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Problem – Integral Khusus | 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 $g(a-x)=k-g(x)$. Dari sini bisa kita peroleh bahwa:
$$
begin{aligned}
int_0^a f(x)g(x)dx + int_0^a f(a-x)g(a-x)dx &= int_0^a f(x)g(x)dx + int_0^a f(x)left[k-g(x)right]dx\
I+I&= int_0^a f(x)g(x)dx + int_0^a k f(x)dx -int_0^a f(x)g(x)dx \
2I&= kint_0^a  f(x)dx \
I&=frac{1}{2}kint_0^a f(x)dx
end{aligned}
$$

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