5.5: Kubadilisha
- Page ID
- 178512
- Tumia mbadala ili kutathmini integrals isiyojulikana.
- Matumizi badala ya kutathmini integrals uhakika.
Theorem ya msingi ya Calculus alitupa njia ya kutathmini integrals bila kutumia Riemann kiasi. Upungufu wa njia hii, ingawa, ni kwamba tunapaswa kupata antiderivative, na hii si rahisi kila wakati. Katika sehemu hii tunachunguza mbinu, inayoitwa ushirikiano na badala, kutusaidia kupata antiderivatives. Hasa, njia hii inatusaidia kupata antiderivatives wakati integrand ni matokeo ya derivative ya utawala wa mnyororo.
Mara ya kwanza, njia ya utaratibu wa kubadilisha inaweza kuonekana dhahiri sana. Hata hivyo, kimsingi ni kazi ya kuona - yaani, integrand inaonyesha nini cha kufanya; ni suala la kutambua fomu ya kazi. Kwa hiyo, tunatakiwa kuona nini? Tunatafuta integrand ya fomu\(f\big[g(x)\big]g′(x)\,dx\). Kwa mfano, katika muhimu
\[ ∫(x^2−3)^3 \, 2x \, dx. \label{eq1} \]
tuna
\[ f(x)=x^3 \nonumber \]
na
\[g(x)=x^2−3.\nonumber \]
Kisha
\[ g'(x)=2x.\nonumber \]
na
\[ f[g(x)]g′(x)=(x^2−3)^3(2x),\nonumber \]
na tunaona kwamba integrand yetu iko katika fomu sahihi. Mbinu inaitwa badala kwa sababu sisi badala sehemu ya integrand na kutofautiana\(u\) na sehemu ya integrand na\(du\). Pia inajulikana kama mabadiliko ya vigezo kwa sababu tunabadilisha vigezo ili kupata usemi ambao ni rahisi kufanya kazi na kwa kutumia sheria za ushirikiano.
Hebu\(u=g(x)\), ambapo\(g′(x)\) ni kuendelea juu ya muda, basi\(f(x)\) kuendelea juu ya mbalimbali sambamba ya\(g\), na basi\(F(x)\) kuwa antiderivative ya\(f(x).\) Kisha,
\[ \begin{align*} ∫f[g(x)]g′(x)\,dx &=∫f(u)\,du \\[4pt] &=F(u)+C \\[4pt] &= F(g(x))+C \end{align*}\]
Hebu\(f\),\(g\)\(u\), na\(F\) uwe kama ilivyoelezwa katika theorem. Kisha
\[ \dfrac{d}{dx}\big[F(g(x))\big]=F′(g(x))g′(x)=f[g(x)]g′(x). \nonumber \]
Kuunganisha pande zote mbili kwa heshima na\(x\), tunaona kwamba
\[ ∫f[g(x)]g′(x)\,dx=F(g(x))+C. \nonumber \]
Kama sisi sasa mbadala\(u=g(x)\), na\(du=g'(x)\,dx\), sisi kupata
\[ ∫f[g(x)]g′(x)\,dx=∫f(u)\,du=F(u)+C=F(g(x))+C. \nonumber \]
□
Kurudi kwenye tatizo tuliloangalia awali, tunaruhusu\(u=x^2−3\) na kisha\(du=2x\,dx\).
Andika upya muhimu (Equation\ ref {eq1}) katika suala la\(u\):
\[ ∫(x^2−3)^3(2x\,dx)=∫u^3\,du. \nonumber \]
Kutumia utawala wa nguvu kwa integrals, tuna
\[ ∫u^3\,du=\dfrac{u^4}{4}+C. \nonumber \]
Badilisha maneno ya awali kwa\(x\) nyuma katika suluhisho:
\[ \dfrac{u^4}{4}+C=\dfrac{(x^2−3)^4}{4}+C.\nonumber \]
Tunaweza kuzalisha utaratibu katika Mkakati wa kutatua matatizo yafuatayo.
- Angalia kwa makini katika integrand na kuchagua kujieleza\(g(x)\) ndani ya integrand kuweka sawa na u Hebu\(g(x)\) uchaguzi. vile kwamba pia\(g′(x)\) ni sehemu ya integrand.
- Mbadala\(u=g(x)\) na\(du=g′(x)dx.\) katika muhimu.
- Tunapaswa sasa kuwa na uwezo wa kutathmini muhimu kwa heshima na\(u\). Kama muhimu hawezi kuwa tathmini tunahitaji kurudi nyuma na kuchagua kujieleza tofauti kutumia kama\(u\).
- Tathmini muhimu katika suala la\(u\).
- Andika matokeo kwa maneno\(x\) na maneno\(g(x).\)
Matumizi badala ya kupata antiderivative ya\(\displaystyle ∫6x(3x^2+4)^4\,dx.\)
Suluhisho
Hatua ya kwanza ni kuchagua maneno kwa\(u\). Sisi kuchagua\(u=3x^2+4\) kwa sababu basi\(du=6x\,dx\) na sisi tayari tuna\(du\) katika integrand. Andika muhimu katika suala la\(u\):
\[ ∫6x(3x^2+4)^4\,dx=∫u^4\,du. \nonumber \]
Kumbuka kwamba\(du\) ni derivative ya kujieleza waliochaguliwa kwa\(u\), bila kujali ni nini ndani ya integrand. Sasa tunaweza kutathmini muhimu kwa heshima na\(u\):
\[ ∫u^4\,du=\dfrac{u^5}{5}+C=\dfrac{(3x^2+4)^5}{5}+C.\nonumber \]
Uchambuzi
Tunaweza kuangalia jibu letu kwa kuchukua derivative ya matokeo ya ushirikiano. Tunapaswa kupata integrand. Kuchukua thamani kwa\(C\) ya\(1\),\(y=\dfrac{1}{5}(3x^2+4)^5+1.\) sisi basi Tuna
\[ y=\dfrac{1}{5}(3x^2+4)^5+1,\nonumber \]
kwa hivyo
\[ \begin{align*} y′ &=\left(\dfrac{1}{5}\right)5(3x^2+4)^46x \\[4pt] &=6x(3x^2+4)^4.\end{align*}\]
Hii ni hasa kujieleza tulianza na ndani ya integrand.
Matumizi badala ya kupata antiderivative ya\(\displaystyle ∫3x^2(x^3−3)^2\,dx.\)
- Kidokezo
-
Hebu\(u=x^3−3.\)
- Jibu
-
\(\displaystyle ∫3x^2(x^3−3)^2\,dx=\dfrac{1}{3}(x^3−3)^3+C \)
Wakati mwingine tunahitaji kurekebisha constants katika muhimu yetu kama hawana mechi up hasa na maneno sisi ni substituting.
Matumizi badala ya kupata antiderivative ya\[ ∫z\sqrt{z^2−5}\,dz. \nonumber \]
Suluhisho
Andika upya muhimu kama\(\displaystyle ∫z(z^2−5)^{1/2}\,dz.\) Hebu\(u=z^2−5\) na\(du=2z\,dz.\) Sasa tuna tatizo kwa sababu\(du=2z\,dz\) na kujieleza awali ina tu\(z\,dz.\) Tuna kubadilisha kujieleza yetu kwa\(du\) au muhimu katika\(u\) itakuwa mara mbili kubwa kama ni lazima kuwa. Kama sisi kuzidisha pande zote mbili za\(du\) equation\(\dfrac{1}{2}\) na. tunaweza kutatua tatizo hili. Hivyo,
\[ u=z^2−5\nonumber \]
\[ du=2z\,dz \nonumber \]
\[ \dfrac{1}{2}du=\dfrac{1}{2}(2z)\,dz=z\,dz. \nonumber \]
Andika muhimu katika suala la\(u\), lakini kuvuta\(\dfrac{1}{2}\) nje ishara ya ushirikiano:
\[ ∫z(z^2−5)^{1/2}\,dz=\dfrac{1}{2}∫u^{1/2}\,du.\nonumber \]
Unganisha maneno katika\(u\):
\[ \begin{align*} \dfrac{1}{2}∫u^{1/2}\,du &= \left(\dfrac{1}{2}\right)\dfrac{u^{3/2}}{\dfrac{3}{2}}+C \\[4pt] &= \left(\dfrac{1}{2}\right)\left(\dfrac{2}{3}\right)u^{3/2}+C \\[4pt] &=\dfrac{1}{3}u^{3/2}+C \\[4pt] &=\dfrac{1}{3}(z^2−5)^{3/2}+C \end{align*}\]
Matumizi badala ya kupata antiderivative ya\(\displaystyle ∫x^2(x^3+5)^9\,dx.\)
- Kidokezo
-
Kuzidisha du equation na\(\dfrac{1}{3}\).
- Jibu
-
\(\displaystyle ∫x^2(x^3+5)^9\,dx = \dfrac{(x^3+5)^{10}}{30}+C \)
Matumizi badala ya kutathmini muhimu\(\displaystyle ∫\dfrac{\sin t}{\cos^3t}\,dt.\)
Suluhisho
Tunajua derivative ya\(\cos t\) ni\(−\sin t\), hivyo sisi kuweka\(u=\cos t\). Kisha\(du=−\sin t\,dt.\)
Kubadilisha katika muhimu, tuna
\[ ∫\dfrac{\sin t}{\cos^3t}\,dt=−∫\dfrac{du}{u^3}.\nonumber \]
Kutathmini muhimu, tunapata
\[ −∫\dfrac{du}{u^3}=−∫u^{−3}\,du=−\left(−\dfrac{1}{2}\right)u^{−2}+C.\nonumber \]
Kuweka jibu nyuma katika suala la t, tunapata
\[ ∫\dfrac{\sin t}{\cos^3t}\,dt=\dfrac{1}{2u^2}+C=\dfrac{1}{2\cos^2t}+C.\nonumber \]
Matumizi badala ya kutathmini muhimu\( \displaystyle ∫\dfrac{\cos t}{\sin^2t}\,dt.\)
- Kidokezo
-
Tumia mchakato kutoka Mfano\(\PageIndex{3}\) ili kutatua tatizo.
- Jibu
-
\(\displaystyle ∫\dfrac{\cos t}{\sin^2t}\,dt = −\dfrac{1}{\sin t}+C\)
Matumizi badala ya kutathmini muhimu kwa muda usiojulikana\(\displaystyle ∫\cos^3t\sin t\,dt. \)
- Kidokezo
-
Tumia mchakato kutoka Mfano\(\PageIndex{3}\) ili kutatua tatizo.
- Jibu
-
\(\displaystyle ∫\cos^3t\sin t\,dt = −\dfrac{\cos^4t}{4}+C \)
Wakati mwingine tunahitaji kuendesha muhimu kwa njia ambazo ni ngumu zaidi kuliko kuzidisha au kugawa kwa mara kwa mara. Tunahitaji kuondoa maneno yote ndani ya integrand ambayo ni katika suala la kutofautiana awali. Wakati sisi ni kosa,\(u\) lazima variable tu katika integrand. Katika hali nyingine, hii inamaanisha kutatua kwa kutofautiana kwa asili kwa suala la\(u\). Mbinu hii inapaswa kuwa wazi katika mfano unaofuata.
Matumizi badala ya kupata antiderivative ya\[ ∫\dfrac{x}{\sqrt{x−1}}\,dx. \nonumber \]
Suluhisho
Kama sisi basi\(u=x−1,\) basi\(du=dx\). Lakini hii haina akaunti kwa ajili ya\(x\) katika nambari ya integrand. Tunahitaji kueleza\(x\) katika suala la\(u.\) Kama\(u=x−1\), basi\(x=u+1.\) Sasa tunaweza kuandika upya muhimu katika suala la\(u:\)
\[ ∫\dfrac{x}{\sqrt{x−1}}\,dx=∫\dfrac{u+1}{\sqrt{u}}\,du=∫\left(\sqrt{u}+\dfrac{1}{\sqrt{u}}\right)\,du=∫\left(u^{1/2}+u^{−1/2}\right)\,du.\nonumber \]
Kisha sisi kuunganisha kwa njia ya kawaida, kuchukua nafasi\(u\) na kujieleza awali, na sababu na kurahisisha matokeo. Hivyo,
\[ \begin{align*} ∫(u^{1/2}+u^{−1/2})\,du &=\dfrac{2}{3}u^{3/2}+2u^{1/2}+C \\[4pt] &= \dfrac{2}{3}(x−1)^{3/2}+2(x−1)^{1/2}+C \\[4pt] &= (x−1)^{1/2}\left[\dfrac{2}{3}(x−1)+2\right]+C \\[4pt] &= (x−1)^{1/2}\left(\dfrac{2}{3}x−\dfrac{2}{3}+\dfrac{6}{3}\right) \\[4pt] &= (x−1)^{1/2}\left(\dfrac{2}{3}x+\dfrac{4}{3}\right) \\[4pt] &= \dfrac{2}{3}(x−1)^{1/2}(x+2)+C. \end{align*}\]
Badala ya Integrals ya uhakika
Badala inaweza kutumika kwa integrals uhakika, pia. Hata hivyo, kutumia mbadala ya kutathmini muhimu ya uhakika inahitaji mabadiliko kwa mipaka ya ushirikiano. Kama sisi kubadilisha vigezo katika integrand, mipaka ya mabadiliko ya ushirikiano pia.
Hebu\(u=g(x)\) na basi\(g'\) kuwa kuendelea juu ya muda\([a,b]\), na basi\(f\) kuendelea juu ya aina mbalimbali ya\(u=g(x).\) Kisha,
\[∫^b_af(g(x))g′(x)\,dx=∫^{g(b)}_{g(a)}f(u)\,du. \nonumber \]
Ingawa hatuwezi kuthibitisha theorem hii rasmi, tunaihalalisha kwa mahesabu fulani hapa. Kutoka utawala badala kwa integrals muda usiojulikana, kama\(F(x)\) ni antiderivative ya\(f(x),\) tuna
\[ ∫f(g(x))g′(x)\,dx=F(g(x))+C. \nonumber \]
Kisha
\[\begin{align*} ∫^b_af[g(x)]g′(x)\,dx &= F(g(x))\bigg|^{x=b}_{x=a} \\[4pt] &=F(g(b))−F(g(a)) \\[4pt] &= F(u) \bigg|^{u=g(b)}_{u=g(a)} \\[4pt] &=∫^{g(b)}_{g(a)}f(u)\,du \end{align*}\]
na tuna matokeo ya taka.
Tumia mbadala ili kutathmini\[ ∫^1_0x^2(1+2x^3)^5\,dx. \nonumber \]
Suluhisho
Hebu\(u=1+2x^3\), hivyo\(du=6x^2\,dx\). Kwa kuwa kazi ya awali ni pamoja na sababu moja ya\(x^2\) na\(du=6x^2\,dx\), kuzidisha pande zote mbili za\(du\) equation na\(1/6.\) Kisha,
\[ \begin{align*} du &=6x^2\,dx \\[4pt] \text{becomes}\quad \dfrac{1}{6}du &=x^2\,dx. \end{align*}\]
Kurekebisha mipaka ya ushirikiano, kumbuka kuwa wakati\(x=0,u=1+2(0)=1,\) na wakati\(x=1,\;u=1+2(1)=3.\)
Kisha
\[ ∫^1_0x^2(1+2x^3)^5dx=\dfrac{1}{6}∫^3_1u^5\,du. \nonumber \]
Kutathmini maneno haya, tunapata
\[ \begin{align*} \dfrac{1}{6}∫^3_1u^5\,du &=\left(\dfrac{1}{6}\right)\left(\dfrac{u^6}{6}\right)\Big|^3_1 \\[4pt] &=\dfrac{1}{36}\big[(3)^6−(1)^6\big] \\[4pt] &=\dfrac{182}{9}. \end{align*}\]
Tumia mbadala ili kutathmini muhimu ya uhakika\(\displaystyle ∫^0_{−1}y(2y^2−3)^5\,dy. \)
- Kidokezo
-
Tumia hatua kutoka Mfano\(\PageIndex{5}\) ili kutatua tatizo.
- Jibu
-
\(\displaystyle ∫^0_{−1}y(2y^2−3)^5\,dy = \dfrac{91}{3}\)
Tumia mbadala ili kutathmini\(\displaystyle ∫^1_0x^2 \cos \left(\dfrac{π}{2}x^3\right)\,dx. \)
- Kidokezo
-
Tumia mchakato kutoka Mfano\(\PageIndex{5}\) ili kutatua tatizo.
- Jibu
-
\(\displaystyle ∫^1_0x^2 \cos \left(\dfrac{π}{2}x^3\right)\,dx = \dfrac{2}{3π}≈0.2122\)
Tumia mbadala ili kutathmini\[ ∫^1_0xe^{4x^2+3}\,dx. \nonumber \]
Suluhisho
Basi\(u=4x^3+3.\) basi,\(du=8x\,dx.\) Kurekebisha mipaka ya ushirikiano, tunaona kwamba wakati\(x=0,\,u=3\), na wakati\(x=1,\,u=7\). Hivyo badala yetu inatoa
\[\begin{align*} ∫^1_0xe^{4x^2+3}\,dx &= \dfrac{1}{8}∫^7_3e^u\,du \\[4pt] &=\dfrac{1}{8}e^u\Big|^7_3 \\[4pt] &=\dfrac{e^7−e^3}{8} \\[4pt] &≈134.568 \end{align*}\]
Kubadilishwa inaweza kuwa moja tu ya mbinu zinazohitajika kutathmini muhimu. Mali yote na sheria za ushirikiano zinatumika kwa kujitegemea, na kazi za trigonometric zinaweza kuhitajika kuandikwa upya kwa kutumia utambulisho wa trigonometric kabla hatuwezi kuomba badala. Pia, tuna fursa ya kuchukua nafasi ya kujieleza awali kwa\(u\) baada ya kupata antiderivative, ambayo ina maana kwamba hatuna mabadiliko ya mipaka ya ushirikiano. Mbinu hizi mbili zinaonyeshwa katika Mfano\(\PageIndex{7}\).
Tumia mbadala ili kutathmini\[∫^{π/2}_0\cos^2θ\,dθ. \nonumber \]
Suluhisho
Hebu kwanza tutumie utambulisho wa trigonometric ili uandike upya muhimu. utambulisho trig\(\cos^2θ=\dfrac{1+\cos 2θ}{2}\) inaruhusu sisi kuandika upya muhimu kama
\[∫^{π/2}_0\cos^2θ\,dθ=∫^{π/2}_0\dfrac{1+\cos2θ}{2}\,dθ. \nonumber \]
Kisha,
\[ \begin{align*} ∫^{π/2}_0\left(\dfrac{1+\cos2θ}{2}\right)\,dθ &=∫^{π/2}_0\left(\dfrac{1}{2}+\dfrac{1}{2}\cos 2θ\right)\,dθ \\[4pt] &=\dfrac{1}{2}∫^{π/2}_0\,dθ+∫^{π/2}_0\cos2θ\,dθ. \end{align*}\]
Tunaweza kutathmini muhimu ya kwanza kama ilivyo, lakini tunahitaji kufanya badala ya kutathmini muhimu ya pili. Basi\(u=2θ.\) basi,\(du=2\,dθ,\) au\(\dfrac{1}{2}\,du=dθ\). Pia, wakati\(θ=0,\,u=0,\) na wakati\(θ=π/2,\,u=π.\) Expressing muhimu ya pili katika suala la\(u\), tuna
\[ \begin{align*}\dfrac{1}{2}∫^{π/2}_0\,dθ+\dfrac{1}{2}∫^{π/2}_0 \cos 2θ\,dθ &=\dfrac{1}{2}∫^{π/2}_0\,dθ+\dfrac{1}{2}\left(\dfrac{1}{2}\right)∫^π_0 \cos u \,du \\[4pt] &=\dfrac{θ}{2}\,\bigg|^{θ=π/2}_{θ=0}+\dfrac{1}{4}\sin u\,\bigg|^{u=θ}_{u=0} \\[4pt] &=\left(\dfrac{π}{4}−0\right)+(0−0)=\dfrac{π}{4} \end{align*}\]
Dhana muhimu
- Kubadilishwa ni mbinu ambayo inaeleza ushirikiano wa kazi ambazo ni matokeo ya derivative ya utawala wa mnyororo. Neno 'badala' linamaanisha kubadilisha vigezo au kubadilisha variable\(u\) na\(du\) kwa maneno sahihi katika integrand.
- Wakati wa kutumia badala ya muhimu ya uhakika, tunapaswa pia kubadili mipaka ya ushirikiano.
Mlinganyo muhimu
- Kubadilishwa na Integrals isiyojulikana\[∫f[g(x)]g′(x)\,dx=∫f(u)\,du=F(u)+C=F(g(x))+C \nonumber \]
- Kubadilisha na Integrals ya uhakika\[∫^b_af(g(x))g'(x)\,dx=∫^{g(b)}_{g(a)}f(u)\,du \nonumber \]
faharasa
- mabadiliko ya vigezo
- badala ya kutofautiana, kama vile\(u\), kwa kujieleza katika integrand
- ushirikiano na badala
- mbinu ya ushirikiano ambayo inaruhusu ushirikiano wa kazi ambazo ni matokeo ya derivative ya utawala wa mnyororo