Hasil dari [tex]\displaystyle{\int\limits {x^2\sqrt{1-x}} \, dx }[/tex] adalah [tex]\displaystyle{\boldsymbol{-\frac{2}{3}\left ( 1-x \right )^{\frac{3}{2}}+\frac{4}{5}\left ( 1-x \right )^{\frac{5}{2}}-\frac{2}{7}\left ( 1-x \right )^{\frac{7}{2}}+C} }[/tex].
PEMBAHASAN
Integral merupakan operasi yang menjadi kebalikan dari operasi turunan/diferensial. Sehingga integral sering juga disebut sebagai antiturunan.
[tex]\displaystyle{f(x)=\int\limits {\left [ \frac{df(x)}{dx} \right ]} \, dx}[/tex]
Sifat - sifat operasi pada integral adalah sebagai berikut :
[tex](i)~\displaystyle{\int\limits {ax^n} \, dx=\frac{a}{n+1}x^{n+1}+C},~~~dengan~C=konstanta[/tex]
[tex](ii)~\displaystyle{\int\limits {kf(x)} \, dx=k\int\limits {f(x)} \, dx}[/tex]
[tex](iii)~\displaystyle{\int\limits {\left [ f(x)\pm g(x) \right ]} \, dx=\int\limits {f(x)} \, dx\pm\int\limits {g(x)} \, dx}[/tex]
[tex](iv)~\displaystyle{\int\limits^b_a {f(x)} \, dx=F(b)-F(a)}[/tex]
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DIKETAHUI
[tex]\displaystyle{\int\limits {x^2\sqrt{1-x}} \, dx= }[/tex]
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DITANYA
Tentukan hasilnya dengan :
a. Substitusi u = 1 - x
b. Substitusi u = √(1-x)
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PENYELESAIAN
Soal a.
Misal :
[tex]u=1-x~\to~x=1-u[/tex]
[tex]du=-dx[/tex]
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Maka :
[tex]\displaystyle{\int\limits {x^2\sqrt{1-x}} \, dx }[/tex]
[tex]\displaystyle{=\int\limits {(1-u)^2\sqrt{u}} \, (-du) }[/tex]
[tex]\displaystyle{=-\int\limits {(1-2u+u^2)u^{\frac{1}{2}}} \, du }[/tex]
[tex]\displaystyle{=-\int\limits {\left ( u^{\frac{1}{2}}-2u^{\frac{3}{2}}+u^{\frac{5}{2}} \right )} \, du }[/tex]
[tex]\displaystyle{=-\left ( \frac{1}{1+\frac{1}{2}}u^{1+\frac{1}{2}}-\frac{2}{1+\frac{3}{2}}u^{1+\frac{3}{2}}+\frac{1}{1+\frac{5}{2}}u^{1+\frac{5}{2}} \right )+C }[/tex]
[tex]\displaystyle{=-\left ( \frac{2}{3}u^{\frac{3}{2}}-\frac{4}{5}u^{\frac{5}{2}}+\frac{2}{7}u^{\frac{7}{2}} \right )+C }[/tex]
[tex]\displaystyle{=-\frac{2}{3}u^{\frac{3}{2}}+\frac{4}{5}u^{\frac{5}{2}}-\frac{2}{7}u^{\frac{7}{2}}+C }[/tex]
[tex]\displaystyle{=-\frac{2}{3}(1-x)^{\frac{3}{2}}+\frac{4}{5}(1-x)^{\frac{5}{2}}-\frac{2}{7}(1-x)^{\frac{7}{2}}+C }[/tex]
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Soal b.
Misal :
[tex]u=\sqrt{1-x}[/tex]
[tex]u^2=1-x~\to~x=1-u^2[/tex]
[tex]2udu=-dx[/tex]
[tex]dx=-2udu[/tex]
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Maka :
[tex]\displaystyle{\int\limits {x^2\sqrt{1-x}} \, dx }[/tex]
[tex]\displaystyle{=\int\limits {(1-u^2)^2u} \, (-2udu) }[/tex]
[tex]\displaystyle{=-2\int\limits {(1-2u^2+u^4)u^2} \, du }[/tex]
[tex]\displaystyle{=-2\int\limits {\left ( u^2-2u^4+u^6 \right )} \, du }[/tex]
[tex]\displaystyle{=-2\left ( \frac{1}{1+2}u^{1+2}-\frac{2}{1+4}u^{1+4}+\frac{1}{1+6}u^{1+6} \right )+C }[/tex]
[tex]\displaystyle{=-2\left ( \frac{1}{3}u^3-\frac{2}{5}u^5+\frac{1}{7}u^7 \right )+C }[/tex]
[tex]\displaystyle{=-\frac{2}{3}u^3+\frac{4}{5}u^5-\frac{2}{7}u^7+C }[/tex]
[tex]\displaystyle{=-\frac{2}{3}\left ( \sqrt{1-x} \right )^3+\frac{4}{5}\left ( \sqrt{1-x} \right )^5-\frac{2}{7}\left ( \sqrt{1-x} \right )^7+C }[/tex]
[tex]\displaystyle{=-\frac{2}{3}\left ( 1-x \right )^{\frac{3}{2}}+\frac{4}{5}\left ( 1-x \right )^{\frac{5}{2}}-\frac{2}{7}\left ( 1-x \right )^{\frac{7}{2}}+C }[/tex]
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KESIMPULAN
Hasil dari [tex]\displaystyle{\int\limits {x^2\sqrt{1-x}} \, dx }[/tex] adalah [tex]\displaystyle{\boldsymbol{-\frac{2}{3}\left ( 1-x \right )^{\frac{3}{2}}+\frac{4}{5}\left ( 1-x \right )^{\frac{5}{2}}-\frac{2}{7}\left ( 1-x \right )^{\frac{7}{2}}+C} }[/tex].
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PELAJARI LEBIH LANJUT
- Integral fungsi : https://brainly.co.id/tugas/30176534
- Integral fungsi : https://brainly.co.id/tugas/30067184
- Luas daerah kurva : https://brainly.co.id/tugas/30113906
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DETAIL JAWABAN
Kelas : 11
Mapel: Matematika
Bab : Integral
Kode Kategorisasi: 11.2.10
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