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The Lebesgue integral and convergence theorems
Integrating by slicing the range instead of the domain; monotone and dominated convergence.
Riemann slices the x-axis; Lebesgue slices the y-axis and asks how much of the domain sits at each height. The payoff is the convergence theorems: under mild conditions, the limit of the integrals is the integral of the limit. Picture it: horizontal strips under the curve instead of vertical ones. Think it: L¹ and L² become complete spaces, the Hilbert spaces of functional analysis.
हल किया गया उदाहरण · integrate e^(-x) dx from 0 to oo
चरण दर चरण
- \int_{0}^{\infty} e^{- x}\, dx
First find an antiderivative F, then evaluate F(b) − F(a).
- u = - x,\quad du = -1\, dx
Substitute u = - x.
- \int e^{- x}\, dx = \int - e^{u}\, d_u
Rewrite the integral in terms of u.
- \int - e^{u}\, d_u = -1 \int e^{u}\, d_u
Pull the constant -1 out of the integral.
- \int e^{u}\, d_u = e^{u}
∫ aᵘ du = aᵘ / ln a (for eˣ that is just eˣ).
- = - e^{- x}
Substitute back u = - x.
- F(\infty) - F(0) = \left(0\right) - \left(-1\right)
Fundamental theorem of calculus: plug in the limits.
- = 1
Simplify.
जवाब दिखाएँ
अब आप कोई प्रश्न चुनें, या अपना प्रश्न टाइप करें या बनाएं। हर चरण, एक चित्र, और उत्तर तब तक छिपा रहता है जब तक आप न पूछें।
एक स्वतंत्र वृत्तान्त प्रत्येक पाठ पर आगे कहता है, जो आपने पूरा किया है, आपके हल किये गये समस्याओं का रिकॉर्ड, और एक शिक्षक जिसके बारे में आप पूछ सकते हैं ।
साइन अप करें लॉगिनयहाँ उपयोग में संकेत
पूरी परिभाषा, चित्र, और इसमें हर अक्षर का अर्थ क्या है?
कैसे: The Lebesgue integral and convergence theorems
- First find an antiderivative F, then evaluate F(b) − F(a).
- Substitute u = - x.
- Rewrite the integral in terms of u.
- Pull the constant -1 out of the integral.
- ∫ aᵘ du = aᵘ / ln a (for eˣ that is just eˣ).
- Substitute back u = - x.
- Fundamental theorem of calculus: plug in the limits.
- Simplify.
लोग जो सवाल पूछते हैं
What is wrong with the Riemann integral?
It fails on functions that oscillate too much, and it does not interact well with limits: the limit of integrable functions need not be integrable. Lebesgue's integral fixes both.