Integral Fungsi Aljabar
Secara umum integral sanggup dibedakan menjadi dua, yaitu integral tak tentu dan integral tentu.
Integral tak tentu fungsi f(x) dinyatakan oleh :
dengan :
f(x) = integran/fungsi yang diintegralkan
F(X) = anti turunan dari f(x)
C = konstanta
$$\mathrm{\mathbf{\int a\:dx=ax+C}}$$ Contoh
1. ∫ 2 dx = 2x + C
2. ∫ \(\frac{1}{2}\) dx = \(\frac{1}{2}\)x + C
Untuk f(x) = axn , n ≠ −1 maka :
$$\mathrm{\mathbf{\int ax^{n}\:dx=\frac{a}{n+1}x^{n+1}+C}}$$ Contoh
1. ∫ 2x4 dx = ...
Jawab :
⇒ \(\mathrm{\frac{2}{4+1}}\)x4+1 + C
⇒ \(\mathrm{\frac{2}{5}}\)x5 + C
2. ∫ x-6 dx = ...
Jawab :
⇒ \(\mathrm{\frac{1}{-6+1}}\)x-6+1 + C
⇒ \(\mathrm{-\frac{1}{5}}\)x-5 + C
Untuk f(x) = (ax + b)n , n ≠ −1 maka :
⇒ \(\mathrm{\frac{1}{10}}\)(2x − 1)5 + C
2. ∫ (x + 1)-7 dx = ...
Jawab :
⇒ \(\mathrm{\frac{1}{1(-7+1)}}\)(x + 1)-7+1 + C
⇒ \(\mathrm{-\frac{1}{6}}\)(x + 1)-6 + C
Untuk f(x) = \(\mathrm{\mathbf{\frac{1}{x}}}\), maka :
$$\mathrm{\int \mathbf{\frac{1}{x}\:dx=ln|x|+C}}$$
Untuk memilih integral yang integrannya memuat bentuk akar atau pecahan, langkah awal yang harus dilakukan ialah mengubah terlebih dahulu integran tersebut ke bentuk eksponen (pangkat).
Berikut beberapa sifat akar dan pangkat yang sering dipakai :
1. \(\mathrm{\int \sqrt{x}\:dx=}\)
Jawab :
\(\mathrm{\Rightarrow \int x^{\frac{1}{2}}\:dx}\)
\(\mathrm{=\frac{1}{\frac{1}{2}+1}x^{\frac{1}{2}+1}+C}\)
\(\mathrm{=\frac{2}{3}x^{\frac{3}{2}}+C}\)
\(\mathrm{=\frac{2}{3}x\sqrt{x}+C}\)
2. \(\mathrm{\int \frac{1}{x^{2}}\:dx=}\)
Jawab :
\(\mathrm{\Rightarrow \int x^{-2}\:dx}\)
\(\mathrm{=\frac{1}{-2+1}x^{-2+1}+C}\)
\(\mathrm{=-x^{-1}+C}\)
\(\mathrm{=-\frac{1}{x}+C}\)
3. \(\mathrm{\int x\sqrt[3]{\mathrm{x^{2}}}\:dx=}\)
Jawab :
\(\mathrm{\Rightarrow \int x.x^{\frac{2}{3}}\:dx}\)
\(\mathrm{\Rightarrow \int x^{\frac{5}{3}}\:dx}\)
\(\mathrm{=\frac{1}{\frac{5}{3}+1}x^{\frac{5}{3}+1}+C}\)
\(\mathrm{=\frac{3}{8}x^{\frac{8}{3}}+C}\)
\(\mathrm{=\frac{3}{8}\sqrt[3]{x^{8}}+C}\) atau
\(\mathrm{=\frac{3}{8}x^{2}\sqrt[3]{x^{2}}+C}\)
Contoh
∫ 3x4 dx = 3 ∫ x4 dx
∫ 3x4 dx = 3 . \(\mathrm{\frac{1}{5}x^{5}+C}\)
∫ 3x4 dx = \(\mathrm{\frac{3}{5}x^{5}+C}\)
Contoh
a. ∫ (3x7 − π) dx = ...
Jawab :
= \(\mathrm{\frac{3}{7+1}}\)x7+1 − πx + C
= \(\mathrm{\frac{3}{8}}\)x8 − πx + C
b. ∫ (6x5 + 2x3 − x2) dx = ...
Jawab :
\(\mathrm{= \frac{6}{5+1}x^{5+1}+\frac{2}{3+1}x^{3+1}-\frac{1}{2+1}x}^{2+1}+C\)
\(\mathrm{= x^{6}+\frac{1}{2}x^{4}-\frac{1}{3}x}^{3}+C\)
c. \(\mathrm{\int \frac{6x^{5}-2x^{4}+9}{x^{4}}\:dx=...}\)
Jawab :
\(\mathrm{\Rightarrow \int \left (\frac{6x^{5}}{x^{4}}-\frac{2x^{4}}{x^{4}}+\frac{9}{x^{4}} \right )\:dx}\)
\(\mathrm{\Rightarrow \int \left (6x-2+9x^{-4} \right )dx}\)
\(\mathrm{=\frac{6}{1+1}x^{1+1}-2x+\frac{9}{-4+1}x^{-4+1}+C}\)
\(\mathrm{=3x^{2}-2x-3x^{-3}+C}\)
\(\mathrm{=3x^{2}-2x-\frac{3}{x^{3}}+C}\)
d. \(\mathrm{\int \left (\sqrt{x}+\frac{2}{\sqrt{x}} \right )\:dx=...}\)
Jawab :
\(\mathrm{\Rightarrow \int \left ( x^{\frac{1}{2}}+2x^{-\frac{1}{2}} \right )dx}\)
\(\mathrm{=\frac{1}{\frac{1}{2}+1}x^{\frac{1}{2}+1}+\frac{2}{-\frac{1}{2}+1}x^{-\frac{1}{2}+1}+C}\)
\(\mathrm{=\frac{2}{3}x^{\frac{3}{2}}+4x^{\frac{1}{2}}+C}\)
\(\mathrm{=\frac{2}{3}x\sqrt{x}+4\sqrt{x}+C}\)
e. \(\mathrm{\int \left ( x\sqrt{x}-\frac{x}{\sqrt{x}} \right )dx=...}\)
Jawab :
\(\mathrm{\Rightarrow \int \left (x^{\frac{3}{2}}-x^{\frac{1}{2}} \right )dx}\)
\(\mathrm{=\frac{2}{5}x^{\frac{5}{2}}-\frac{2}{3}x^{\frac{3}{2}}+C}\)
\(\mathrm{=\frac{2}{5}x^{2}\sqrt{x}-\frac{2}{3}x\sqrt{x}+C}\)
f. \(\mathrm{\int \left ( \sqrt{x}+\frac{1}{ \sqrt{x}} \right )^{2}\:dx=}\)
Jawab :
\(\mathrm{\Rightarrow \int \left (x+2+\frac{1}{x} \right )\:dx}\)
\(\mathrm{=\frac{1}{1+1}x^{1+1}+2x+ln|x|+C}\)
\(\mathrm{=\frac{1}{2}x^{2}+2x+ln|x|+C}\)
g. \(\mathrm{\int \frac{1}{\sqrt[3]{(3x+1)^{2}}}\:dx}\)
Jawab :
\(\mathrm{\Rightarrow \int (3x+1)^{-\frac{2}{3}}\:dx}\)
\(\mathrm{=\frac{1}{3\left ( -\frac{2}{3}+1 \right )}(3x+1)^{-\frac{2}{3}+1}+C}\)
\(\mathrm{=(3x+1)^{\frac{1}{3}}+C}\)
\(\mathrm{=\sqrt[3]{3x+1}+C}\)
Contoh 2
Tentukan f(x) jikalau diketahui :
a. f '(x) = 2x + 2 ; f(0) = −1
Jawab :
f(x) = ∫ f '(x) dx
f(x) = ∫ (2x + 2) dx
f(x) = x2 + 2x + C
f(0) = −1
⇔ (0)2 + 2(0) + C = −1
⇔ C = −1
Jadi, f(x) = x2 + 2x − 1
b. f ''(x) = 12x − 2 ; f(0) = 2 dan f '(1) = 4
Jawab :
f '(x) = ∫ f ''(x) dx
f '(x) = ∫ (12x − 2) dx
f '(x) = 6x2 − 2x + C
f '(1) = 4
⇔ 6(1)2 − 2(1) + C = 4
⇔ C = 0
diperoleh : f '(x) = 6x2 − 2x
f(x) = ∫ f '(x) dx
f(x) = ∫ (6x2 − 2x) dx
f(x) = 2x3 − x2 + C
f(0) = 2
⇔ 2(0)3 − (0)2 + C = 2
⇔ C = 2
Jadi, f(x) = 2x3 − x2 + 2
Sumber http://smatika.blogspot.com
Integral tak tentu fungsi f(x) dinyatakan oleh :
∫ f(x) dx = F(x) + C
dengan :
f(x) = integran/fungsi yang diintegralkan
F(X) = anti turunan dari f(x)
C = konstanta
Rumus-Rumus Dasar Integral
Untuk f(x) = a dengan a konstan, maka :$$\mathrm{\mathbf{\int a\:dx=ax+C}}$$ Contoh
1. ∫ 2 dx = 2x + C
2. ∫ \(\frac{1}{2}\) dx = \(\frac{1}{2}\)x + C
Untuk f(x) = axn , n ≠ −1 maka :
$$\mathrm{\mathbf{\int ax^{n}\:dx=\frac{a}{n+1}x^{n+1}+C}}$$ Contoh
1. ∫ 2x4 dx = ...
Jawab :
⇒ \(\mathrm{\frac{2}{4+1}}\)x4+1 + C
⇒ \(\mathrm{\frac{2}{5}}\)x5 + C
2. ∫ x-6 dx = ...
Jawab :
⇒ \(\mathrm{\frac{1}{-6+1}}\)x-6+1 + C
⇒ \(\mathrm{-\frac{1}{5}}\)x-5 + C
Untuk f(x) = (ax + b)n , n ≠ −1 maka :
$$\mathrm{\mathbf{\int (ax+b)^{n}\:dx=\frac{1}{a(n+1)}x^{n+1}+C}}$$ Contoh
1. ∫ (2x − 1) 4 dx = ... Jawab :
⇒ \(\mathrm{\frac{1}{2(4+1)}}\)(2x − 1)4+1 + C⇒ \(\mathrm{\frac{1}{10}}\)(2x − 1)5 + C
2. ∫ (x + 1)-7 dx = ...
Jawab :
⇒ \(\mathrm{\frac{1}{1(-7+1)}}\)(x + 1)-7+1 + C
⇒ \(\mathrm{-\frac{1}{6}}\)(x + 1)-6 + C
Untuk f(x) = \(\mathrm{\mathbf{\frac{1}{x}}}\), maka :
$$\mathrm{\int \mathbf{\frac{1}{x}\:dx=ln|x|+C}}$$
Untuk memilih integral yang integrannya memuat bentuk akar atau pecahan, langkah awal yang harus dilakukan ialah mengubah terlebih dahulu integran tersebut ke bentuk eksponen (pangkat).
Berikut beberapa sifat akar dan pangkat yang sering dipakai :
- \(\mathrm{x^{m}.\;x^{n}=x^{m+n}}\)
- \(\mathrm{\frac{x^{m}}{x^{n}}=x^{m-n}}\)
- \(\mathrm{\frac{1}{x^{n}}=x^{-n}}\)
- \(\mathrm{\sqrt{x}=x^{\frac{1}{2}}}\)
- \(\mathrm{x\sqrt{x}=x^{\frac{3}{2}}}\)
- \(\mathrm{\sqrt[\mathrm{n}]{\mathrm{x^{m}}}=x^{\frac{m}{n}}}\)
1. \(\mathrm{\int \sqrt{x}\:dx=}\)
Jawab :
\(\mathrm{\Rightarrow \int x^{\frac{1}{2}}\:dx}\)
\(\mathrm{=\frac{1}{\frac{1}{2}+1}x^{\frac{1}{2}+1}+C}\)
\(\mathrm{=\frac{2}{3}x^{\frac{3}{2}}+C}\)
\(\mathrm{=\frac{2}{3}x\sqrt{x}+C}\)
2. \(\mathrm{\int \frac{1}{x^{2}}\:dx=}\)
Jawab :
\(\mathrm{\Rightarrow \int x^{-2}\:dx}\)
\(\mathrm{=\frac{1}{-2+1}x^{-2+1}+C}\)
\(\mathrm{=-x^{-1}+C}\)
\(\mathrm{=-\frac{1}{x}+C}\)
3. \(\mathrm{\int x\sqrt[3]{\mathrm{x^{2}}}\:dx=}\)
Jawab :
\(\mathrm{\Rightarrow \int x.x^{\frac{2}{3}}\:dx}\)
\(\mathrm{\Rightarrow \int x^{\frac{5}{3}}\:dx}\)
\(\mathrm{=\frac{1}{\frac{5}{3}+1}x^{\frac{5}{3}+1}+C}\)
\(\mathrm{=\frac{3}{8}x^{\frac{8}{3}}+C}\)
\(\mathrm{=\frac{3}{8}\sqrt[3]{x^{8}}+C}\) atau
\(\mathrm{=\frac{3}{8}x^{2}\sqrt[3]{x^{2}}+C}\)
Sifat-Sifat Integral
1. ∫ k f(x) dx = k ∫ f(x) dx (k = konstan)Contoh
∫ 3x4 dx = 3 ∫ x4 dx
∫ 3x4 dx = 3 . \(\mathrm{\frac{1}{5}x^{5}+C}\)
∫ 3x4 dx = \(\mathrm{\frac{3}{5}x^{5}+C}\)
2. ∫{f(x) ± g(x)} dx = ∫ f(x) dx ± ∫ g(x) dx
∫ (4x2 + 3x − 2) dx = ...
⇒ ∫ 4x2 dx + ∫ 3x dx − ∫ 2 dx
= \(\mathrm{\frac{4}{3}x^{3}+\frac{3}{2}x^{2}-2x+C}\)
Contoh-Contoh Latihan Soal Integral Fungsi Aljabar
Contoh 1
Tentukan integral berikut !
Tentukan integral berikut !
a. ∫ (3x7 − π) dx = ...
Jawab :
= \(\mathrm{\frac{3}{7+1}}\)x7+1 − πx + C
= \(\mathrm{\frac{3}{8}}\)x8 − πx + C
b. ∫ (6x5 + 2x3 − x2) dx = ...
Jawab :
\(\mathrm{= \frac{6}{5+1}x^{5+1}+\frac{2}{3+1}x^{3+1}-\frac{1}{2+1}x}^{2+1}+C\)
\(\mathrm{= x^{6}+\frac{1}{2}x^{4}-\frac{1}{3}x}^{3}+C\)
c. \(\mathrm{\int \frac{6x^{5}-2x^{4}+9}{x^{4}}\:dx=...}\)
Jawab :
\(\mathrm{\Rightarrow \int \left (\frac{6x^{5}}{x^{4}}-\frac{2x^{4}}{x^{4}}+\frac{9}{x^{4}} \right )\:dx}\)
\(\mathrm{\Rightarrow \int \left (6x-2+9x^{-4} \right )dx}\)
\(\mathrm{=\frac{6}{1+1}x^{1+1}-2x+\frac{9}{-4+1}x^{-4+1}+C}\)
\(\mathrm{=3x^{2}-2x-3x^{-3}+C}\)
\(\mathrm{=3x^{2}-2x-\frac{3}{x^{3}}+C}\)
d. \(\mathrm{\int \left (\sqrt{x}+\frac{2}{\sqrt{x}} \right )\:dx=...}\)
Jawab :
\(\mathrm{\Rightarrow \int \left ( x^{\frac{1}{2}}+2x^{-\frac{1}{2}} \right )dx}\)
\(\mathrm{=\frac{1}{\frac{1}{2}+1}x^{\frac{1}{2}+1}+\frac{2}{-\frac{1}{2}+1}x^{-\frac{1}{2}+1}+C}\)
\(\mathrm{=\frac{2}{3}x^{\frac{3}{2}}+4x^{\frac{1}{2}}+C}\)
\(\mathrm{=\frac{2}{3}x\sqrt{x}+4\sqrt{x}+C}\)
e. \(\mathrm{\int \left ( x\sqrt{x}-\frac{x}{\sqrt{x}} \right )dx=...}\)
Jawab :
\(\mathrm{\Rightarrow \int \left (x^{\frac{3}{2}}-x^{\frac{1}{2}} \right )dx}\)
\(\mathrm{=\frac{2}{5}x^{\frac{5}{2}}-\frac{2}{3}x^{\frac{3}{2}}+C}\)
\(\mathrm{=\frac{2}{5}x^{2}\sqrt{x}-\frac{2}{3}x\sqrt{x}+C}\)
f. \(\mathrm{\int \left ( \sqrt{x}+\frac{1}{ \sqrt{x}} \right )^{2}\:dx=}\)
Jawab :
\(\mathrm{\Rightarrow \int \left (x+2+\frac{1}{x} \right )\:dx}\)
\(\mathrm{=\frac{1}{1+1}x^{1+1}+2x+ln|x|+C}\)
\(\mathrm{=\frac{1}{2}x^{2}+2x+ln|x|+C}\)
g. \(\mathrm{\int \frac{1}{\sqrt[3]{(3x+1)^{2}}}\:dx}\)
Jawab :
\(\mathrm{\Rightarrow \int (3x+1)^{-\frac{2}{3}}\:dx}\)
\(\mathrm{=\frac{1}{3\left ( -\frac{2}{3}+1 \right )}(3x+1)^{-\frac{2}{3}+1}+C}\)
\(\mathrm{=(3x+1)^{\frac{1}{3}}+C}\)
\(\mathrm{=\sqrt[3]{3x+1}+C}\)
Contoh 2
Tentukan f(x) jikalau diketahui :
a. f '(x) = 2x + 2 ; f(0) = −1
Jawab :
f(x) = ∫ f '(x) dx
f(x) = ∫ (2x + 2) dx
f(x) = x2 + 2x + C
f(0) = −1
⇔ (0)2 + 2(0) + C = −1
⇔ C = −1
Jadi, f(x) = x2 + 2x − 1
b. f ''(x) = 12x − 2 ; f(0) = 2 dan f '(1) = 4
Jawab :
f '(x) = ∫ f ''(x) dx
f '(x) = ∫ (12x − 2) dx
f '(x) = 6x2 − 2x + C
f '(1) = 4
⇔ 6(1)2 − 2(1) + C = 4
⇔ C = 0
diperoleh : f '(x) = 6x2 − 2x
f(x) = ∫ f '(x) dx
f(x) = ∫ (6x2 − 2x) dx
f(x) = 2x3 − x2 + C
f(0) = 2
⇔ 2(0)3 − (0)2 + C = 2
⇔ C = 2
Jadi, f(x) = 2x3 − x2 + 2
Sumber http://smatika.blogspot.com
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