Cov Lus Nug Piv Txwv thiab Kev Sib Tham Txog Kev Siv Integrals Hauv Kev Xam Thaj Chaw ntawm Lub Dav Hlau Tiaj
Hauv kev qhia lej, feem ntau ntsib cov integrals hauv calculus. Ib qho ntawm cov ntawv thov paub zoo tshaj plaws ntawm integrals yog xam thaj tsam hauv qab ib daim duab nkhaus lossis daim duab dav hlau. Tsab xov xwm no yuav tham txog ntau qhov teeb meem piv txwv thiab tham txog kev siv cov integrals rau kev xam thaj tsam ntawm daim duab dav hlau.
Kev Taw Qhia Txog Kev Tshawb Fawb
Ua ntej mus rau qhov teeb meem piv txwv, cia peb rov xyuas lub tswv yim yooj yim ntawm kev xam thaj tsam hauv qab ib daim nkhaus siv cov integrals. Yog tias peb muaj ib qho function f(x) uas txuas ntxiv mus rau ntawm lub sijhawm [a, b], ces thaj tsam hauv qab daim nkhaus y = f(x) txij x = a txog x = b yog:
\[ L = \int_{a}^{b} f(x) \, dx \]
Geometrically, qhov no txhais tau tias peb tab tom suav qhov chaw ntawm lub duab plaub nyias nyias los ntawm x = a txog x = b.
Piv txwv lus nug 1
Lo lus nug
Xam thaj tsam hauv qab tus nkhaus y = x² hauv lub sijhawm [1, 3].
Kev Sib Tham
Yuav kom xam tau thaj tsam, peb siv qhov sib npaug:
\[ L = \int_{1}^{3} x^2 dx \]
Peb pib los ntawm kev nrhiav tus antiderivative ntawm \( x^2 \). Tus antiderivative ntawm \( x^2 \) yog \( \frac{x^3}{3} \). Tom qab ntawd qhov integral dhau los ua:
\[ L = \left[ \frac{x^3}{3} \right]_{1}^{3} \]
Nco ntsoov tias peb yuav tsum soj ntsuam qhov antiderivative ntawm cov kev txwv ntawm qhov integral:
\[ L = \sab laug ( \frac{3^3}{3} \right) – \sab laug ( \frac{1^3}{3} \right) \]
\[ L = \sab laug ( \frac{27}{3} \right) – \sab laug ( \frac{1}{3} \right) \]
\[ L = 9 – \frac{1}{3} \]
\[ L = \frac{27}{3} – \frac{1}{3} \]
\[ L = \frac{26}{3} \]
Yog li, thaj tsam hauv qab nkhaus y = x² ntawm x = 1 txog x = 3 yog:
\[ \frac{26}{3} \, \text{chav tsev cheeb tsam} \]
Piv txwv lus nug 2
Lo lus nug
Txheeb xyuas thaj tsam ntawm thaj tsam uas ciam teb los ntawm kab nkhaus y = x³ thiab cov kab x = 1 thiab x = 2.
Kev Sib Tham
Yuav kom xam tau thaj tsam, peb siv qhov sib npaug:
\[ L = \int_{1}^{2} x^3 dx \]
Ib yam li niaj zaus, peb pib los ntawm kev nrhiav tus antiderivative ntawm \( x^3 \). Tus antiderivative ntawm \( x^3 \) yog \( \frac{x^4}{4} \). Tus integral dhau los ua:
\[ L = \left[ \frac{x^4}{4} \right]_{1}^{2} \]
Soj ntsuam cov kev txwv ntawm qhov sib xyaw ua ke:
\[ L = \sab laug ( \frac{2^4}{4} \right) – \sab laug ( \frac{1^4}{4} \right) \]
\[ L = \sab laug ( \frac{16}{4} \right) – \sab laug ( \frac{1}{4} \right) \]
\[ L = 4 – \frac{1}{4} \]
\[ L = \frac{16}{4} – \frac{1}{4} \]
\[ L = \frac{15}{4} \]
Yog li, thaj tsam hauv qab nkhaus y = x³ ntawm x = 1 txog x = 2 yog:
\[ \frac{15}{4} \, \text{chav tsev cheeb tsam} \]
Piv txwv lus nug 3
Lo lus nug
Txheeb xyuas thaj tsam ntawm thaj tsam uas ciam teb los ntawm cov kab nkhaus y = x² + 1 thiab y = 2x + 2 nyob rau hauv lub sijhawm x = 0 txog x = 1.
Kev Sib Tham
Ua ntej, peb yuav tsum nrhiav cov ntsiab lus sib tshuam los txiav txim siab qhov txwv ntawm kev sib koom ua ke. Cov lus teb rau \( x^2 + 1 = 2x + 2 \):
\[ x^2 + 1 = 2x + 2 \]
\[ x^2 – 2x – 1 = 0 \]
Siv cov qauv quadratic:
\[ x = \frac{2 \pm \sqrt{4 + 4}}{2} \]
\[ x = \frac{2 \pm \sqrt{8}}{2} \]
\[ x = \frac{2 \pm 2\sqrt{2}}{2} \]
\[ x = 1 \pm \sqrt{2} \]
Txawm li cas los xij, rau qhov txwv sab saud thiab qis dua ntawm 0 thiab 1, peb tsis tas yuav siv cov kev daws teeb meem quadratic, tsuas yog qhov txwv integral ib txwm muaj los ntawm 0 txog 1. Tom ntej no, xam thaj tsam ntawm qhov nkhaus y sab saud rho tawm qhov nkhaus y hauv qab raws li cov kev txwv no:
\[ L = \int_{0}^{1} [(2x + 2) – (x^2 + 1)] \, dx \]
Kev ua kom yooj yim rau kev ua haujlwm:
\[ L = \int_{0}^{1} (2x + 2 – x^2 – 1) \, dx \]
\[ L = \int_{0}^{1} (-x^2 + 2x + 1) \, dx \]
Tom ntej no, peb pom cov antiderivative:
Tus antiderivative ntawm \((-x^2) \) yog \(-\frac{x^3}{3} \),
Tus antiderivative ntawm \((2x) \) yog \(x^2 \),
Tus antiderivative ntawm \((1) \) yog \(x \).
Yog li ntawd,
\[ L = \left. \left(-\frac{x^3}{3} + x^2 + x \right) \right|_0^1 \]
Kev ntsuam xyuas tom ntej:
\[ L = \sab laug[ -\frac{1^3}{3} + 1^2 + 1 \right] – \sab laug[ -\frac{0^3}{3} + 0^2 + 0 \right] \]
\[ L = \sab laug[ -\frac{1}{3} + 1 + 1 \right] – \sab laug[ 0 \right] \]
\[ L = -\frac{1}{3} + 2 \]
\[ L = \frac{6}{3} – \frac{1}{3} \]
\[ L = \frac{5}{3} \]
Yog li, thaj tsam ntawm thaj tsam uas raug txwv los ntawm cov kab y = x² + 1 thiab y = 2x + 2 ntawm lub sijhawm [0, 1] yog:
\[ \frac{5}{3} \, \text{chav tsev cheeb tsam} \]
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Los ntawm cov piv txwv saum toj no, peb tuaj yeem pom tias cov integrals tuaj yeem siv los xam thaj chaw hauv qab ib txoj kab nkhaus lossis ntawm ob txoj kab nkhaus. Nrog kev nkag siab zoo txog cov ntsiab lus yooj yim ntawm integrals thiab cov txheej txheem antiderivative, kev suav cov cheeb tsam no dhau los ua qhov systematic thiab ua haujlwm tau zoo. Vam tias, tsab xov xwm no tau ua rau peb nkag siab ntau ntxiv txog kev siv cov integrals hauv lub ntiaj teb tiag tiag, tshwj xeeb tshaj yog hauv kev ntsuas thaj chaw ntawm cov nto dav hlau.
