6.4 Second Fundamental Theorem of Calculus¶

${\displaystyle {\int }_a^{b}f(t)dt=F(b)-F(a)}$ where $F^{\prime }(x)=f(x)$. ${\displaystyle {\int }_a^{x}f(t)dt=f(x)-F(a)}$ ie: $\frac{d}{dx}(x^x-x)=\frac{d}{dx}\left(x^2\right)-\frac{d}{dx}\left(x\right)=2x-1$ $\Rightarrow {\displaystyle \frac{d}{dx}\left({\int }_a^{x}f(t)dt\right)=\frac{d}{dx}(f(x)-F(a))}$ $\Rightarrow {\displaystyle \frac{d}{dx}\left({\int }_a^{x}f(t)dt\right)=\frac{d}{dx}(F(x))-\frac{d}{dx}(F(a))}$ $\Rightarrow {\displaystyle \frac{d}{dx}\left({\int }_a^{x}f(t)dt\right)=f(x)}$ So ${\displaystyle {\int }_a^{x}f(t)dt}$ is an antiderivative of $f$

ie: $\frac{d}{dx}\left({\int }_{-3}^x{t}^2-2tdt\right)=x^2-x$ $\frac{d}{dx}\left({\int }_{\pi }^x\sin (t^2)\cos (t)dt\right)=\sin (x^2)\cos (x)$

TLDR: ${\displaystyle \frac{d}{dx}\left({\int }_{a}xf(t)dt\right)=f(x)}$

Hokora's TLDR: its a function to find the antiderivative of literally any function

Let ${\displaystyle F(x)={\int }_0^x{e}^{-t^2}dt}$ ${\displaystyle \frac{d}{dx}\left(\frac{F(x)}{x}\right)=\frac{x\frac{d}{dx}(F(x))-F(x)\frac{d}{dx}\left(x\right)}{x^2}}$

"Low dee high minus high dee low square the denominator down below"

${\displaystyle =\frac{x{e}^{-x^2}-F(x)}{x^2}=\frac{e^{-x^2}-{\int }_0^x{e}^{-t^2}}{x^2}}$

${\displaystyle \frac{d}{dx}\left({\int }_0^{x^2}{e}^{-t^2}dt\right)=\frac{d}{dx}(F(x^2))}$ $=F^{\prime }(x^2)\ast \frac{d}{dx}\left(x^2\right)$ ${\displaystyle e^{-(x^2{)}^2}(2x)}$ $=2x{e}^{-x^4}$

Recall ${\displaystyle \underset{t\to 0}{lim}\frac{\sin (t)}{t}=1}$. ${\displaystyle f(t)=\{\begin{array}{cc}\frac{\sin t}{t}& ,t\ne 0\\ 1& ,t=0\end{array}}$ Scientists call $Si(x)={\int }_0^{x}f(t)dt$

Estimate: $Si(0),Si(1),Si(2),Si(3)$. using Riemann sums with $\mathrm{\Delta }t=0.5$ $Si(0)={\int }_0^0\frac{\sin (t)}{dt}dt=0$ $Si(1)={\int }_0^1\frac{\sin t}{t}dt=$ ! LHS: $f(0)\ast 0.5+f(0.5)\ast 0.5=(1\ast 0.5)+\left(\left(\frac{\sin 0.5}{0.5}\right)0.5\right)$ ! RHS: $f(0.5)\ast 0.5+f(1)\ast 0.5=\left(\left(\frac{\sin 0.5}{0.5}\right)0.5\right)+\left(\left(\frac{\sin 1}{1}\right)0.5\right)$ ! AVG: $(0.97945\cdots +0.9002\dots )/2=0.94\dots$

$Si(2)={\int }_0^2\frac{\sin t}{t}dt\approx 1.60$ $Si(3)={\int }_0^3\frac{\sin t}{t}dt\approx 1.84$

$\frac{d}{dx}(Si(x))=\frac{\sin x}{x}$

7.1 Integration by Substitution¶

$\frac{d}{dx}(sin(x^3))=\underset{\text{derivative of outer function}}{\underbrace{\cos (x^3)}}\ast \underset{\text{derivative of inner function}}{\underbrace{3x^2}}$ $\frac{d}{dx}(f(g(x))=f^{\prime }(g(x))\ast g^{\prime }(x)$ $\frac{d}{dx}(f(u))=f^{\prime }(u)\ast \frac{du}{dx}$ $\int \frac{df}{dx}dx=\int \frac{df}{du}\ast \int \frac{du}{dx}dx$ $f(x)=\int f^{\prime }(u)\ast \frac{du}{dx}dx=\int f^{\prime }(u)du$ $\int f^{\prime }(g(x))g^{\prime }(x)dx=\int f^{\prime }(u)du=f(u)+C=f(g(x))+C\to$ let $u=g(x)$ $\frac{du}{dx}=g^{\prime }(x)$ $du=g^{\prime }(x)dx$

$u=x^3$ $\frac{du}{dx}=3x^2$ $du=3x^{2}dx$ $\underset{=\int \cos ()}{\int \cos }\underset{u}{\underbrace{(x^3)}}\ast \underset{du}{\underbrace{3x^{2}dx}}$ $=\sin (u)+C$ $=\sin (x^3)+C$

Ex: $\int x^3\sqrt{x^4+5}dx$

$u=x^4+5$ $\frac{du}{4}=\frac{4x^{3}dx}{4}$ $\frac{1}{4}du=x^{3}dx$

${\displaystyle =\int \sqrt{4}(\frac{1}{4}du)}$ ${\displaystyle =\frac{1}{4}\int u^{\frac{1}{2}}du}$ ${\displaystyle =\frac{1}{4}\left(\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right)+C}$ ${\displaystyle =\frac{1}{2\xcancel{4}}\ast \frac{x^1}{3}{u}^{\frac{3}{2}}+C}$ ${\displaystyle =\frac{1}{6}{(x^4+5)}^{\frac{3}{2}}+C}$

Ex: $\int e^{\cos \mathrm{\Theta }}\sin \mathrm{\Theta }d\mathrm{\Theta }$

$u=\cos \mathrm{\Theta }$ $du=-\sin \mathrm{\Theta }d\mathrm{\Theta }$ $-du=\sin \mathrm{\Theta }d\mathrm{\Theta }$

$=\int e^u(-du)$ $=-\int e^{u}du$ $=-e^u+C$ $=-e^{\cos \mathrm{\Theta }}+C$

Ex: ${\displaystyle \int \frac{e^t}{1+e^t}dt}$ $=\int \frac{e^{t}dt}{1+e^t}=\int \frac{du}{u}$

$u=1+e^t$ $du=e^{t}dt$

$=\int \frac{1}{u}du$ $=\ln |u|+C$ $=\ln |1+e^t|+C$ $=\ln (1+e^t)+C$

$\frac{d}{dt}(\ln (1+e^t))=\frac{1}{1+e^t}(1+e^t{)}^{\prime }=\frac{e^t}{1+e^t}$

Ex: ${\displaystyle \int \tan \mathrm{\Theta }d\mathrm{\Theta }=\int \frac{\sin \mathrm{\Theta }}{\cos \mathrm{\Theta }}d\mathrm{\Theta }}$

$\xcancel{\begin{array}{c}u=\sin \mathrm{\Theta }\\ du=\cos \mathrm{\Theta }d\mathrm{\Theta }\end{array}}$ $u=\cos \theta$ $du=-\sin \mathrm{\Theta }d\mathrm{\Theta }$ $-du=\sin \mathrm{\Theta }d\mathrm{\Theta }$

$=\int \frac{-du}{u}$ $=\int -\frac{1}{u}du$ $=-\ln |u|+C$ $=-\ln |\cos \mathrm{\Theta }|+C$

Ex: ${\displaystyle \int e^x{\cos }^3(e^x)\sin (e^x)dx}$

$u=e^x$ $du=e^{x}dx$

$=\int \underset{(\cos u{)}^3}{\underbrace{{\cos }^3(u)}}\sin (u)du$

$w=cosu$ $dw=\sin (u)du$ $-dw=sin(u)du$

${\displaystyle =\int -w^{3}dw=-\frac{w^4}{4}+C=\frac{{\cos }^{4}u}{4}+C=-\frac{{\cos }^4(e^x)}{4}+C}$

Second way of doing it:

$\int e^x{\cos }^3(e^x)\sin (e^x)dx$

$u=\cos (e^x)$ $du=-\sin (e^x)(e^x)dx$ $-du=\sin (e^x)e^{x}dx$

$=\int u^3(-du)=-\frac{u^4}{4}+C$ $=-\frac{{\cos }^4(e^x)}{4}+C$

Ex:${\displaystyle {\int }_0^{2}x{e}^{x^2}dx}$

$u=x^2$ $\underset{2}{\underset{―}{du}}=\underset{2}{\underset{―}{2xdx}}$ $\frac{du}{2}=xdx$

$={\int }_0^2{e}^u$ $=\frac{1}{2}{\int }_{x=0}^{x=2}{e}^{u}du$ $=\frac{1}{2}{\int }_{x=0}^{x=2}{e}^{u}du$ $\xcancel{\begin{array}{c}=\frac{1}{2}{e}^u{|}_0^2\\ =\frac{1}{2}{e}^2-\frac{1}{2}{e}^0\\ =\frac{1}{2}{e}^2-\frac{1}{2}\end{array}}$

$\int x{e}^{x^2}dx$

$u=x^2$ $\frac{du}{2}=xdx$

$=\frac{1}{2}\int e^{u}du$ $=\frac{1}{2}{e}^u+C$ $=\frac{1}{2}{e}^{x^2}+C$