Matematika Ari

Sabtu, 21 September 2013

Integral

Posting pertama kali ini adalah mengenai soal Matematika kelas XII / IPA, yaitu soal Integral Tak Tentu. Ada sepuluh contoh soal Integral Tak Tentu beserta pembahasan dan jawabannya. Selamat berlatih dan semoga bermanfaat.

$\int (2x^2 + 4x - 5) \: \mathrm{d}x = \dots$

Jawaban

$\frac{2}{3} x^3 + 2x^2 - 5x + C$
$\int 5x \sqrt[3]{x^2} \: \mathrm{d}x = \dots$

Jawaban

$\begin{align*} \int 5x \sqrt[3]{x^2} \: \mathrm{d}x &= \int 5x \cdot x^\frac{2}{3} \: \mathrm{d}x \\ &= \int 5x^\frac{5}{3} \: \mathrm{d}x \\ &= 5 \cdot \frac{3}{8} \cdot x^\frac{8}{3} + C \\ &= \frac{15}{8} x^2 \sqrt[3]{x^2} + C \\ \end{align*}$
$\int x(2x-1)^2 \: \mathrm{d}x = \dots$

Jawaban

$\begin{align*} \int x(2x-1)^2 \: \mathrm{d}x &= \int x(4x^2 - 4x + 1) \: \mathrm{d}x \\ &= \int (4x^3 - 4x^2 + x) \: \mathrm{d}x \\ &= x^4 - \frac{4}{3}x^3 + \frac{1}{2}x^2 + C \end{align*}$
$\int \frac{x^3 - 1}{\sqrt{x^3} - \sqrt{x}} \: \mathrm{d}x = \dots$

Jawaban

$\begin{align*} \int \frac{x^3 - 1}{\sqrt{x^3} - \sqrt{x}} \: \mathrm{d}x &= \int \frac{(x-1)(x^2+x+1)}{(x-1)\sqrt{x}} \: \mathrm{d}x \\ &= \int \frac{\cancel{(x-1)}(x^2+x+1)}{\cancel{(x-1)}\sqrt{x}} \: \mathrm{d}x \\ &= \int x^{-\frac{1}{2}}(x^2+x+1) \: \mathrm{d}x \\ &= \int x^\frac{3}{2} + x^\frac{1}{2} + x^{-\frac{1}{2}} \: \mathrm{d}x \\ &= \frac{2}{5}x^\frac{5}{2} + \frac{2}{3}x^\frac{3}{2} + 2x^\frac{1}{2} + C \\ &= \frac{2}{5}x^2\sqrt{x} + \frac{2}{3}x\sqrt{x} + 2\sqrt{x} + C \end{align*}$
Sebuah kurva mempunyai turunan $\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 2x$ . Kurva tersebut melewati titik $(2, 5)$ . Tentukan persamaan kurva tersebut.

Jawaban
- Pertama cari dahulu integral dari turunan
  $\int 3x^2 - 2x \: \mathrm{d}x = x^3 - x^2 + C$
- Selanjutnya cari nilai C dengan memasukkan titik $(2, 5)$ ke persamaan
  $\begin{align*} y &= x^3 - x^2 + C \\ 5 &= 2^3 - 2^2 + C \\ 5 &= 8 - 4 + C \\ 5 &= 4 + C \\ C &= 1 \end{align*}$
  
  Jadi Persamaan kurva tersebut adalah $y = x^3 - x^2 + 1$
$\int \frac{\mathrm{d}x}{4x^3} = \dots$

Jawaban

$\begin{align*} \int \frac{\mathrm{d}x}{4x^3} &= \frac{1}{4} \int x^{-3} \: \mathrm{d}x \\ &= \frac{1}{4} (\frac{x^{-2}}{-2}) + C \\ &= \frac{x^{-2}}{-8} + C \\ &= - \frac{1}{8x^2} + C \end{align*}$
$\int \frac{x^2 - 4x + 3}{x^2 - x} \: \mathrm{d}x = \dots$

Jawaban

$\begin{align*} \int \frac{x^2 - 4x + 3}{x^2 - x} \: \mathrm{d}x &= \int \frac{(x-1)(x-3)}{x(x-1)} \: \mathrm{d}x \\ &= \int \frac{\cancel{(x-1)}(x-3)}{x\cancel{(x-1)}} \: \mathrm{d}x \\ &= \int \frac{x-3}{x} \: \mathrm{d}x \\ &= \int 1 - \frac{3}{x} \: \mathrm{d}x \\ &= \int 1 \: \mathrm{d}x - \int \frac{3}{x} \: \mathrm{d}x \\ &= x - 3 \ln{|x|} + C \end{align*}$
$\int (a^\frac{1}{3} - x^\frac{1}{3})^3 \: \mathrm{d}x = \dots$

Tutup Jawaban

Ingat bahwa : $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$

$\begin{align*} \int (a^\frac{1}{3} - x^\frac{1}{3})^3 \: \mathrm{d}x &= \int (a^\frac{1}{3})^3 - 3(a^\frac{1}{3})^2x + 3a(x^\frac{1}{3})^2 - (x^\frac{1}{3})^3\: \mathrm{d}x \\ &= \int a - 3a^\frac{2}{3}x + 3ax^\frac{2}{3} + x \: \mathrm{d}x \\ &= ax - 3a^\frac{2}{3} \cdot \frac{1}{2}x^2 + 3a \cdot \frac{3}{5} \cdot x^\frac{5}{3} - \frac{1}{2}x^2 + C \\ &= ax - \frac{3}{2}a^\frac{2}{3}x^2 + \frac{9}{5}ax^\frac{5}{3} + C \\ &= ax - \frac{3}{2}\sqrt[3]{a^2}x^2 + \frac{9}{5}ax\sqrt[3]{x^2} + C \end{align*}$
$\int \frac{4x^6 - 3x^5 - 8}{x^7} \: \mathrm{d}x = \dots$

Jawaban

$\begin{align*} \int \frac{4x^6 - 3x^5 - 8}{x^7} \: \mathrm{d}x &= \int \frac{4}{x} - \frac{3}{x^2} - \frac{8}{x^7} \: \mathrm{d}x \\ &= 4 \ln{|x|} - 3 (-1) (x^{-1}) - 8 (-\frac{1}{6})(x^{-6}) + C \\ &= 4 \ln{|x|} + \frac{3}{x} + \frac{8}{6x^6} + C \\ \end{align*}$
$\int \frac{\sqrt{x^3}-x^3}{\sqrt{x}-x} \: \mathrm{d}x = \dots$

Jawaban

Ingat bahwa : $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$

$\begin{align*} \int \frac{\sqrt{x^3}-x^3}{\sqrt{x}-x} \: \mathrm{d}x &= \int \frac{(x^3)^\frac{1}{2} - x^3}{x^\frac{1}{2} - x} \: \mathrm{d}x \\ &=\int \frac{(x^\frac{1}{2})^3 - x^3}{x^\frac{1}{2} - x} \: \mathrm{d}x \\ &= \int \frac{(x^\frac{1}{2} - x)\left((x^\frac{1}{2})^2 + (x^\frac{1}{2})(x) + (x)^2\right)}{(x^\frac{1}{2} - x)} \: \mathrm{d}x \\ &= \int \frac{\cancel{(x^\frac{1}{2} - x)}\left((x^\frac{1}{2})^2 + (x^\frac{1}{2})(x) + (x)^2\right)}{\cancel{(x^\frac{1}{2} - x)}} \: \mathrm{d}x \\ &= \int x + x^\frac{3}{2} + x^2 \: \mathrm{d}x \\ &= \frac{1}{2} x^2 + \frac{2}{5}x^\frac{5}{2} + \frac{1}{3}x^3 + C \\ &= \frac{1}{2} x^2 + \frac{2}{5}x^2\sqrt{x} + \frac{1}{3}x^3 + C \end{align*}$