Mengintegralkan Menggunakan Integrasi Parsial
Diperkenalkan| English | Makna |
|---|---|
| Integration by parts/ˌɪntɪˈɡreɪʃn baɪ pɑːts/ | integrasi parsial |
The product rule, run backward
- The chain rule reversed gave u-substitution. What reverses the product rule? Integration by parts 分部积分法.
- It handles integrals of a product that substitution can't crack — like $\int x e^x\,dx$.
- The trick: trade a hard integral for an easier one by moving the derivative from one factor to the other.
- It's the BC workhorse for products of unlike functions.
Integrasi parsial merupakan kebalikan dari aturan mana?
Kesulitan 1/5 · 100 koin + bonus kecil pada penyelesaian pertama hari ini
Ini membatalkan aturan perkalian.
The formula
- From the product rule, integrated:
-
$$\int u\,dv = uv - \int v\,du$$
- You pick part of the integrand to be $u$ (which you'll differentiate) and the rest to be $dv$ (which you'll integrate).
- Then $du=u'\,dx$ and $v=\int dv$, and you assemble $uv-\int v\,du$.
Sebuah perkalian untuk diintegrasikan
y = a·e^{bx}
Integrasi parsial menukar $\int x e^x$ dengan $\int e^x$ yang lebih mudah — memindahkan turunan dari faktor aljabar.
Rumus integrasi parsial adalah $\int u\,dv=$
Kesulitan 1/5 · 100 koin + bonus kecil pada penyelesaian pertama hari ini
$\int u\,dv=uv-\int v\,du$ (perhatikan tanda minus).
Choosing $u$ and $dv$ wisely
- Aim to make the new integral $\int v\,du$ simpler than the original.
- A helpful priority for $u$ (LIATE): Logarithm, Inverse trig, Algebraic, Trig, Exponential — pick $u$ from earliest on this list.
- So in $\int x e^x\,dx$: $u=x$ (algebraic) and $dv=e^x\,dx$.
- A good choice collapses the problem; a bad choice makes it worse.
Untuk $\int x e^x\,dx$, pilihan yang tepat adalah $u=$
Kesulitan 1/5 · 100 koin + bonus kecil pada penyelesaian pertama hari ini
Menurut LIATE, aljabar $x$ adalah $u$; $dv=e^x\,dx$.
Mnemoteknik umum untuk memilih $u$ adalah ____ (L-I-A-T-E).
Kesulitan 1/5 · 100 koin + bonus kecil pada penyelesaian pertama hari ini
Logaritma, Trigonometri Invers, Aljabar, Trigonometri, Eksponensial.
Anda harus memilih $u$ dan $dv$ sehingga integral baru $\int v\,du$ menjadi lebih sederhana daripada yang asli.
Kesulitan 1/5 · 100 koin + bonus kecil pada penyelesaian pertama hari ini
Itulah tujuan utama pemilihan tersebut.
Sometimes repeat, sometimes loop
- If $\int v\,du$ is still a product, apply integration by parts again.
- For $\int x^2 e^x\,dx$ you'd use it twice, peeling one power of $x$ each time.
- A few integrals (like $\int e^x\sin x\,dx$) loop back to the original — solve algebraically for it.
- Keep $u,du,v,dv$ neatly labeled to avoid sign errors.
$\displaystyle\int x e^x\,dx=$
Kesulitan 1/5 · 100 koin + bonus kecil pada penyelesaian pertama hari ini
$xe^x-\int e^x\,dx=xe^x-e^x+C$.
Choose $u$ and $dv$ so that $\int v\,du$ is simpler — the wrong pick can make it harder or send you in circles. Don't forget the minus sign: the formula is $uv-\int v\,du$. And $v$ is any antiderivative of $dv$ (drop its $+C$ until the very end of a definite/indefinite result).
Evaluate $\displaystyle\int x e^x\,dx$.
- Let $u=x$, $dv=e^x\,dx$. Then $du=dx$, $v=e^x$.
- $\displaystyle\int x e^x\,dx = uv-\int v\,du = x e^x - \int e^x\,dx$.
- $= x e^x - e^x + C = e^x(x-1)+C$.
Integration by parts reverses the product rule: $\int u\,dv=uv-\int v\,du$. Choose $u$ (to differentiate) and $dv$ (to integrate) — via LIATE — so that $\int v\,du$ is simpler. Mind the minus sign, and apply it repeatedly (or solve the loop) when needed.