Rules summary from Monday¶

\(\displaystyle\int kdx=kx+C\) \(\displaystyle\int x^ndx=\frac{x^{n+1}}{n+1}+C,n\ne-1\) \(\displaystyle\int\frac{1}{x}dx=ln|x|+C\) \(\displaystyle\int e^xdx=e^x+C\) \(\displaystyle\int\sin{x}dx=-\cos{x}+C\) \(\displaystyle\int\cos{x}dx=\sin{x}+C\)

6.2¶

holy shit hokora put the image here please im going to explode D:

\(f(x)=x^2+4x-1\) \(g(x)=x^2-4x+5\) \(\displaystyle A=\int_{1}^{3}{-x^2+4x-1-(x^2-4x+5)}dx\) \(\displaystyle=\int_{1}^{3}{-2x^2+8x-6}dx=F(3)-F(1)\) where \(F'=f\) \(\displaystyle\int-2x^2+7x-6dx=-2\left(\frac{x'}{3}\right)+8\left(\frac{x^2}{2}\right)-6x+C\) \(\displaystyle\left(\frac{x^3}{3}\right)'=\left(\frac{3x^2}{3}\right)'\) \(=x^2\) \(\displaystyle=\frac{2}{3}x^3+4x^2-6x+C\)

Check: \(\left(-\frac{2}{3}x^3+4x^2-6x+C\right)''\) \(\displaystyle=-\frac{2}{3}\left(3x^2\right)+4(2x)-6+0\) \(-2x^2+8x-6\)

\(\displaystyle=\left(-2*\left(\frac{3^3}{3}\right)+4(3)^2-6(3)\right)-\left(-\frac{2}{3}(1)^3+4(1)^2-6(1)\right)\) \(\displaystyle=\left(-18+36-18\right)-\left(-\frac{2}{3}+4-6\right)\) \(\displaystyle=+\left(+\frac{2}{3}+2\right)=2\frac{2}{3}\)

Theorem: Properties of Antiderivatives¶

1) \(\displaystyle\int f(x)\pm g(x)dx=\int f(x)dx\pm\int g(x)dx\) 2) \(\displaystyle\int cf(x)dx=c\int f(x)dx\)

FTC: If \(F'=f\) then \(\displaystyle\int_{a}^{b}{f(x)}dx=F(b)-F(a)\)

Recall: \(\displaystyle\int_{a}^{b}{f(x)}dx==F(b)-F(a)\) \(=F(x)\biggr|_{a}^{b}\)

Ex:¶

\(\displaystyle\int_{1}^{2}3x^3dx=3\left(\frac{x^3}{3}\right)\Biggr|_{1}^{2}=x^3\biggr|_{1}^{2}\) \(=2^3-1^3=8-1=7\) \(\displaystyle\int_{1}^{2}3x^2dx=x^3+C\biggr|_{1}^{2}\) \(=2^3+C-\left(1^3+C\right)\) \(=8+C-1-C\) \(=8-1=\fbox{7}\)

\(\left(\dfrac{1}{\sin^2{\Theta}}\right)''\) \(\left(\csc^2{\Theta}\right)''\) \(=2\csc{\Theta}(\csc{\Theta})'\) \(=2\csc{\Theta}(-\csc{\Theta}\cot{\Theta})'\) \(=\displaystyle2\frac{1}{\sin{\Theta}}*\frac{1}{\sin{\Theta}}*\frac{\cos{\Theta}}{\sin{\Theta}}\) \(\displaystyle=\frac{-2\cos{\Theta}}{\sin^3{\Theta}}\)

\(\displaystyle\int_{0}^{\frac{\pi}{4}}{\frac{1}{\cos^2{}\Theta}d\Theta}=\int_{0}^{\frac{\pi}{4}}{\sec^2{\Theta}d\Theta}=\tan{\Theta}\biggr|_{0}^{\frac{\pi}{4}}=\tan{\frac{\pi}{4}}-\tan{0}=1-0=\fbox{1}\)

\(\left(\frac{1}{t^2}\right)'=\left(t^{-2}\right)'\) \(=2t^{-3}\) \(=-\dfrac{2}{t^3}\)

Ex:¶

\(\displaystyle\int_{1}^{e}{\frac{3}{t}-\frac{1}{t^2}}dt=\int_{1}^{e}{3*\frac{1}{t}-t^{-2}}dt\) \(=3\ln{|t|}+\dfrac{t^{-1}}{+1}\Biggr|_{1}^{e}\) \(=3\ln{|t|}+\dfrac{1}{t}\biggr|_{1}^{e}\) \(\displaystyle=3\ln{e^1}+\frac{1}{e}-\left(3\ln{\frac{e^0}{1}}+\frac{1}{1}\right)\) \(\displaystyle=3(1)+\frac{1}{e}-\left(\cancel{3(0)}+1\right)\) \(\displaystyle=3+\frac{1}{e}-1\) \(\displaystyle=2+\frac{1}{e}\)

Ex:¶

6.4 Sound Fundamental Theorem of Calculus¶

We say \(f\) is an elementary function if it can be written using sums, products, roots, scaling, and compositions (function in function) of finitely many polynomial, trigonometric, and inverse trigonometric, exponential, and logarithmic functions.

Examples of elementary functions:¶

\(\displaystyle\frac{3}{t}-\frac{1}{t^2}\) \(2x^2+4x-7\) \(\sqrt{x^3}\) \(\displaystyle\frac{e^{x^2}}{\tan{(\ln{x})}}\) \(\pi\arcsin{\left(\sqrt[4]{e^{-\sin{x}}+1}\right)}\) We are actually very lucky if we can find an antiderivative \(F\) of \(f\) which is also an elementary function

\(\displaystyle\int{2t}dt=t^2+C\) Q: What about \(\displaystyle\int{e^{-t^2}}dt\)?

hokora please put the fuckin image here oh my god im gonna explod

Q: Is \(\DeclareMathOperator\erf{erf}\erf{(x)}\) an elementary function? (\(\erf\) is error function) A: No: proven by Liouville in 1835.

Let's say \(\erf{(x)}\) does exist. \(\displaystyle\erf{(b)}-\erf{(a)}=\int_{a}^{b}e^{-t^2}dt\) \(\displaystyle\erf{(x)}-\erf{(0)}=\int_{0}^{x}{e^{-t^2}}dt\) \(\displaystyle\erf(x)=\int_{0}^{x}{e^{-t^2}}dt\)

Thm: Sound Fundamentals Theorem of Calculus¶

(Construction Thm for Antiderivatives) If \(f\) is a continuous function on an interval, and \(a\) is any number in that interval, then the function \(F\) defined on that interval as \(F(x)=\int_{a}^{x}{f(t)}dt\) is an antiderivative of \(f\).

\(\displaystyle F'(x)=\frac{d}{dx}{\left(\int_{a}^{x}{f(t)}dt\right)}=f(x)\)

Let's do a more "informal" way to prove it

hokoraput the fUCKING image!!!!!!!!

"Proof": We want to show \(F'(x)=f(x)\). \(\displaystyle F'(x)=\lim_{h\rightarrow0}{\frac{F(x+h)-F(x)}{h}}\) Let's break apart what this means! \(\displaystyle F(x+h)=\int_{a}^{x+h}{f(t)}dt\) \(\displaystyle F(x)=\int_{a}^{x}{f(t)}dt\) \(\displaystyle F(x+h)-F(x)=\int_{a}^{x+h}{f(t)}dt-\int_{a}^{x}{f(t)dt}\) \(\displaystyle=\int_{x}^{x+h}{f(t)}dt\underset{\underset{\text{as }h\rightarrow0}{\uparrow}}{\approx}\underbrace{f(x)h}_{\underset{\text{rectangle}}{\text{Area of}}}\) \(\displaystyle F'(x)=\lim_{h\rightarrow0}{\frac{F(x+h)-F(x)}{h}}\) \(\displaystyle\approx\lim_{h\rightarrow0}{\frac{f(x)\cancel{h}}{\cancel{h}}}=f(x)\)

Proof of 1st FTC using 2nd FTC¶

Let \(F(x)=\int_{a}^{x}{f(t)}dt\). So let \(G\) be any antiderivative of \(f\). So \(G'=f\). But \(F'=f\Rightarrow F'=G'\Rightarrow F'-G'=0\Rightarrow(F-G)'=0\Rightarrow F(x)=G(x)+C\)

Note \(\displaystyle F(a)=\int_{a}^{a}{f(t)}dt=0\) \(\displaystyle\int_{a}^{b}{f(t)}dt=F(b)\) \(=F(b)-F(a)\) \(=G(b)+C-\left(G(a)+C\right)\) \(=G(b)-G(a)\) So for any antiderivative \(G\), \(\int_{a}^{b}{f(t)}ft=G(b)-G(a)\).

Ex:¶

\(\displaystyle\erf{(x)}=\int_{0}^{x}{e^{-t^2}}dt\) \(\displaystyle\frac{d}{dx}\left(\erf{(x)}\right)=e^{-x^2}\) \(\displaystyle\frac{d}{dx}\int_{3}^{x}{t^2+3}dt=x^2+3\) \(\displaystyle\frac{d}{dx}\int_{-\pi}^{x}{\sin{t}}dt=\sin{x}\)

\(F(x)=\int_{a}^{x}{f(t)}dt\) is an antiderivative of \(f\). \(\displaystyle\frac{d}{dx}\int_{a}^{x}{f(t)}dt=f(x)\)

\(\displaystyle\frac{d}{dx}(x\erf{(x)})=\frac{d}{dx}(x)\erf{(x)}+x\frac{d}{dx}(\erf{(x)})\) \(=\erf(x)+xe^{-x^2}\) \(\displaystyle\frac{d}{dx}\int_{0}^{x^2}{\cos{t}}dt\) \(\displaystyle\frac{d}{du}\int_{0}^{u}{\cos{t}}dt=\cos{u}\) \(\displaystyle u=x^2\Rightarrow\frac{dy}{dx}=2x\) \(\displaystyle\frac{d}{dx}\int_{0}^{u}{\cos{t}}dt\) \(\displaystyle=\frac{d}{du}\left(\int_{0}^{u}{\cos{t}}dt\right)\frac{dy}{dx}\) \(=\cos{u}(2x)\) \(=2x\cos{\left(x^2\right)}\) \(u=x^2\) \(\displaystyle\frac{d}{dx}\left(e^{x^2}\right)=\frac{d}{dx}\left(e^u\right)\) \(=\frac{d}{du}\left(e^u\right)\frac{dy}{dx}\) \(=e^u(2x)\) \(e^{x^2}(2x)\)

Power rule¶

\(\displaystyle\frac{d}{dx}\left(x*e^x\right)\) \(\displaystyle=\frac{d}{dx}(x)e^x+x\frac{d}{dx}(e^x)\) \(=e^x+xe^x\)