Jawaban:
itu jawabannya ya, semoga bermanfaat ya
Diketahui f'(x) = x² – 4x + 1 dan f(–3)= –28, maka:[tex]\large\text{$\begin{aligned}\boxed{\vphantom{\bigg|}\,f(x)=\frac{x^3}{3}-2x^2+x+2\,}\end{aligned}$}[/tex]
Integral
[tex]\begin{aligned}f(x)&=\int f'(x)\,dx\\&=\int \left(x^2-4x+1\right)dx\\&=\frac{x^3}{3}-\frac{4x^2}{2}+x+C\\\vphantom{\Bigg|}f(x)&=\frac{x^3}{3}-2x^2+x+C\\\end{aligned}[/tex]
Dengan nilai f(–3)= –28:
[tex]\begin{aligned}f(-3)&=\frac{(-3)^3}{3}-2(-3)^2+(-3)+C\\-28&=(-1)(-3)^2-2(9)-3+C\\&=-9-18-3+C\\&=-30+C\\C&=30-28=\bf2\end{aligned}[/tex]
∴ Dengan demikian kita peroleh:[tex]\boxed{\vphantom{\Bigg|}\,f(x)=\frac{x^3}{3}-2x^2+x+2\,}[/tex][tex]\blacksquare[/tex]
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Jawaban:
itu jawabannya ya, semoga bermanfaat ya
Diketahui f'(x) = x² – 4x + 1 dan f(–3)= –28, maka:
[tex]\large\text{$\begin{aligned}\boxed{\vphantom{\bigg|}\,f(x)=\frac{x^3}{3}-2x^2+x+2\,}\end{aligned}$}[/tex]
Penjelasan dengan langkah-langkah:
Integral
[tex]\begin{aligned}f(x)&=\int f'(x)\,dx\\&=\int \left(x^2-4x+1\right)dx\\&=\frac{x^3}{3}-\frac{4x^2}{2}+x+C\\\vphantom{\Bigg|}f(x)&=\frac{x^3}{3}-2x^2+x+C\\\end{aligned}[/tex]
Dengan nilai f(–3)= –28:
[tex]\begin{aligned}f(-3)&=\frac{(-3)^3}{3}-2(-3)^2+(-3)+C\\-28&=(-1)(-3)^2-2(9)-3+C\\&=-9-18-3+C\\&=-30+C\\C&=30-28=\bf2\end{aligned}[/tex]
∴ Dengan demikian kita peroleh:
[tex]\boxed{\vphantom{\Bigg|}\,f(x)=\frac{x^3}{3}-2x^2+x+2\,}[/tex]
[tex]\blacksquare[/tex]