Całki cz.2

[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) \(y=x^3-6x^2+9x-5\)

[domino21]

zad 3.

\(y=x^3-6x^2+9x-5, \ x\in R
y'=3x^2-12x+9=3(x-3)(x-1), \ x\in R\)

\(f'(x)=0 \ \Leftrightarrow \ 3(x-1)(x-3)=0 \ \Rightarrow \ x=1 \ \vee \ x=3
f'(x)>0 \ \Leftrightarrow \ 3(x-1)(x-3)>0 \ \Rightarrow \ x\in (-\infty;1) \cup (3;+\infty)
f'(x)<0 \ \Leftrightarrow \ 3(x-1)(x-3)<0 \ \Rightarrow \ x\in (1;3)\)

\(f_{max}=f(1)=1-6+9-5=-1
f_{min}=f(3)=27-54+27-5=-5\)

[domino21]

zad 2.

\(\int \sqrt{x+4} dx =\left( x+4=t \\ dx=dt \right)= \int \sqrt{t} dt = \int t^{\frac{1}{2}} dt =\frac{1}{\frac{3}{2}} \cdot t^{\frac{3}{2}}+C=\frac{2}{3} t\sqrt{t} +C=\frac{2}{3}(x+4)\sqrt{x+4}+C\)

\(\int_{0}^{1} \sqrt{x+4} dx=\left[\frac{2}{3}(x+4)\sqrt{x+4}\right] ^1_0=\frac{2}{3} (1+4)\sqrt{1+4}-\frac{2}{3}(0+4)\sqrt{0+4}=\frac{10\sqrt{5}}{3}-\frac{16}{3}=\frac{10\sqrt{5}-16}{3}\)

[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.
\(\int \frac{x^4+3x^3+2}{x} dx=\int x^3+3x^2+\frac{2}{x}dx=\int x^3dx +3\int x^2 dx+2\int \frac{1}{x} dx=\\=\frac{1}{4}x^4+3\cdot \frac{1}{3}x^3+2\ln|x|+C=\frac{1}{4}x^4+x^3+2\ln |x|+C\)

[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.
\(\int \sqrt{5x+1} dx =\left(5x+1=t \\ 5dx=dt \\ dx=\frac{1}{5}dt\right)=\int \sqrt{t} \cdot \frac{1}{5}dt=\frac{1}{5}\int \sqrt{t} dt=\frac{1}{5} \int t^{\frac{1}{2}} dt= \\ =\frac{1}{5} \cdot \frac{1}{\frac{3}{2}}t^{\frac{3}{2}}+C=\frac{2}{15}t\sqrt{t}+C=\frac{2}{15}(5x+1)\sqrt{5x+1}+C\)