MATEMATIKA INDONESIA: Maret 2012

Buktikan bahwa luas daerah yang dibatasi oleh parabola y = ax² + bx + c dengan sumbu x adalah

$L=\frac{D\sqrt{D}}{6a^{2}}$

Jawab :
Jika kita anggap parabola membuka ke atas dan titik potong dengan sumbu x adakah di 2 titik.

Jika parabola kita anggap membuka ke atas sesuai gambar maka
$L=-\int_{x_{1}}^{x_{2}}\left ( ax^{2}+bx+c \right )dx$

$L=-\int_{x_{1}}^{x_{2}}\left a(x-x_{1} \right )\left ( x-x_{2} \right )dx$
untuk lebih mudahnya bisa kita buat pemisalan
y= x - x₁ maka dy = dx atau dx = dy, sehingga
jika x = x₁ maka y = 0
jika x = x₂ maka y = x₂ - x₁
Jadi
$L=-\int_{0}^{x_{2}-x_{1}}ay\left ( y+x_{1}-x_{2} \right )dy$
$L=-\int_{0}^{x_{2}-x_{1}}a\left ( y^{2} +(x_{1}-x_{2})y\right )dy$
$L=-\int_{0}^{x_{2}-x_{1}}a\left ( y^{2} -(x_{2}-x_{1})y\right )dy$
$L=-a\left ( \frac{1}{3}y^{3}-\frac{1}{2}\left ( x_{2}-x_{1} \right )y^{2} \right )\left | \begin{matrix} ^{x_{2}-x_{1}}\\ _{0} \end{matrix} \right$
$L=-a\left [ \frac{1}{3}(x_{2}-x_{1})^{3} -\frac{1}{2}(x_{2}-x_{1})(x_{2}-x_{1})^{2}\right ]$
$L=-a\left [ \frac{1}{3}(x_{2}-x_{1})^{3} -\frac{1}{2}(x_{2}-x_{1})^{3}\right ]$
$L=-a\left [ -\frac{1}{6}(x_{2}-x_{1})^{3} \right ]=\frac{1}{6}a(x_{2}-x_{1})^{3}$
Karena $x_{2}-x_{1}=\frac{\sqrt{D}}{a}$ maka
$L=\frac{1}{6}a\left ( \frac{\sqrt{D}}{a} \right )^{3}=\frac{1}{6}a\frac{D\sqrt{D}}{a^{3}}=\frac{D\sqrt{D}}{6a^{2}}$