MATH 409 Lecture 20

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Lecture Slides

Exam next Thursday

Homework problems just require definition of improper Riemann definition; look at hint for problem 5.3.12

Integral with Variable Limit

(See Integral with a variable limit→)

Theorem. If $f$ is continuous at a point $x\in [a,b]$, then $F$ is differentiable at $x$ and $F'(x)=f(x)$

Proof. For any $x,y\in [a,b]$ such that $x<y$, we have

$\int _{a}^{y}f(t)\,\mathrm {d} t=\int _{a}^{x}f(t)\,\mathrm {d} t+\int _{x}^{y}f(t)\,\mathrm {d} t\,\!$

Then

$F(y)-F(x)-f(x)(y-x)=\int _{x}^{y}f(t)\,\mathrm {d} t-\int _{x}^{y}f(x)\,\mathrm {d} t$

So that

$\left|F(y)-F(x)-f(x)(y-x)\right|=\left|\int _{x}^{y}\left(f(t)-f(x)\right)\,\mathrm {d} t\right|\leq \int _{x}^{y}\left|f(t)-f(x)\right|\,\mathrm {d} t\leq \sup _{t\in [x,y]}\left|f(t)-f(x)\right|\,(y-x)$

Dividing both sides by $(y-x)$ gives

${\frac {F(y)-F(x)}{y-x}}-f(x)\leq \sup _{t\in [x,y]}\left|f(t)-f(x)\right|$

If the function $f$ is right continuous at $x$ (i.e. $f(y)\to f(x)$ as $y\to x^{+}$), then $\sup _{t\in [x,y]}\left|f(t)-f(x)\right|\to 0$ as $y\to x^{+}$. It follows that $f(x)$ is the right-hand derivative of $F$ at $x$. Likewise, one can prove that left continuity of f at x implies $f(x)$ is the left derivative at $x$.

quod erat demonstrandum

The Fundamental Theorem of Calculus

Theorem. [Part I]. If a function $f$ is continuous on an interval $[a,b]$, then the function

$F(x)=\int _{a}^{x}f(t)\,\mathrm {d} t\qquad x\in [a,b]$

is continuously differentiable on $[a,b]$. Moreover, $F'(x)=f(x)$ for all $x\in [a,b]$.

Proof. Since $f$ is continuous, it is also integrable on $[a,b]$. As proved earlier, integrability of $f$ implies that the function $F$ is well-defined on $[a,b]$. Moreover, $F'(x)=f(x)$ whenever $f$ is continuous at the point $x$. Therefore, the continuoity of $f$ on $[a,b]$ implies that $F'(x)=f(x)$ for all $x\in [a,b]$. In particular, $F$ is continuously differentiable on $[a,b]$.

quod erat demonstrandum

Theorem. [Part II]. If a function $F$ is differentiable on $[a,b]$ and the derivative $F'$ is integrable on $[a,b]$, then

$\int _{a}^{x}F'(t)\,\mathrm {d} t=F(x)-F(a)$ for all $x\in [a,b]$

Proof. The case $x=a$ is trivial: $0=0$. Assume $x\in (a,b]$. Since $F'$ is integrable on $[a,b]$, it is also integrable on any subinterval $[a,x]$, $x\in (a,b)$. Therefore, it is no loss to assume that $x=b$.

Consider an arbitrary partition $P=\left\{x_{0},x_{1},\ldots ,x_{n}\right\}$ of $[a,b]$. Let us choose samples $t_{j}\in [x_{j-1},x_{j}]$ for the Riemann sum ${\mathcal {S}}(F',P,t_{j})$ so that $F(x_{j})-F(x_{j-1})=F'(t_{j})\,(x_{j}-x_{j-1})$ (this is possible due to the Mean Value Theorem). Then

${\begin{aligned}{\mathcal {S}}(F',P,t_{j})&=\sum _{j=1}^{n}F'(t_{j})\,(x_{j}-x_{j-1})\\&=\sum _{j=1}^{n}\left(F(x_{j})-F(x_{j-1})\right)\\&=F(x_{n})-F(x_{0})\\&=F(b)-F(a)\end{aligned}}$

Since the sums ${\mathcal {S}}(F',P,t_{j})$ converge to $\int _{a}^{b}F'(t)\,\mathrm {d} t$ as $\left\|P\right\|\to 0$, the theorem follows: $\int _{a}^{b}F'(t)\,\mathrm {d} t=F(b)-F(a)$.

quod erat demonstrandum

The Indefinite Integral

Given a function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f : [a,b] \to \mathbb{R}} , a function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F : [a,b] \to \mathbb{R}} is called the indefinite integral (or antiderivative, primitive integral, or the primitive) of $f$ if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F'(x) = f(x)} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in [a,b]} . Notation:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int f(x) \, \mathrm{d}x}

If the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is continuous on $[a,b]$, then the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(x) = \int_a^x f(t) \, \mathrm{d}t} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in [a,b]} is an indefinite integral of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} due to the Fundamental Theorem of Calculus.

Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F} is an antiderivative of $f$. If $G$ is another antiderivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} , then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G' = F'} on $[a,b]$. Hence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (G-F)' = G' - F' = 0} on $[a,b]$. It follows that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G - F} is a constant function. Conversely, for any constant Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} , the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G(x) = F(x) + C} is also an antiderivative of $f$.

$\int f(x)\,\mathrm {d} x=F(x)+C$

Where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} is an arbitrary constant.

Examples

• Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^\alpha \,\mathrm{d}x = \frac{x^{\alpha+1}}{\alpha+1} + C} on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (0, \infty)} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha \ne 1} .
• $\int {\frac {1}{x}}\,\mathrm {d} x=\log {x}+C$ on $(0,\infty )$.
• Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \sin{x} = -\cos{x} + C}
• Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \cos{x} = \sin{x} + C}

Integration by Parts

Theorem. Suppose that functions Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and $g$ are differentiable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} with derivatives $f'$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g'} integrable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} . Then

$\int _{a}^{b}f(x)\,g'(x)\,\mathrm {d} x=f(b)\,g(b)-f(a)\,g(a)-\int _{a}^{b}f'(x)\,g(x)\,\mathrm {d} x$

Proof. By the product rule, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (f\,g)' = f' \, g + f \, g'} on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} . Since $f,f',g,g'$ are integrable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} by the hypothesis, so are the products $f'\,g$ and $f\,g'$. Then $(f\,g)'$ is integrable on $[a,b]$ as well. By the Fundamental Theorem of Calculus,

${\begin{aligned}f(b)\,g(b)-f(a)\,g(a)&=\int _{a}^{b}(f\,g)'(x)\,\mathrm {d} x\\&=\int _{a}^{b}f'(x)\,g(x)\,\mathrm {d} x+\int _{a}^{b}f(x)\,g'(x)\,\mathrm {d} x\end{aligned}}$

quod erat demonstrandum

Corollary. Suppose that functions $f,g$ are continuously differentiable on $[a,b]$. Then

$\int f(x)\,g'(x)\,\mathrm {d} x=f(x)\,g(x)-\int f'(x)\,g(x)\,\mathrm {d} x$ on $[a,b]$

To simplify notation, it is convenient to use the Leibniz differential Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{d}f} of a function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} defined by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{d}f(x) = f'(x) \,\mathrm{d}x = \frac{\mathrm{d}f}{\mathrm{d}x}\,\mathrm{d}x} . Another convenient notation is $\left.f(x)\right|_{x=a}^{b}$ or simply $\left.f(x)\right|_{a}^{b}$, which denotes the difference $f(b)-f(a)$.

Now the formula of integration by parts can be rewritten as

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{a}^{b} f(x) \,\mathrm{d}g(x) = \left. f(x) \, g(x) \right|_a^b - \int_{a}^{b} g(x) \,\mathrm{d}f(x)}

for definite integrals, and as

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int f \,\mathrm{d}g = f \, g - \int g \,\mathrm{d}f}

for indefinite integrals

Examples

• Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \log{x} \,\mathrm{d}x = x \, \log{x} - \int x \,\mathrm{d}(\log{x}) = x \, \log{x} - \int \mathrm{d}x = x \, \log{x} - x + C}
• Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^{\frac{\pi}{2}} x \, \sin{x} \,\mathrm{d}x = \left. -x \, \cos{x} \right|_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} (-\cos{x}) \,\mathrm{d}x = \int_0^{\frac{\pi}{2}} \cos{x} \,\mathrm{d}x = \left. \sin{x} \right|_0^{\frac{\pi}{2}} = 1} .

Change of the variable in an integral

(commonly called u-substitution as $\phi$ is often replaced by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} in practice.)

Theorem. if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} is continuously differentiable on a closed, nondegenerate interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} and $f$ is continuous on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi([a,b])} , then

$\int _{\phi (a)}^{\phi (b)}f(t)\,\mathrm {d} t=\int _{a}^{b}f(\phi (x))\,\phi '(x)\,\mathrm {d} x=\int _{a}^{b}f(\phi (x))\,\mathrm {d} \phi (x)$

Be aware that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t = \phi(x)} is a proper change of variable only if the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} is strictly monotone. However, the theorem holds even without this assumption.

Proof. Let us define two functions

${\begin{aligned}F(u)&=\int _{\phi (a)}^{u}f(t)\,\mathrm {d} t\qquad u\in \phi ([a,b])\\G(u)&=\int _{a}^{x}f(\phi (s))\,\phi '(s)\,\mathrm {d} s\qquad x\in [a,b]\end{aligned}}$

It follows from the Fundamental Theorem of Calculus that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F'(u) = f(u)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G'(x) = f(\phi(x)) \, \phi'(x)} . By the Chain Rule, $(F\circ \phi )'(x)=F'(\phi (x))\,\phi '(x)=f(\phi (x))\,\phi '(x)=G'(x)$.

Therefore, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( F(\phi(x)) - G(x) \right)' = 0} for all $x\in [a,b]$. It follows that the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(\phi(x)) - G(x)} is constant on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} . In particular, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(\phi(b)) - G(b) = F(\phi(a)) - G(a) = 0 - 0 = 0} . Hence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(\phi(b)) = G(b)} .

quod erat demonstrandum

Note: It is possible that $\phi (a)\geq \phi (b)$. To make sense of this case, we set Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{c}^{d} f(t) \,\mathrm{d}t {{=}} -\int_{d}^{c} f(t) \,\mathrm{d}t} if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c > d} . Also we set the integral to be 0 if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c = d} .