frac{1}{3}- 2\frac{1}{2}+ 1 \right) - \left( \frac{0^{3}}{3} - 2 \frac{0}{2}+ 0 \right) = \frac{1}{3}\)

  5. \(f: \mathbb{R}\rightarrow \mathbb{R}, f(x) = \sin(x)\)

    \(F(x) = -\cos(x)\)

    \(\displaystyle\int_{0}^{\pi}\sin(x) dx = -\cos(x) \Big|_{0}^{\pi}= -\cos(\pi) - (-\cos(0)) = - (- 1 ) - (-1) = 2\)

  6. \(f(x) = \exp(x) = \sum_{n=0}^{\infty}\frac{x^{n}}{n!}\) and \(F(x)=\exp(x)=e^{x}\)

    \(\displaystyle\int_{0}^{x} e^{t} dt = e^{t}\Big|_{0}^{x} = e^{x} - e^{0} = e^{x} - 1\) and \(\displaystyle\int_{0}^{1} e^{t} dt = e - 1\)

  7. \(\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n , \quad x \in (-1, 1)\)

    In \([0, x]\), \(0 \leq x < 1\), the convergence is uniform

    \(\displaystyle\int_{0}^{x}\frac{1}{1-t}dt = \int_{0}^{x}(\sum_{n=0}^{\infty}t^{n} ) dt = \sum_{n=0}^{\infty}\int_{0}^{x}t^{n}dt = \sum_{n=0}^{\infty}\frac{t^{n+1}}{n+1} \Big |_{0}^{x}\)

    \(= \sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}= \sum_{k=n+1}^{\infty}\frac{x^{k}}{k}= -\ln(1-x) = -\log(1-x)\)​

  8. \(\frac{1}{1+x} = \sum_{n=0}^{\infty} (-x)^n = \sum_{n=0}^{\infty} (-1)^n x^n\)

    \(\displaystyle\int_{0}^{x} \frac{1}{1+t}dt = \sum_{n=0}^{\infty}(-1)^{n} \int_{0}^{x} t^{n} dt\)

    \(= \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}\) \(= \sum_{k=1}^{\infty}(-1)^{k-1}\frac{x^{k}}{k}\) \(= -\sum_{k=1}^{\infty} (-1)^k \frac{x^k}{k}\) \(= -\sum_{k=1}^{\infty} \frac{(-x)^k}{k}\)

    \(= -(- \ln(1+x))\) \(= \ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{n}}{n}\)

  9. Only prop: \(\ln(1+x) '= \frac{1}{1+x}\)

    Exercise: \(e^{\ln(1+x)} = 1+x\)

    Hint: Compute \(\left( \frac{e^{\ln(1+x)}}{1+x} \right)'\)

Theorem

Let \(f: [a, b] \rightarrow \mathbb{R}\) be a continuous function. Then \(f\) has a primitive on \([a, b]\).

Proof

By fundamental theorem of calculus, we know \(F(x)=\displaystyle \int^x_af(t)dt\), then \(F\) is differentiable in \([a,b]\) and \(F'(x)=f(x),\forall x\in [a,b]\)

So, \(f\) has a primitive

The following version of the FTC is for integrable functions. Here we know the result if the function has a primitive (which is always the case if the function is integrable.

Theorem (Fundamental Theorem of Calculus for integrable functions)

Let \(f: [a, b] \rightarrow \mathbb{R}\) be an integrable function on \([a, b]\). If there exists a differentiable function such that \(g: [a, b] \rightarrow \mathbb{R}\) such that \(g'(x) = f(x)\) for all \(x\) in \([a, b]\), then \(\displaystyle\int_{a}^{b}f(x) dx = g(b) - g(a)\)

Note that we don't have \(f\) is continuous, thus we cannot apply FTC directly

Proof

Idea: Consider upper and lower sum and use mean value theorem

We give a direct proof using partitions and the mean value theorem.

Let \(P = [x_0, x_1, ..., x_n]\) be a partition of \([a, b]\). Consider an interval of the partition \([x_{i-1}, x_i]\).

Then, since \(g\) is differentiable on \([a, b]\), by the Mean Value Theorem \(\exists c_i \in [x_{i-1}, x_i]\) such that

\(g(x_i) - g(x_{i-1}) = g'(c_i)(x_i - x_{i-1})\) \(= f(c_i)(x_i - x_{i-1})\)

For this partition \(P\), we define as usual \(m_i = \inf \{ f(x) \mid x \in [x_{i-1}, x_i]\}\), \(M_i = \sup \{ f(x) \mid x \in [x_{i-1}, x_i]\}\)

Then, we have that \(m_i(x_i - x_{i-1}) \leq f(c_i)(x_i - x_{i-1}) \leq M_i(x_i - x_{i-1})\)

\(\Rightarrow m_i(x_i - x_{i-1}) \leq g'(c_i)(x_i - x_{i-1}) \leq M_i(x_i - x_{i-1})\)

\(\Rightarrow m_i(x_i - x_{i-1}) \leq g(x_i) - g(x_{i-1}) \leq M_i(x_i - x_{i-1})\)

Thus, as this holds for any interval \([x_{i-1}, x_i]\), we have

\(L(f, P) = \sum_{i=1}^{n} m_i(x_i - x_{i-1}) \leq \sum_{i=1}^{n} (g(x_i) - g(x_{i-1}))\) \(= g(x_n) - g(x_0)\) \(= g(b) - g(a)\)

\(U(f, P) = \sum_{i=1}^{n} M_i(x_i - x_{i-1}) \geq \sum_{i=1}^{n} (g(x_i) - g(x_{i-1}))\) \(= g(x_n) - g(x_0)\) \(= g(b) - g(a)\)

Hence, \(L(f,P) \leq g(b) - g(a) \leq U(f,P)\) and this holds for any partition \(P\).

Thus \(L(f) \leq g(b) - g(a) \leq U(f)\).

Since \(L(f) = U(f)\), it follows that \(\displaystyle L(f) = U(f) = \int_{a}^{b}f(x) dx = g(b) - g(a)\)