Integrals of inverse functions!

Proof without words (see image; credit: Jonathan Steinbuch, CC BY-SA 3.0, via Wikimedia Commons)...

For any montonic and invertible function \(f(x)\) in the interval \([a,b]\):
\[\displaystyle\int_a^bf(x)~ \mathrm dx+\int_{f(a)=c}^{f(b)=d}f^{-1}(x)~\mathrm dx=b\cdot f(b)-a\cdot f(a)=bd-ac\]

If \(F\) is an antiderivative of \(f\), then the antiderivatives of \(f^{-1}\) are:
\[\boxed{\displaystyle\int f^{-1}(y)~\mathrm dy=yf^{-1}(y)-F\circ f^{-1}(y)+C}\]
where \(C\) is an arbitrary constant (of integration), and \(\circ\) is the composition operator (function composition).

For example:
\[\begin{align*}\displaystyle\int \sin^{-1}(y) \, \mathrm dy &= y\sin^{-1}(y) - (-\cos(\sin^{-1}(y)))+C\\ &=y\sin^{-1}(y)+\sqrt{1-y^2}+C\end{align*}\]

\[\displaystyle\int \ln(y) \, dy = y\ln(y)-\exp(\ln(y)) + C= y\ln(y)-y + C.\]

#Function #InverseFunction #InverseFunctions #Functions #Integral #Integrals #Antiderivative #Integration #Calculus #FunctionComposition #CompositeFunction)

I drew the said curve, y=x², starting with x= -32 , ending with x=0, with this ingenious toy: www.matheretter.de/rechner/gfplot #Februarty2025, day 8, #puzzle: There is a #square 32*32=1024 and a #parabola. Find the area under the under the curve #calculus #antiderivative

RE: https://bsky.app/profile/did:plc:omyr27fmzj3phbagch4sqyub/post/3lhoq76ar322q

Plotter für Polynomfunktionen ...
Plotter für Polynomfunktionen - Matheretter

Plotter für Polynomfunktionen inklusive Nullstellenberechnung im Reellen und Komplexen. Im Funktionsplotter sind beliebige Eingaben möglich.

A beautiful curve y=x² interval [-32;0] Higher boundary minus lower boundary: 0--32=32 #puzzle = #antiderivative = area under the curve =0³/3--1³/3= 1/3=341+1/3 square units of 1024=32² 1x² = 1/3 of half the boundaries of the cube x³ 6/2=3squares x² #sharingisthenewlearning #parabola #calculus
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