[tex] \bf{ \int\limits^{3}_{2} {x}^{3} - {3x}^{2} + 6x \: dx }[/tex]
Integral~
hasil dari [tex]\bf{\int_{2}^{3}x^{3}-3x^{2}+6x\ dx}[/tex] ialah
[tex]\boxed{\bf{\int_{2}^{3}x^{3}-3x^{2}+6x\ dx=\boxed{\bf{12\frac{1}{4}}}}}[/tex]
[tex] \: [/tex]
Pendahuluan
[tex]\boxed{\boxed{\mathbf{A.}} \ \boxed{\mathbf{Pengertian \ Singkat}}}[/tex]
Integral => lawan dari turunan. Jika f(x) turunan pertama dari F(x), maka :
[tex]\boxed{\mathbf{\int_{ }^{ }f\left(x\right)dx=F\left(x\right)+C}}[/tex]
Rumus yang sering dipakai :
[tex]\boxed{\mathbf{\int_{ }^{ }ax^{n}\ dx=\frac{a}{n+1}x^{n+1}+C}}[/tex]
[tex] \: [/tex]
[tex]\boxed{\boxed{\mathbf{B.}} \ \boxed{\mathbf{Integral \ Tak \ Tentu}}}[/tex]
ada 6 integral tak tentu yang perlu anda ketahui, diantaranya :
[tex]\mathbf{1.\ \ \int_{ }^{ }ax^{n}\ dx=\frac{a}{n+1}x^{n+1}+C;n\ne1}[/tex]
[tex]\mathbf{2.\ \ \int_{ }^{ }\frac{1}{x}\ dx=\ln\ | x |+C}[/tex]
[tex]\mathbf{3.\ \ \int_{ }^{ }\sin x\ dx=-\cos x+C}[/tex]
[tex]\mathbf{4.\ \ \int_{ }^{ }\cos x\ dx=\sin x+C}[/tex]
[tex]\mathbf{5.\ \ \int_{ }^{ }e^{x}\ dx=e^{x}+C}[/tex]
[tex]\mathbf{6.\ \ \int_{ }^{ }a^{x}\ dx=\frac{a^{x}}{\ln a}+C}[/tex]
[tex] \: [/tex]
[tex]\boxed{\boxed{\mathbf{C.}} \ \boxed{\mathbf{Integral \ Tentu}}}[/tex]
ada 6 integral tentu juga yang perlu anda pahami, diantaranya :
[tex]\mathbf{1.\ \ \int_{a}^{b}kf\left(x\right)dx=k\int_{a}^{b}f\left(x\right)dx}[/tex]
[tex]\footnotesize\mathbf{2.\ \ \int_{a}^{b}f\left(x\right)\pm g\left(x\right)dx=\int_{a}^{b}f\left(x\right)dx\pm\int_{a}^{b}g\left(x\right)dx}[/tex]
[tex]\mathbf{3.\ \ \int_{a}^{b}f\left(x\right)\ dx=-\int_{b}^{a}f\left(x\right)\ dx}[/tex]
[tex]\small\mathbf{4.\ \ \int_{a}^{b}f\left(x\right)dx+\int_{b}^{c}f\left(x\right)dx=\int_{a}^{c}f\left(x\right)dx}[/tex]
[tex]\mathbf{5.\ \ \int_{a}^{a}f\left(x\right)\ dx=0}[/tex]
[tex]\footnotesize\mathbf{6.\ \ \int_{a}^{b}f\left(x\right)dx=\int_{a+k}^{b+k}f\left(x-k\right)dx=\int_{a-k}^{b-k}f\left(x+k\right)dx}[/tex]
[tex] \: [/tex]
[tex] \: [/tex]
Pembahasan
Diketahui :
[tex]\bf{\int_{2}^{3}x^{3}-3x^{2}+6x\ dx}[/tex]
Ditanya :
hasil dari integral tersebut adalah ....
Jawaban :
[tex]\bf{\int_{2}^{3}x^{3}-3x^{2}+6x\ dx}[/tex]
[tex]\bf{=\left[\frac{1}{4}x^{4}-\frac{3}{3}x^{3}+\frac{6}{2}x^{2}\right]_{2}^{3}}[/tex]
[tex]\bf{=\left[\frac{x^{4}}{4}-x^{3}+3x^{2}\right]_{2}^{3}}[/tex]
[tex]\small\bf{=\left(\frac{\left(3\right)^{4}}{4}-\left(3\right)^{3}+3\left(3\right)^{2}\right)-\left(\frac{\left(2\right)^{4}}{4}-\left(2\right)^{3}+3\left(2\right)^{2}\right)}[/tex]
[tex]\bf{=\left(\frac{81}{4}-27+27\right)-\left(\frac{16}{4}-8+12\right)}[/tex]
[tex]\bf{=20\frac{1}{4}-8}[/tex]
[tex]\boxed{\bf{=12\frac{1}{4}}}[/tex]
[tex] \: [/tex]
[tex] \: [/tex]
Pelajari Lebih Lanjut :
- Contoh soal integral tentu (1) : brainly.co.id/tugas/50510100
- Contoh soal integral tentu (2) : brainly.co.id/tugas/50454066
- Tentukan ∫(2x^{2} + 5x)^{2} dx : brainly.co.id/tugas/50364777
- Integral dari (x^3 +√x) dx : https://brainly.co.id/tugas/50722822
[tex] \: [/tex]
[tex] \: [/tex]
Detail Jawaban :
Kelas : 12 SMA
Bab : 1
Sub Bab : Bab 1 - Integral
Kode kategorisasi : 12.2.1
Kata Kunci : Integral.
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