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Integral Arsip | Catatan Nurkholis

Problem – Integral Khusus

Problem: Misalkan $f$ dan $g$ adalah fungsi kontinyu pada $[0,a]$, dan misalkan pula $f(x)=f(a-x)$ dan $g(x)+g(a-x)=k$ untuk semua $x$ pada $[0,a]$. Buktikan bahwa $$int_0^a f(x)g(x)dx=frac{1}{2}kint_0^a f(x)dx$$ Solusi:Sebut saja $$I=int_0^a f(x)g(x)dx$$ Dengan sedikit modifikasi $y=a-x$, didapat $dy=-dx$. dan $x=a-y$ dan$$ begin{aligned} I&=int_0^a f(x)g(x)dx\ &=int_{a-0}^{a-a} f(a-y)g(a-y)dx\ &=int_a^0 f(a-y)g(a-y)(-dy)\ &=int_0^a f(a-y)g(a-y)dy\ &=int_0^a f(a-x)g(a-x)dx\ end{aligned}$$ Sementara itu $f(a-x)=f(x)$ dan $g(a-x)=k-g(x)$. …

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Integral Tak Tentu ~ Matematika SMA

Pada materi Turunan/diferensial, kita telah mengetahui bahwa:$$begin{aligned}f(x)=x^2+5xquad&Rightarrowquad f'(x)=2x+5\f(x)=x^2+5x+2quad&Rightarrowquad f'(x)=2x+5\f(x)=x^2+5x+9quad&Rightarrowquad f'(x)=2x+5\f(x)=x^2+5x+Cquad&Rightarrowquad f'(x)=2x+5end{aligned}$$Berapapun nilai $C$ yang kita berikan pada suatu fungsi $f(x)$ akan tetap menghasilkan turunan/diferensial yang sama. Sehingga anti-turunan dari fungsi $f'(x)=2x+5$ dapat dinyatakan dalam bentuk:$$f(x)=int f'(x)dx=int 2x+5dx = x^2+5x+C$$ Notasi Secara umum dapat kita tuliskan$$int x^n dx = frac{x^{n+1}}{n+1}+C,quad nneq-1$$ Sifat Sifat-sifat Integral tak tentu: $int …

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