Problemlösning med Integraler och volymintegraler

${x}^{2}+px+q=0$ ${x}_{1,2}=-\frac{p}{2}\pm\sqrt{{\left(\frac{p}{2}\right)}^{2}-q}$ $k=\frac{\bigtriangleup y}{\bigtriangleup x}=\frac{y_2-y_1}{x_2-x_1}$ $k=\frac{y-y}{x-x}$ $f´(x)=\lim _{h\to 0 }\frac{f(x+h)-f(x)}{h}$ $f´(a)=\lim _{h\to 0 }\frac{f(x+a)-f(a)}{h}$ ${a}^{2}+{b}^{2}={c}^{2}$ ${a}^{2}={b}^{2}+{c}^{2}-2bc\cos A$ $V_{\text{kon}}=\frac{\pi r^2h}{3}$ $a^x\cdot a^y=a^{x+y}$ $\frac{a^x}{a^y}=a^{x-y},(a≠0)$ $(a^x)^y=a^{x\cdot y}$ $(a\cdot b)^x=a^x\cdot b^x$ $\frac{a }{b })^x=\frac{a^x}{b^x}$ $a^{-x}=\frac{1}{a^x}$ $\text{lg }A+\text{lg }B=\text{lg }A\cdot B$ $\text{lg }A-\text{lg }B=\text{lg } \frac{A}{B}$ $\text{lg }x\text{ }^p=p \cdot \text{lg } x$ $\sqrt{\frac{(x_1-\overline{x})^{2}+(x_2-\overline{x})^{2}+...+(x_n-\overline{x})^{2}}{n-1}}$ $(a+b)^2=a^2+2ab+b^2$ $(a-b)^2=a^2-2ab+b^2$ $(a+b)(a-b)=a^2-b^2$ $f(x)=k(x-x_1)(x-x_2)$ $x_{sym}=\frac{x_1+x_2}{2}$ $\int _a^bf(x)dx=[F(x)]^b_a=F(b)-F(a)$ $A_\text{rektangel}=\text{basen}\cdot \text{höjden}$ $A_\text{rektangel}= b \cdot h$ $A_\text{triangel}=\frac{\text{basen} \cdot \text{höjden}}{2}$ $A_{\text{triangel}}=\frac{b\cdot h}{2}$ $A_\text{cirkel}=\pi r^2$ $O_\text{cirkel}=2\pi r=\pi d$ $V_{\text{cylinder}}=\pi r^2h$ $V_{\text{klot}}=\frac{4\pi r^3}{3}$ ${z}_{1}\cdot{z}_{2}={r}_{1}\cdot{r}_{2}(cos({v}_{1}+{v}_{2})+i\cdot sin({v}_{1}+{v}_{2}))$ $\frac{z_1}{z_2}=\frac{r_1}{r_2}\left(\text{cos }\left(v_1-v_2\right)+i\text{ sin }\left(v_1-v_2\right)\right)$ $z^n=\left(r\left(\text{cos }v+i\text{ sin }v\right)\right)^n=r^n\left(\text{cos }nv+i\text{ sin }nv\right)$ $y'=f'\left(x\right)\cdot g\left(x\right)+f\left(x\right)\cdot g'\left(x\right)$ $y'=f'\left(g\left(x\right)\right)\cdot g'\left(x\right)$ $y'=\frac{f'\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot g'\left(x\right)}{\left(g\left(x\right)\right)^2}$ $d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$