6.4 Second Fundamental Theorem of Calculus¶

\(\displaystyle \int_{a}^{b}{f(t)}dt=F(b)-F(a)\) where \(F'(x)=f(x)\). \(\displaystyle\int_{a}^{x}{f(t)}dt=f(x)-F(a)\) ie: \(\frac{d}{dx}\left(x^x-x\right)=\frac{d}{dx}\left(x^2\right)-\frac{d}{dx}\left(x\right)=2x-1\) \(\Rightarrow\displaystyle\frac{d}{dx}\left(\int_{a}^{x}{f(t)}dt\right)=\frac{d}{dx}\left(f(x)-F(a)\right)\) \(\Rightarrow\displaystyle\frac{d}{dx}\left(\int_{a}^{x}{f(t)}dt\right)=\frac{d}{dx}\left(F(x)\right)-\frac{d}{dx}\left(F(a)\right)\) \(\Rightarrow\displaystyle\frac{d}{dx}\left(\int_{a}^{x}{f(t)}dt\right)=f(x)\) So \(\displaystyle\int_{a}^{x}{f(t)}dt\) is an antiderivative of \(f\)

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

TLDR: \(\displaystyle\frac{d}{dx}\left(\int_{a}{x}{f(t)}dt\right)=f(x)\)

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

Let \(\displaystyle F(x)=\int_{0}^{x}{e^{-t^2}}dt\) \(\displaystyle\frac{d}{dx}\left(\frac{F(x)}{x}\right)=\frac{x\frac{d}{dx}\left(F(x)\right)-F(x)\frac{d}{dx}\left(x\right)}{x^2}\)

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

\(\displaystyle=\frac{xe^{-x^2}-F(x)}{x^2}=\frac{e^{-x^2}-\int_{0}^{x}e^{-t^2}}{x^2}\)

\(\displaystyle\frac{d}{dx}\left(\int_{0}^{x^2}{e^{-t^2}}dt\right)=\frac{d}{dx}\left(F(x^2)\right)\) \(=F'(x^2)*\frac{d}{dx}\left(x^2\right)\) \(\displaystyle e^{-(x^2)^2}(2x)\) \(=2xe^{-x^4}\)

Recall \(\displaystyle\lim_{t\rightarrow0}{\frac{\sin{(t)}}{t}}=1\). \(\displaystyle f(t)=\biggr\{\begin{matrix}{\frac{\sin t}{t}}&{,t\ne0}\\{1}&{,t=0}\end{matrix}\) Scientists call \(Si(x)=\int_{0}^{x}{f(t)}dt\)

Estimate: \(Si(0),Si(1),Si(2),Si(3)\). using Riemann sums with \(\Delta t=0.5\) \(Si(0)=\int_{0}^{0}{\frac{\sin{(t)}}{dt}}dt=0\) \(Si(1)=\int_{0}^{1}{\frac{\sin t}{t}}dt=\) ! LHS: \(f(0)*0.5+f(0.5)*0.5=\left(1*0.5\right)+\left(\left(\frac{\sin0.5}{0.5}\right)0.5\right)\) ! RHS: \(f(0.5)*0.5+f(1)*0.5=\left(\left(\frac{\sin0.5}{0.5}\right)0.5\right)+\left(\left(\frac{\sin1}{1}\right)0.5\right)\) ! AVG: \((0.97945\dots+0.9002\dots)/2=0.94\dots\)

\(Si(2)=\int_{0}^{2}\frac{\sin t}{t}dt\approx1.60\) \(Si(3)=\int_{0}^{3}\frac{\sin t}{t}dt\approx 1.84\)

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

7.1 Integration by Substitution¶

\(\frac{d}{dx}(sin(x^3))=\underbrace{\cos(x^3)}_{\text{derivative of outer function}}*\underbrace{3x^2}_{\text{derivative of inner function}}\) \(\frac{d}{dx}(f(g(x))=f'(g(x))*g'(x)\) \(\frac{d}{dx}(f(u))=f'(u)*\frac{du}{dx}\) \(\int\frac{df}{dx}dx=\int\frac{df}{du}*\int\frac{du}{dx}dx\) \(f(x)=\int f'(u)*\frac{du}{dx}dx=\int f'(u)du\) \(\int f'(g(x))g'(x)dx=\int f'(u)du=f(u)+C=f(g(x))+C\rightarrow\) let \(u=g(x)\) \(\frac{du}{dx}=g'(x)\) \(du=g'(x)dx\)

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

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

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

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

Ex: \(\int e^{\cos\Theta}\sin\Theta d\Theta\)

\(u=\cos\Theta\) \(du=-\sin\Theta d\Theta\) \(-du=\sin\Theta d\Theta\)

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

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

\(u=1+e^t\) \(du=e^tdt\)

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

\(\frac{d}{dt}(\ln(1+e^t))=\frac{1}{1+e^t}(1+e^t)'=\frac{e^t}{1+e^t}\)

Ex: \(\displaystyle\int\tan\Theta d\Theta=\int\frac{\sin\Theta}{\cos\Theta}d\Theta\)

\(\xcancel{\begin{matrix}u=\sin\Theta \\ du=\cos\Theta d\Theta\end{matrix}}\) \(u=\cos\theta\) \(du=-\sin\Theta d\Theta\) \(-du=\sin\Theta d\Theta\)

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

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

\(u=e^x\) \(du=e^xdx\)

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

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

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

Second way of doing it:

\(\int e^x\cos^3(e^x)\sin(e^x)dx\)

\(u=\cos(e^x)\) \(du=-\sin(e^x)(e^x)dx\) \(-du=\sin(e^x)e^xdx\)

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

Ex:\(\displaystyle\int_{0}^{2}xe^{x^2}dx\)

\(u=x^2\) \(\underset{2}{\underline{du}}=\underset{2}{\underline{2xdx}}\) \(\frac{du}{2}=xdx\)

\(=\int_{0}^{2}e^u\) \(=\frac{1}{2}\int_{x=0}^{x=2}e^udu\) \(=\frac{1}{2}\int_{x=0}^{x=2}e^udu\) \(\xcancel{\begin{matrix}{=\frac{1}{2}e^u\biggr|_{0}^{2}} \\ {=\frac{1}{2}e^2-\frac{1}{2}e^0} \\ {=\frac{1}{2}e^2-\frac{1}{2}}\end{matrix}}\)

\(\int xe^{x^2}dx\)

\(u=x^2\) \(\frac{du}{2}=xdx\)

\(=\frac{1}{2}\int e^udu\) \(=\frac{1}{2}e^u+C\) \(=\frac{1}{2}e^{x^2}+C\)