Soal-Jawab Turunan dan Integral Fungsi

Soal:
Tentukan y=f(x) yang memenuhi
$ \frac {dy}{dx} = x \sqrt {x^2+9}$
y(-4)= 0

Pembahasan:
 $\frac {dy}{dx} = x \sqrt {x^2+9} \\ dy =    x \sqrt {x^2+9} dx  \\ \int dy = \int   x \sqrt {x^2+9} dx \\ y = \int   x \sqrt {x^2+9} dx $
Misalkan
 $ m= x^2+9 \\ dm = 2x dx \\ dx= \frac {1}{2x} dm$
 $ y =\int   x \sqrt {x^2+9} dx \\ y = \int x  \sqrt m \frac {1}{2x} dm \\ y = \int \frac {1}{2} m^{\frac {1}{2}} dm \\ y = \frac {1}{3} m^{ \frac {3}{2}}+C \\ y = \frac {1}{3} m \sqrt m +C \\ y = \frac {1}{3} (x^2+9) \sqrt {x^2+9} +C$

Sekarang cari nilai C, dari y(-4)=0
$y = \frac {1}{3} (x^2+9) \sqrt {x^2+9}+C \\  0 =  \frac {1}{3} ((-4)^2+9) \sqrt {(-4)^2+9} +C \\ C = \frac {-125}{3}$
Jadi fungsi tersebut,
$y= y = \frac {1}{3} (x^2+9) \sqrt {x^2+9}- \frac {125}{3}$