Materi: Matematika (Integral)
A. Sifat dan Rumus Integral
Sifat integral
\(\displaystyle \int f(ax)\,dx=a\int f(x)\,dx + c\) (bila sesuai), \(c\) konstanta.
\(\displaystyle \int (f(x)\pm g(x))\,dx=\int f(x)\,dx \pm \int g(x)\,dx\).
Jika \(f(x),g(x)\) kontinu pada \([a,b]\):
\(\displaystyle \int_a^a f(x)\,dx=0\)
\(\displaystyle \int_a^b f(x)\,dx=-\int_b^a f(x)\,dx\);
\(\displaystyle \int_a^b f(x)\,dx=\int_a^p f(x)\,dx+\int_p^b f(x)\,dx\) untuk \(a\le p\le b\).
Rumus dasar
\(\displaystyle \int x^n dx=\frac{x^{n+1}}{n+1}+c\) (\(n\ne -1\))
\(\displaystyle \int \frac{1}{x}\,dx=\ln|x|+c\)
\(\displaystyle \int a^x dx=\frac{a^x}{\ln a}+c\),
\(\displaystyle \int e^x dx=e^x+c\)
\( \int \frac{P(x)}{Q(x)} \, dx \quad \text{dengan } \deg(P(x)) \geq \deg(Q(x)) \)
\( \text{Lakukan pembagian aljabar:} \quad \frac{P(x)}{Q(x)} = H(x) + \frac{R(x)}{Q(x)} \)
\( \text{dengan:} \quad \begin{cases} H(x) & \text{hasil bagi (fungsi polinomial)} \\ R(x) & \text{sisa, dengan } \deg(R(x)) < \deg(Q(x)) \end{cases} \)
\( \text{Maka integral menjadi:} \quad \int \frac{P(x)}{Q(x)} \, dx = \int H(x) \, dx + \int \frac{R(x)}{Q(x)} \, dx \)
\( \text{Bagian pertama: } \int H(x) \, dx \quad \text{(langsung diintegralkan sebagai polinomial)} \)
\( \text{Bagian kedua: } \int \frac{R(x)}{Q(x)} \, dx \quad \text{(diselesaikan dengan pecahan parsial atau substitusi)} \)
CONTOH SOAL \[ f(x) = \frac{x^3 + 2x^2 + 3x + 4}{x + 1} \]
\[ \textbf{Langkah 1: Turunan Fungsi Pecahan} \] Gunakan aturan turunan pembagian: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
Dengan: \(u(x) = x^3 + 2x^2 + 3x + 4, \quad v(x) = x + 1\)
\(u'(x) = 3x^2 + 4x + 3, \quad v'(x) = 1\)
Maka: \[ f'(x) = \frac{(3x^2 + 4x + 3)(x + 1) - (x^3 + 2x^2 + 3x + 4)}{(x + 1)^2} \]
Hitung pembilang: \[ (3x^2 + 4x + 3)(x + 1) = 3x^3 + 7x^2 + 7x + 3 \]
\[ (3x^3 + 7x^2 + 7x + 3) - (x^3 + 2x^2 + 3x + 4) = 2x^3 + 5x^2 + 4x - 1 \]
Jadi: \[ f'(x) = \frac{2x^3 + 5x^2 + 4x - 1}{(x + 1)^2} \]
\[ \textbf{Langkah 2: Integral dari Hasil Turunan} \]
\[ \int f'(x)\, dx = \int \frac{2x^3 + 5x^2 + 4x - 1}{(x + 1)^2} \, dx \]
Lakukan dekomposisi pecahan parsial: \[ \frac{2x^3 + 5x^2 + 4x - 1}{(x + 1)^2} = 2x + 1 - \frac{2}{(x + 1)^2} \]
Integralkan satu per satu: \[ \int \left(2x + 1 - \frac{2}{(x + 1)^2} \right) dx = \int 2x \, dx + \int 1 \, dx - \int \frac{2}{(x + 1)^2} dx \]
*********************************************** \[ \text{Diketahui:} \quad f(x) = \frac{-2}{(x + 1)^2} \]
\[ \text{Kita gunakan rumus:} \quad \int \frac{1}{(x + a)^n} \, dx = \frac{-(x + a)^{1 - n}}{n - 1} + C, \quad n \ne 1 \]
\[ \text{Dengan } a = 1, \; n = 2 \Rightarrow \int \frac{-2}{(x + 1)^2} \, dx = -2 \cdot \int \frac{1}{(x + 1)^2} \, dx = -2 \cdot \left( \frac{-1}{x + 1} \right) + C \]
\[ \text{Hasil akhir:} \quad \int \frac{-2}{(x + 1)^2} \, dx = \frac{2}{x + 1} + C \]
*********************************************** \[ = x^2 + x + \frac{2}{x + 1} + C \]
\( \textbf{Langkah 3: Kembalikan ke Fungsi Asal} \)
\[ x^2 + x + \frac{2}{x + 1} = \frac{x^3 + 2x^2 + 3x + 4}{x + 1} = f(x) \]
\( \text{Sehingga:} \quad \int f'(x)\, dx = f(x) + C \)
Rumus pengembangan
\(\displaystyle \int \sqrt{a^2-u^2}\,du=\tfrac12\!\left(u\sqrt{a^2-u^2}+a^2\arcsin\!\frac{u}{a}\right)+c\)
\(\displaystyle \int \sqrt{u^2+a^2}\,du=\tfrac12\!\left(u\sqrt{u^2+a^2}+a^2\ln|u+\sqrt{u^2+a^2}|\right)+c\)
\(\displaystyle \int \sqrt{u^2-a^2}\,du=\tfrac12\!\left(u\sqrt{u^2-a^2}-a^2\ln|u+\sqrt{u^2-a^2}|\right)+c\)
\(\displaystyle \int \frac{du}{a^2-u^2}=\frac{1}{2a}\ln\left|\frac{a+u}{a-u}\right|+c\)
\(\displaystyle \int \frac{du}{a^2+u^2}=\frac{1}{a}\arctan\!\frac{u}{a}+c\)
\(\displaystyle \int \frac{du}{\sqrt{a^2+u^2}}=\ln|u+\sqrt{u^2+a^2}|+c\)
\(\displaystyle \int \frac{du}{\sqrt{a^2-u^2}}=\arcsin\!\frac{u}{a}+c\)
\(\displaystyle \int \frac{du}{u^2-a^2}=\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|+c\)
\(\displaystyle \int \frac{du}{u\sqrt{u^2-a^2}}=\frac{1}{a}\operatorname{arcsec}\!\left|\frac{u}{a}\right|+c\)
Rumus integral trigonometri
\(\displaystyle \int \sin x\,dx=-\cos x+c\), \(\displaystyle \int \cos x\,dx=\sin x+c\)
\(\displaystyle \int \sin(ax+b)\,dx=-\frac{1}{a}\cos(ax+b)+c\)
\(\displaystyle \int \cos(ax+b)\,dx=\frac{1}{a}\sin(ax+b)+c\)
\(\displaystyle \int e^{ax}\sin bx\,dx=\frac{e^{ax}}{a^2+b^2}(a\sin bx-\!b\cos bx)+c\)
\(\displaystyle \int e^{ax}\cos bx\,dx=\frac{e^{ax}}{a^2+b^2}(a\cos bx+\!b\sin bx)+c\)
\(\displaystyle \int \sin^n x\cos x\,dx=\frac{1}{n+1}\sin^{\,n+1}x+c\)
\(\displaystyle \int \cos^n x\sin x\,dx=-\frac{1}{n+1}\cos^{\,n+1}x+c\)
B. Metode Substitusi
Bentuk: \(\displaystyle \int f(a\,b^x)\,dx\) atau \(\int f(g(x))g'(x)\,dx\).
Langkah: pilih substitusi \(u=g(x)\), ubah \(dx\) menjadi \(du\), integralkan dalam \(u\), lalu kembalikan ke \(x\).
C. Integral Parsial (By Parts)
Bentuk: \(\displaystyle \int u\,dv\). Rumus: \(\displaystyle \int u\,dv = u\,v - \int v\,du\).
Pilih \(u\) (mudah diturunkan) dan \(dv\) (mudah diintegralkan).
D. Integral Trigonometri (ringkas)
Tabel turunan dasar
\( (\sin x)'=\cos x\),
\( (\cos x)'=-\sin x\),
\( (\tan x)'=\sec^2 x\),
\( (\cot x)'=-\csc^2 x\),
\( (\sec x)'=\sec x\tan x\),
\( (\csc x)'=-\csc x\cot x\).
Integralnya
\(\displaystyle \int \sec^2 x\,dx=\tan x+c\),
\(\displaystyle \int \csc^2 x\,dx=-\cot x+c\),
\(\displaystyle \int \sec x\tan x\,dx=\sec x+c\),
\(\displaystyle \int \csc x\cot x\,dx=-\csc x+c\).
Rumus-rumus umum
\(\displaystyle \int \sin(ax+b)\,dx=-\frac{1}{a}\cos(ax+b)+c\),
\(\displaystyle \int \cos(ax+b)\,dx=\frac{1}{a}\sin(ax+b)+c\),
\(\displaystyle \int \sin^n x\cos^m x\,dx\) gunakan identitas trig sesuai paritas \(n,m\).
Identitas sering dipakai
\(2\sin A\cos B=\sin(A+B)+\sin(A-B)\)
\(2\cos A\cos B=\cos(A+B)+\cos(A-B)\)
\(2\cos A\sin B=\sin(A+B)-\sin(A-B)\)
\(2\sin A\sin B=\cos(A-B)-\cos(A+B)\)
Luas Daerah
