Strona 1 z 1
Całki cz.2
: 17 wrz 2010, 17:15
autor: PriCe
1.Oblicz Całki :
a)
\(\int_{}^{} (6x^5+ \frac{3}{x^4} dx\)
b)
\(\int_{}^{} \frac{x^4+3x^3+2}{x} dx\)
c)
\(\int_{}^{} \frac{dx}{3x+2}\)
d)
\(\int_{}^{} \sqrt{5x+1}dx\)
e)
\(\int_{}^{} xe^-^x^2dx\)
f)
\(\int_{}^{} xsinxdx\)
g)
\(\int_{}^{} x^3lnxdx\)
w przykładzie e) tam jest -x do potęgi 2
2.Oblicz Całkę Oznaczona:
a)
\(\int_{0}^{1} \sqrt{x+4} dx\)
3.Określ Ekstrema funkcji :
a)
\(y=x^3-6x^2+9x-5\)
: 17 wrz 2010, 17:32
autor: domino21
zad 3.
\(y=x^3-6x^2+9x-5, \ x\in R
y'=3x^2-12x+9=3(x-3)(x-1), \ x\in R\)
\(f'(x)=0 \ \Leftrightarrow \ 3(x-1)(x-3)=0 \ \Rightarrow \ x=1 \ \vee \ x=3
f'(x)>0 \ \Leftrightarrow \ 3(x-1)(x-3)>0 \ \Rightarrow \ x\in (-\infty;1) \cup (3;+\infty)
f'(x)<0 \ \Leftrightarrow \ 3(x-1)(x-3)<0 \ \Rightarrow \ x\in (1;3)\)
\(f_{max}=f(1)=1-6+9-5=-1
f_{min}=f(3)=27-54+27-5=-5\)
: 17 wrz 2010, 19:04
autor: domino21
zad 2.
\(\int \sqrt{x+4} dx =\left( x+4=t \\ dx=dt \right)= \int \sqrt{t} dt = \int t^{\frac{1}{2}} dt =\frac{1}{\frac{3}{2}} \cdot t^{\frac{3}{2}}+C=\frac{2}{3} t\sqrt{t} +C=\frac{2}{3}(x+4)\sqrt{x+4}+C\)
\(\int_{0}^{1} \sqrt{x+4} dx=\left[\frac{2}{3}(x+4)\sqrt{x+4}\right] ^1_0=\frac{2}{3} (1+4)\sqrt{1+4}-\frac{2}{3}(0+4)\sqrt{0+4}=\frac{10\sqrt{5}}{3}-\frac{16}{3}=\frac{10\sqrt{5}-16}{3}\)
: 17 wrz 2010, 19:13
autor: domino21
a.
\(\int 6x^5+\frac{3}{x^4} dx=\int 6x^5dx+\int \frac{3}{x^4} dx=6\int x^5 dx+3\int x^{-4}dx=\\=6\cdot \frac{1}{6} x^6 +3\cdot \frac{1}{-3}x^{-3}+C=x^6-x^{-3}+C\)
b.
\(\int \frac{x^4+3x^3+2}{x} dx=\int x^3+3x^2+\frac{2}{x}dx=\int x^3dx +3\int x^2 dx+2\int \frac{1}{x} dx=\\=\frac{1}{4}x^4+3\cdot \frac{1}{3}x^3+2\ln|x|+C=\frac{1}{4}x^4+x^3+2\ln |x|+C\)
: 17 wrz 2010, 19:24
autor: domino21
c.
\(\int \frac{dx}{3x+2}=\left(3x+2=t \\ 3dx=dt \\ dx=\frac{1}{3}dt\right)=\int \frac{\frac{1}{3}dt}{t}=\int \frac{1}{3} \cdot \frac{1}{t} dt=\frac{1}{3} \int \frac{1}{t} dt=\frac{1}{3} \ln |t|+C=\frac{1}{3}\ln |3x+2|+C\)
d.
\(\int \sqrt{5x+1} dx =\left(5x+1=t \\ 5dx=dt \\ dx=\frac{1}{5}dt\right)=\int \sqrt{t} \cdot \frac{1}{5}dt=\frac{1}{5}\int \sqrt{t} dt=\frac{1}{5} \int t^{\frac{1}{2}} dt= \\ =\frac{1}{5} \cdot \frac{1}{\frac{3}{2}}t^{\frac{3}{2}}+C=\frac{2}{15}t\sqrt{t}+C=\frac{2}{15}(5x+1)\sqrt{5x+1}+C\)