>>7\(\Huge{😂😂😂}\)dude summation lmao
dude calculus lmao
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\(\Large {\LaTeX}\) lmao
dude derivatives lmao
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dude integrals lmao
\(\;\;\;\;\;\;\;\;\;\; \big \langle \Big \langle \bigg \langle \Bigg \langle \Huge{U \;\; MENA \;\; HASKAL?} \normalsize \Bigg \rangle \bigg \rangle \Big \rangle \big \rangle\)λλλλλλ
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λλλλnot the same thing but some similar and fun stuff:
Evaluating
\(\int^{2}_{0}(x^{2}+x)dx\) as a Riemann sum.
\(\sum\limits_{i=1}^n=\frac{n(n+1)}{2}\)\(\sum\limits_{i=1}^ni^{2}=\frac{n(n+1)(2n+1))}{6}\)\(\Delta x = \frac{b-a}{n}\)\(a = 0, b = 2\)\(\Delta x = \frac{2-0}{n} = \frac{2}{n}\)\(x_{i}=\frac{2i}{n}\)\(\int_{0}^{2}(x^{2}+x)dx=\displaystyle{\lim_{n \to \infty}} \sum\limits_{i=1}^n\Bigg[\Big(\frac{2i}{n}\Big)^{2}+\frac{2i}{n}\Bigg]\Delta x\)\(=\displaystyle{\lim_{n \to \infty}}\frac{2}{n}\sum\limits_{i=1}^n\Bigg[\frac{4i^{2}}{n^{2}}+\frac{2i}{n}\Bigg]\)\(=\displaystyle{\lim_{n \to \infty}}\Bigg(\frac{2}{n}\sum\limits_{i=1}^n\frac{4i^{2}}{n^{2}}+\frac{2}{n}\sum\limits_{i=1}^n\frac{2i}{n}\Bigg)\)\(=\displaystyle{\lim_{n \to \infty}}\Bigg(\frac{8}{n^{3}}\sum\limits_{i=1}^n i^{2} + \frac{4}{n^{2}}\sum\limits_{i=1}^n i\Bigg)\)\(=\displaystyle{\lim_{n \to \infty}}\Bigg(\frac{8}{n^{3}} * \frac{n(n+1)(2n+1)}{6} + \frac{4}{n^{2}} * \frac{n(n+1)}{2}\Bigg)\)\(=\displaystyle{\lim_{n \to \infty}}\Bigg(\frac{4(2n^{2}+3n+1)}{3n^{2}}+\frac{2(n+1)}{n}\Bigg)\)\(=\displaystyle{\lim_{n \to \infty}}\Bigg(\frac{8n^{2} + 12n + 4}{3n^{2}} + \frac{2n+2}{n}\Bigg)\)Apply L'Hospital's Rule because
\(\frac{\infty}{\infty}\) is indeterminate.
Differentiate the numerator and denominator until it's no longer indeterminate.
Additionally, using the laws of limits, we can break up the limit into two separate limits then solve separately and add together at the end.
1st part:
\(=\displaystyle{\lim_{n \to \infty}}\Bigg(\frac{8n^{2}+12n+4}{3n^{2}}\Bigg) = \displaystyle{\lim_{n \to \infty}}\Bigg(\frac{\frac{d}{dx}(8n^{2}+12n+4)}{\frac{d}{dx}(3n^{2})}\Bigg)\)\(=\displaystyle{\lim_{n \to \infty}}\Bigg(\frac{16n+12}{6n}\Bigg)\)Still indeterminate if you check with direct substitution, therefore L'Hospital's Rule must be applied again. Taking the second derivative of the original, basically.
\(=\displaystyle{\lim_{n \to \infty}}\Big(\frac{16}{6}\Big)\)Limit of a constant is just a constant
\(=\frac{16}{6} = \frac{8}{3}\)Second part:
\(=\displaystyle{\lim_{n \to \infty}}\frac{2n+2}{n} = \frac{\infty}{\infty} = \) indeterminate again, so apply L'Hospital's Rule once more.
\(=\displaystyle{\lim_{n \to \infty}}\frac{2}{1} = 2\)\(\frac{8}{3} + 2 = \frac{8}{3} + \frac{6}{3} = \frac{14}{3}\)Therefore:
\(\int^{2}_{0}(x^{2}+x)dx = \frac{14}{3}\)*DJ Khaled voice* Anotha one.\(A(x+h)-A(x)\approx hf(x)\)\(A(x)=\int_{a}^{x}f(x)dt\)\(f(x)\approx\frac{A(x+h)-A(x)}{h}\)\(=\displaystyle{\lim_{h \to 0}}f(x)=\displaystyle{\lim_{h \to 0}}\Big(\frac{A(x+h)-A(x)}{h}\Big)\)\(f(x)=A'(x)\)💯💯💯💯💯
Let F(x) be any antiderivative of f(x):
\(F(x)=A(x)+c\)\(F(a)=A(a)+c\)\(F(b)=A(b)+c\)\(\;\;\;\;\;\;\;\;\;\; \big \langle \Big \langle \bigg \langle \Bigg \langle \Huge{WEW \;\;LAD} \normalsize \Bigg \rangle \bigg \rangle \Big \rangle \big \rangle\)To be integrable,
\(f\) must be continuous. Suppose
\(f\) is continuous on the interval
\([a,b]\). Then,
(i)
\(\frac{d}{dx}\int_{a}^{x}f(t)dt=f(x), a\leq x \leq b\)and
(ii)
\(\int_{a}^{b}f(t)dt=F(b)-F(a)\), where F is any antiderivative
\(F(b)=A(b)+c\)\(F(b)=\int_{a}^{b}f(t)dt+F(a)\)\(\int_{a}^{b}f(t)dt=F(b)-F(a)\)console.log("anudda one");\(\int_{3}^{x}e^{t^{2}}dt=?\)\(\int_{3}^{x}e^{t^{2}}dt = e^{x^{2}}\)document.writeln("<p>anudda one</p>");\(\frac{d}{dx}\int_{x^{4}}^{1}\sqrt{t^{3}+1}dt=???\)\(=\frac{d}{dx}(-\int_{1}^{x^{4}}\sqrt{t^{3}+1}dt)\)\(=- \frac{d}{dx}\int_{1}^{x^{4}}\sqrt{t^{3}+1}=-\sqrt{(x^{4}+1(4x^{3}))}\)\(=-4x^{3}\sqrt{x^{12}+1}\)// _____ _ _ ___ _ _ ______ _____
// | __ \ | | | |/ / | | | /\ | | | ____| __ \
// | | | | | | | ' /| |__| | / \ | | | |__ | | | |
// | | | |_ | | | < | __ | / /\ \ | | | __| | | | |
// | |__| | |__| | | . \| | | |/ ____ \| |____| |____| |__| |
// |_____/ \____/ |_|\_\_| |_/_/ \_\______|______|_____/ \(\frac{d}{dx}\int_{e^{x}}^{\ln x}\frac{\sin t}{t}dt=?\)\(=\frac{d}{dx}\Big( \int_{e^{x}}^{1}\frac{\sin t}{t}dt + \int_{1}^{\ln x}\frac{\sin t}{t}dt \Big)\)I'll switch the bounds of your definite integral, if you know what I mean ( ≖‿≖)\(=\frac{d}{dx}\Big(-\int_{1}^{e^{x}}\frac{\sin t}{t}dt + \int_{1}^{\ln x} \frac{\sin t}{t}dt\Big)\)NOEHGSANTAEPD\(=-\frac{\sin(e^{x})}{e^{x}}*e^{x}+\frac{\sin(\ln x)}{\ln x} * \frac{1}{x}\)\(=\frac{\sin(\ln x)}{x\ln x} -\sin(e^{x})\)We out here derivin' integrals my dude 😎
\(\frac{d}{dx}(\int_{0}^{\tan^{3}x}\sin(t^{2})dt)=???\)\(\frac{d}{dx}\int_{0}^{\tan^{3}x}\sin(t^{2})dt=\sin(\tan^{3}x)^{2}(3\tan^{2}x\sec^{2}x)\)System.out.println("Anudda one!");Evaluate dat shit:
\(\int_{0}^{2\pi}\mid\cos x\mid dx\)\(=\Large \int_{\normalsize 0}^{\normalsize\frac{\pi}{2}}\normalsize \cos x \; {dx} - \Large \int_{\normalsize\frac{\pi}{2}}^{\normalsize\frac{3\pi}{2}}\normalsize \cos x \; {dx} + \Large\int_{\normalsize\frac{3\pi}{2}}^{\normalsize 2\pi}\normalsize \cos x \; {dx}\)\(\sin x\Huge\vert_{\normalsize 0}^{\normalsize\frac{\pi}{2}}\Large - \normalsize \sin x \Huge\vert_{\normalsize \frac{\pi}{2}}^{\normalsize \frac{3\pi}{2}}\Large + \normalsize \sin x \Huge\vert_{\normalsize\frac{3\pi}{2}}^{\normalsize 2\pi}\)\(=(1-0) - (1-1) + (0+1)\)\(=4\)BBCode Spoiler Pong
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