My Integral Notes

by SomeonecoolLovesMaths, Nov 19, 2024, 1:15 PM

Here is a collection of all the notes I have, without the proves. There are a set of theorems basically to take a look upon if you are forgetting one, definitely not a handout of any sort.

$$\Huge{\text{Integrals}}$$$a = x_0 < x_1 < x_2 < x_3 < \cdots < x_{n-1} < x_n = b$ describes a partition of $[a,b]$, call it $P$.
$$m_i = \text{inf} \; f \; [x_{i-1},x_i] $$$$L(P,f) = \sum_{i=1}^{n} m_i(x_i - x_{i-1})\; \; \text{Worst possible underestimation}$$$$M_i = \text{sup} \; f \; [x_{i-1},x_i]$$$$U(P,f) = \sum_{i=1}^{n} M_i(x_i - x_{i-1})\; \; \text{Worst possible overestimation}$$$$L(P,f) \leq \text{desired area} \leq U(P,f)$$As we add new points to $P$ & make it refined, our approximations get better, i.e., $L(P,f)$ increases & $U(P,f)$ decreases.

$\text{\underline{Our version of integrability:}}$
We say that a bounded $f: [a,b] \longrightarrow \mathbb{R}$ is integrable if $$\text{sup} \; L(P,f) = \text{inf} \; U(P,f) = \text{desired area}$$
$\text{\underline{Another version of integrability:}}$
We say that a bounded $f : [a,b] \longrightarrow \mathbb{R}$ is integrable if $\forall \; \epsilon$, $\exists$ $\delta > 0 $ such that $\forall$ partitions finer than $\delta$, $$U(P,f) - L(P,f) < \epsilon$$
$\text{\underline{Riemann's version of integrability:}}$
Suppose $\exists$ some $Q \in \mathbb{R}$ such that $ \forall \; \epsilon > 0$, $\exists$ $\delta > 0$ such that for all partitions $P$ finer than $\delta$ satisfy $$\left| \sum_{i=1}^{n} f(t_i) (x_i - x_{i-1}) - Q \right| < \epsilon$$for any choice of $t_i$'s. Then $f$ is integrable & $\int_{a}^{b} f(x) \text{d} x = Q$.

$\text{\underline{Some Results:}}$
$$\bullet \; f \; \text{is monotone on } [a,b] \Longrightarrow f \; \text{is integrable}$$$$\bullet \; f \; \text{is continuous on } [a,b] \Longrightarrow f \; \text{is integrable}$$$$\bullet \; f \; \text{is discontinuous at finitely many points then }f \; \text{is integrable}$$
$\text{\underline{Algebra on Integrals:}}$
$$\bullet \; \int_{a}^{b} cf(x) \text{d} x = c\int_{a}^{b} f(x) \text{d} x \; \forall \; c \in \mathbb{R}$$$$\bullet \; \int_{a}^{b} (f(x)+g(x)) \text{d} x = \int_{a}^{b} f(x) \text{d} x + \int_{a}^{b} g(x) \text{d} x $$$$\left[ \text{Provided } \int_{a}^{b} f(x) \text{d} x \text{ and } \int_{a}^{b} g(x) \text{d} x \text{ make sense.} \right]$$
$\text{\underline{Theorem:}}$
$f$ is integrable and $g$ is continuous on the range of $f$ then $g \circ f$ is also integrable.

$\text{\underline{Some Results:}}$
$$\circledast \; f(x) \geq 0 \text{ on } [a,b] \Longrightarrow \int_{a}^{b} f(x) \text{d} x \geq 0$$$$ \circledast \begin{rcases*} \int_{a}^{b} f(x) \text{d} x = 0 \\ f(x) \geq 0 \end{rcases*} \not \Longrightarrow f(x) = 0 \text{ on } [a,b] $$$$ \circledast \; f \text{ continuous }, \; f \geq 0, \int_{a}^{b} f(x) \text{d} x = 0 \iff f \equiv 0 \text{ on } [a,b]$$$$ \circledast \; \int_{a}^{b} f(x) \text{d} x = \int_{a}^{c} f(x) \text{d} x \int_{c}^{b} f(x) \text{d} x$$$$ \circledast \; \int_{a}^{b} f(x) \text{d} x = - \int_{b}^{a} f(x) \text{d} x$$
$\text{\underline{Theorem:}}$
Suppose $f,g$ are integrable, $f$ continuous and $g$ does not change sign on $[a,b]$. Then,
$$\int_{a}^{b} f(x)g(x) \text{d} x = f(c)\int_{a}^{b} f(x) \text{d} x$$for some $c \in (a,b)$.

$\text{\underline{Theorem:}}$
$f,g$ are integrable on $[a,b]$. Then,
$$\left(\int_{a}^{b} f^2(x) \text{d} x \right)\left(\int_{a}^{b} g^2(x) \text{d} x \right) \geq \left(\int_{a}^{b} f(x)g(x) \text{d} x \right)^2$$
$\text{\underline{Fundamental Theorem of Calculus:}}$
Let $\int_{a}^{b} f(x) \text{d} x = A(x)$.
$$f(x) = \lim_{h \longrightarrow 0} \frac{A(x+h) - A(x)}{h} = A'(x)$$$$\int_{a}^{b} f'(x) \text{d} x = f(b) - f(a)$$
$\text{\underline{Theorem:}}$
Let $f$ be integrable on $[a,b]$. Then $F(x) = \int_{a}^{b} f(t) \text{d} t$ is continuous on $[a,b]$.
Further if $f$ is continuous at $c \in [a,b]$, then $F$ be differentiable at $c$ with $F'(c) = f(c)$.

$\text{\underline{Integration by parts:}}$
$$\int_{a}^{b} (f(x)g(x)' \text{d} x = f(b)g(b) - f(a)g(a)$$$$ = \int_{a}^{b} f(x)'g(x) \text{d} x + \int_{a}^{b} f(x)g'(x) \text{d} x$$Let, $g(x) = \int_{0}^{x} h(t) \text{d} t$ ($h$ cont.). ($g(x)$ is some antiderivative of some function) $$\therefore \int_{a}^{b} f(x)g'(x) \text{d} x = \int_{a}^{b} f(x)h(x) \text{d} x = f(b)g(b) - \int_{a}^{b} f'(x) \left(\int h(t) \text{d} t \right) \text{d} x$$$$\int_{a}^{b} f(x)h(x) \text{d} x = \left[ f(x) \int h(x) \text{d}x \right]^{b}_{a} - \int_{a}^{b} \left( f'(x) \int h(x) \text{d}x \right) \text{d} x$$
Note: $\int h(x) \text{d}x$ does not denote any area. It denotes 'one' anti-derivative of $h(x)$ say $g(x)$, i.e., $g'(x) = h(x)$. If $g$ works, $g+C$ also works, but we don't need that $C$.

$f,g$ are differentiable such that $f \circ g$ makes sense $$(f \circ g)'(x) = f'(g(x)) \cdot g'(x)$$by FTC, $$\int_{a}^{b} (f \circ g)'(x) \text{d} x f(g(b)) - f(g(a))$$by chain rule, $$\int_{a}^{b} f'(\underbrace{g(x)}_{=u})\underbrace{g'(x) \text{d} x}_{=\text{d} u} = \int_{g(a)}^{g(b} f'(u) \text{d} u$$.



$\Phi : [a,b] \longrightarrow \mathbb{R}$ is differentiable & $\Phi'$ is integrable.
$f:[c,d] \longrightarrow \mathbb{R}$, $[c,d] \subseteq$ range $\Phi$.
$f$ must be continuous on range of $\Phi$.
$$\int_{a}^{b} f(\Phi (x)) \Phi ' (x) \text{d} x = \int_{\Phi (a)}^{\Phi (b)} f(u) \text{d} u$$
$\text{\underline{Alternate Criterion:}}$
$f$ integrable but $\Phi$ must also be bijective in addition to what is stated above.
[This replaces the critera of $f$ continuous.]

$\text{\underline{Sme Results:}}$
$$\bullet \; \int_{a}^{b} f(x) \text{d} x = \int_{a}^{b} f(a+b-x) \text{d} x $$$$\bullet \; \int_{-a}^{a} f(x) \text{d} x = \int_{0}^{a} (f(x)+f(-x)) \text{d} x $$$$\bullet \; f(x) \text{ odd }, \int_{-a}^{a} f(x) \text{d} x = 0$$
$f(x)$ may not exist at $a$ or $b$ but is continuous on $[c,d] \subset (a,b)$ $\forall$ $c,d$ $$\int_{a}^{b} f(x) \text{d} x = \lim_{t \longrightarrow a^+} \lim_{y \longrightarrow b^-} \int_{t}^{y} f(x) \text{d} x$$
$$\int_{0}^{\infty} f'(t) \text{d} t = \lim_{M \longrightarrow \infty} \int_{0}^{M} f'(t) \text{d}t = \lim_{M \longrightarrow \infty} f(M) - f(0)$$
$\text{\underline{Special Addition:}}$
\[ \boxed{ \lim_{n \to \infty} \frac{b-a}{n} \sum_{k=1}^n f\left(a + k\frac{b-a}{n}\right) = \int_{a}^{b} f(x) \, dx } \]
This post has been edited 5 times. Last edited by SomeonecoolLovesMaths, Nov 22, 2024, 2:08 PM

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add this one new thing to the intergrals
\[ \boxed{ \lim_{n \to \infty} \frac{b-a}{n} \sum_{k=1}^n f\left(a + k\frac{b-a}{n}\right) = \int_{a}^{b} f(x) \, dx } \]

by Levieee, Nov 21, 2024, 8:38 PM

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omegaorzmaxpeerowgawdlvlsaar

by L13832, Nov 25, 2024, 6:07 AM

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thank u so much cool mathematician

by iamahana008, Dec 13, 2024, 11:19 AM

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iamahana008 wrote:
thank u so much cool mathematician

Ma'am when will you upload in your blog?

by SomeonecoolLovesMaths, Dec 15, 2024, 1:04 PM

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  • add this one new thing to the intergrals that is lim n tends to infinity b-a/n summation k=1 to n f(a+k(b-a)/n))= int_{a}^b f(x) dx

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