Question

bantu jawab soal ini kak
\int _ { - 2 } ^ { 1 } ( x - 2 ) ^ { 2 }​

Answer 1

Nilai dari [tex]\bf{\int_{-2}^{1}\left(x-2\right)^{2}dx}[/tex] adalah 21.

[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(x)dx=F(x)+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(x)dx=k\int_{a}^{b}f(x)dx}[/tex]

[tex]\footnotesize\mathbf{2.\ \ \int_{a}^{b}f(x)\pm g(x)dx=\int_{a}^{b}f(x)dx\pm\int_{a}^{b}g(x)dx}[/tex]

[tex]\mathbf{3.\ \ \int_{a}^{b}f(x)\ dx=-\int_{b}^{a}f(x)\ dx}[/tex]

[tex]\small\mathbf{4.\ \ \int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx=\int_{a}^{c}f(x)dx}[/tex]

[tex]\mathbf{5.\ \ \int_{a}^{a}f(x)\ dx=0}[/tex]

[tex]\footnotesize\mathbf{6.\ \ \int_{a}^{b}f(x)dx=\int_{a+k}^{b+k}f(x-k)dx=\int_{a-k}^{b-k}f(x+k)dx}[/tex]

[tex] \: [/tex]

[tex] \: [/tex]

Pembahasan

Diketahui :

[tex]\bf{\int_{-2}^{1}\left(x-2\right)^{2}dx}[/tex]

Ditanya :

Hasil dari integral tersebut adalah...

Jawaban :

Cara 1

Misalkan :

[tex]\bf{\left(x-2\right)=u}[/tex]

[tex]\to[/tex] maka,

[tex]\bf{\int_{-2}^{1}u^{2}\ du}[/tex]

[tex]\bf{=\left[\frac{u^{\left(2+1\right)}}{2+1}\right]_{-2}^{1}}[/tex]

[tex]\bf{=\left[\frac{u^{3}}{3}\right]_{-2}^{1}}[/tex]

[tex]\bf{=\left[\frac{\left(x-2\right)^{3}}{3}\right]_{-2}^{1}}[/tex]

[tex]\bf{=\left(\frac{\left(1-2\right)^{3}}{3}\right)-\left(\frac{\left(-2-2\right)^{3}}{3}\right)}[/tex]

[tex]\bf{=\left(\frac{\left(-1\right)^{3}}{3}\right)-\left(\frac{\left(-4\right)^{3}}{3}\right)}[/tex]

[tex]\bf{=\left(\frac{-1}{3}\right)-\left(\frac{-64}{3}\right)}[/tex]

[tex]\bf{=\frac{63}{3}}[/tex]

[tex]\bf{=\boxed{\bf{21}}}[/tex]

[tex]\to[/tex]

Cara 2

[tex]\bf{\int_{-2}^{1}\left(x-2\right)^{2}dx}[/tex]

[tex]\bf{\int_{-2}^{1}\left(x^{2}-4x+4\right)dx}[/tex]

[tex]\bf{=\left[\frac{x^{3}}{3}-\frac{4x^{2}}{2}+4x\right]_{-2}^{1}}[/tex]

[tex]\bf{=\left(\frac{\left(1\right)^{3}}{3}-\frac{4\left(1\right)^{2}}{2}+4\left(1\right)\right)-\left(\frac{\left(-2\right)^{3}}{3}-\frac{4\left(-2\right)^{2}}{2}+4\left(-2\right)\right)}[/tex]

[tex]\bf{=\left(\frac{1}{3}-\frac{4}{2}+4\right)-\left(\frac{-8}{3}-\frac{16}{2}-8\right)}[/tex]

[tex]\bf{=\left(\frac{1}{3}-2+4\right)-\left(-\frac{8}{3}-8-8\right)}[/tex]

[tex]\bf{=\frac{1}{3}+\frac{8}{3}+2+16}[/tex]

[tex]\bf{=\frac{9}{3}+18}[/tex]

[tex]\bf{=\boxed{\bf{21}}}[/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.