Sunday, September 18, 2016

Integral Tertentu

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Jika f kontinu pada interval [a, b] dan F adalah antiturunan dari f pada interval tersebut, maka : $$\mathrm{\int_{a}^{b}f(x)\;dx=\left [ F(x) \right ]_{a}^{b}=F(b)-F(a)}$$
Contoh :
1.  \(\mathrm{\int_{1}^{3}(2x-1)\;dx}\)
     \(\mathrm{=\left [ x^{2}-x \right ]_{1}^{3}}\)
     \(\mathrm{=\left ( 3^{2}-3 \right )-\left ( 1^{2}-1 \right )}\)
     \(=6-0\)
     \(=6\)

2.  \(\mathrm{\int_{0}^{1}\left ( x^{2}-3x \right )\;dx}\)
     \(\mathrm{=\left [ \frac{1}{3}x^{3}-\frac{3}{2}x^{2} \right ]_{0}^{1}}\)
     \(\mathrm{=\left (\frac{1}{3}(1)^{3}-\frac{3}{2}(1)^{2}  \right )-\left (\frac{1}{3}(0)^{3}-\frac{3}{2}(0)^{2}  \right )}\)
     \(=\frac{1}{3}-\frac{3}{2}\)
     \(=-\frac{7}{6}\)


Sifat-Sifat Integral Tertentu

1.  \(\mathrm{\int_{a}^{b}k\:f(x)\;dx=k\int_{a}^{b}f(x)\:dx}\)

2.  \(\mathrm{\int_{a}^{b}\left [f(x)\pm g(x)  \right ]\;dx=\int_{a}^{b}f(x)\:dx\pm \int_{a}^{b}g(x)\:dx}\)

3.  \(\mathrm{\int_{a}^{a}f(x)\;dx=0}\)

4.  \(\mathrm{\int_{a}^{b}f(x)\;dx=-\int_{b}^{a}f(x)\;dx}\)

5.  \(\mathrm{\int_{a}^{c}f(x)\;dx=\int_{a}^{b}f(x)\;dx+\int_{b}^{c}f(x)\;dx}\)
     dengan : a < b < c



Berikut contoh-contoh latihan soal integral tertentu :

Contoh 1
Jika \(\mathrm{f(x)=3x^{2}-4x}\) dan \(\mathrm{g(x)=2x+1}\), tentukan nilai dari :

a.  \(\mathrm{\int_{1}^{3}f(x)\:dx}\)

     Jawab :
     \(\mathrm{\Rightarrow \int_{1}^{3}\left ( 3x^{2}-4x \right )\:dx}\)
     \(\mathrm{= \left [x^{3}-2x^{2}  \right ]_{1}^{3}}\)
     \(\mathrm{= \left ((3)^{3}-2(3)^{2}  \right )-\left ((1)^{3}-2(1)^{2}  \right )}\)
     \(=9-(-1)\)
     \(\mathrm{=10}\)

b.  \(\mathrm{\int_{-2}^{1}g(x)\:dx}\)

     Jawab :
     \(\mathrm{\Rightarrow\int_{-2}^{1}(2x+1)\:dx}\)
     \(\mathrm{=\left [x^{2}+x  \right ]_{-2}^{1}}\)
     \(\mathrm{=\left ( 1^{2}+1 \right )-\left ( (-2)^{2}+(-2) \right )}\)
     \(=2-2\)
     \(=0\)

c.  \(\mathrm{\int_{0}^{1}\left (f(x)-g(x)  \right )\:dx}\)

Jawab :
\(\mathrm{\Rightarrow \int_{0}^{1}\left (\left ( 3x^{2}-4x \right )-(2x+1)  \right )\:dx}\)
\(\mathrm{\Rightarrow \int_{0}^{1}\left (3x^{2}-6x-1  \right )\:dx}\)
\(\mathrm{=\left [x^{3}-3x^{2}-x  \right ]_{0}^{1}}\)
\(\mathrm{=\left (1^{3}-3(1)^{2}-1  \right )-\left ( 0^{3}-3(0)^{2}-0 \right )}\)
\(=-3-0\)
\(=-3\)


Contoh 2
Tentukan nilai a dari pengintegralan berikut !

a.  \(\mathrm{\int_{a}^{a+1}\left ( 2x+1 \right )\;dx=4}\)

     Jawab :
     \(\mathrm{\Rightarrow \left [ x^{2}+x \right ]_{a}^{a+1}=4}\)
     \(\mathrm{\Rightarrow \left [ \left ( a+1 \right )^{2}+\left ( a+1 \right ) \right ]-\left [ a^{2}+a \right ]=4}\)
     \(\mathrm{\Rightarrow a^{2}+2a+1+a+1-a^{2}-a=4}\)
     \(\mathrm{\Rightarrow 2a+2=4}\)
     \(\mathrm{\Rightarrow 2a=2}\)
     \(\mathrm{\Rightarrow a=1}\)


b.  \(\mathrm{\int_{1}^{a}\left ( 2x-2 \right )\;dx=2a+1\;\;;\,a>0}\)

     Jawab :
     \(\mathrm{\Rightarrow\left [ x^{2}-2x \right ]_{1}^{a}=2a+1}\)
     \(\mathrm{\Rightarrow \left ( a^{2}-2a \right )-\left ( 1^{2}-2.1 \right )=2a+1}\)
     \(\mathrm{\Rightarrow a^{2}-2a+1=2a+1}\)
     \(\mathrm{\Rightarrow a^{2}-4a=0}\)
     \(\mathrm{\Rightarrow a\left ( a-4 \right )=0}\)
     \(\mathrm{\Rightarrow a=0\;atau\;a=4}\)
     Karena a > 0, maka \(\mathrm{a=4}\)


c.  \(\mathrm{\int_{a}^{1}\left ( 3x^{2}+2x \right )\;dx=-10\;\;;\,a\in R}\)

     Jawab ;
     \(\mathrm{\Rightarrow \left [ x^{3}+x^{2} \right ]_{a}^{1}=-10}\)
     \(\mathrm{\Rightarrow \left (1^{3}+1^{2}  \right )-\left ( a^{3}+a^{2} \right )=-10}\)
     \(\mathrm{\Rightarrow 2-a^{3}-a^{2}=-10}\)
     \(\mathrm{\Rightarrow a^{3}+a^{2}-12=0}\)

Nilai a yang mungkin adalah faktor dari 12, yaitu : ±1, ±2, ±3, ±4, ±6, ±12.
Nilai a yang memenuhi adalah a = 2, karena :
\(\mathrm{\Rightarrow 2^{3}+2^{2}-12=0}\)


Contoh 3
Nilai dari \(\mathrm{\int_{0}^{\frac{\pi }{2}}\left (2\,cos\:2x-sin\,x  \right )\;dx}\) adalah...

Jawab :
\(\mathrm{\int_{0}^{\frac{\pi }{2}}\left (2\,cos\:2x-sin\,x  \right )\;dx}\)
\(\mathrm{=\left [ 2\,\frac{1}{2}sin\,2x-(-cos\,x) \right ]_{0}^{\frac{\pi }{2}}}\)
\(\mathrm{=\left [ sin\,2x+cos\,x \right ]_{0}^{\frac{\pi }{2}}}\)
\(\mathrm{=\left ( sin\,2\left ( \frac{\pi }{2} \right )+cos\,\left ( \frac{\pi }{2} \right ) \right )-\left ( sin\,2(0)+cos\,0 \right )}\)\(\mathrm{=\left ( sin\,\pi+cos\left ( \frac{\pi }{2} \right )  \right )-\left ( sin\,0+cos\,0 \right )}\)
\(=\left ( 0+0 \right )-\left ( 0+1 \right )\)
\(=-1\)


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