Misalkan A(t) menyatakan luas daerah di bawah kurva

Misalkan A(t) menyatakan luas daerah di bawah kurva y = bx2, 0 ≤ x ≤ t. Jika titik P(xo, 0) sehingga A(xo) ∶ A(1) = 1 ∶ 8 maka perbandingan luas trapesium ABPQ : DCPQ

A(t)_menyatakan_luas_dibawah_kurva

A
2 : 1
B
3 : 1
C
6 : 1
D
8 : 1
E
9 : 1

B. 3:1

A(t) menyatakan luas daerah di bawah kurva y = bx2, 0 ≤ x ≤ t, maka:
$$\begin{aligned} A(t)&=\int_{o}^{t} bx^{2} dx\\A(x_{o})&= \int_{o}^{x_{o}}bx^{2} dx\\A(1)&=\int_{o}^{1} bx^{2} dx\\\end{aligned}$$

Selanjutnya, menentukan nilai xo :
$$\begin{aligned} A(x_{o}): A(1) &= 1:8\\\frac{ A(x_{o})}{A(1)} &=\frac{1}{8}\\\frac{\int_{o}^{x_{o}}bx^{2} dx}{\int_{o}^{x_{1}}bx^{2} dx} &= \frac{1}{8}\\\frac{\frac{1}{3}.b.x_{o}^{3}}{\frac{1}{3}.b.1^{3}} &= \frac{1}{8}\\x_{o}^{3} &= \frac{1}{8}\\x_{o} &= \frac{1}{2}\end{aligned}$$

Perbandingan trapesium ABPQ dan DCPQ adalah:
trapesium_ABPQ_DCPQ
$$\begin{aligned} \frac{L_{ABPQ}}{L_{DCPQ}} &= \frac{\frac{1}{2}\times (AB+PQ)\times PB}{\frac{1}{2}\times (CD+PQ)\times PC} \:\:\:\: \text{{\color{Red} dimana AB=CD}}\\&= \frac{\frac{1}{2}\times (CD+PQ)\times PB}{\frac{1}{2}\times (CD+PQ)\times PC}\\&= \frac{PB}{PC}\\&= \frac{\frac{3}{2}}{\frac{1}{2}}\\&= \frac{3}{1}\end{aligned}$$

Jadi, perbandingan luas trapesium ABPQ dan DCPQ adalah 3:1

Source:
SBMPTN 2014 | SAINTEK | MATEMATIKA | Kode 512 | 14
Topik: Integral


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