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SMK Teknologi dan Rekayasa

Soal 126 Luas Daerah antara Dua Kurva


Luas\ daerah\ yang\ dibatasi\ oleh\ kurva\ y=3\ dan\ y=4-x^{2}\ adalah\ ....
a.\ \ \frac{8}{6}\ \ \ \ \ \ \ \ \ \ b.\ \ \frac{17}{6}\ \ \ \ \ \ \ \ \ \ c.\ \ \frac{19}{6}\ \ \ \ \ \ \ \ \ \ d.\ \ \frac{25}{6}\ \ \ \ \ \ \ \ \ \ e.\ \ \frac{33}{8}
Jawaban: a

y_{1}=4-x^{2}\\Persamaan\ fungsi\ kuadrat,\ grafiknya\ berupa\ parabola\ terbuka\ ke\ bawah\\(karena\ koefisien\ x^{2}\ bertanda\ negatif)
y_{2}=3\\Persamaan\ fungsi\ konstan,\ grafiknya\ berupa\ garis\ lurus\ mendatar\\(sejajar\ dengan\ sumbu\ X)

Menentukan\ batas\ pengintegralan:
\begin{array}{rcl}y_{1}&=&y_{2}\\4-x^{2}&=&3\\x^{2}&=&1\\x=1&atau&x=-1\end{array}

\begin{array}{rcl}Luas\ daerah&=&\int_{-1}^{1}(y_{1}-y_{2})\ dx\\\\&=&\int_{-1}^{1}(4-x^{2}-3)\ dx\\\\&=&\int_{-1}^{1}(1-x^{2})\ dx\\\\&=&\left [ x-\frac{x^{2+1}}{2+1} \right ]_{-1}^{1}\\\\&=&\left [ x-\frac{x^{3}}{3} \right ]_{-1}^{1}\\\\&=&\left [ (1-\frac{1^{3}}{3})-(-1-\frac{(-1)^{3}}{3}) \right ]\\\\&=&\left [ (1-\frac{1}{3})-(-1-\frac{-1}{3}) \right ]\\\\&=&\left [ \frac{2}{3}-(-1+\frac{1}{3}) \right ]\\\\&=&\left [ \frac{2}{3}+1-\frac{1}{3} \right ]\\\\&=&1\frac{1}{3}\\\\&=&\frac{4}{3}\\\\&=&\frac{8}{6}\end{array}

\mathbf{Trik\ jitu}:
y_{1}-y_{2}=4-x^{2}-3=-x^{2}+1\ \ \ \ \to \ \ \ \ a=-1,\ b=0,\ c=1
Diskriminan:\ \ \ D=b^{2}-4ac=0^{2}-4(-1).1=0+4=4
\begin{array}{rcl}Luas\ daerah&=&\frac{D\sqrt{D}}{6a^{2}}\\\\&=&\frac{4\sqrt{4}}{6.(-1)^{2}}\\\\&=&\frac{4\times 2}{6.1}\\\\&=&\frac{8}{6}\end{array}

About Kalakay

Guru Matematika SMK

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5 thoughts on “Soal 126 Luas Daerah antara Dua Kurva

  1. makasih udh menambah wawasan

    Posted by bautinja | 9 Oktober 2011, 8:41 PM
  2. materi ini sangat membantu sAYa untuk menjawab soal2 freetest besok,,,,

    Posted by erwin nasir | 13 Januari 2012, 1:18 AM
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    Posted by Myp2p Basketball | 19 Juli 2013, 11:21 AM

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