Integralen

Antwoorden bij de opgaven

1. 1. ${\displaystyle F\text{'}\left(x\right)=\frac{1}{ln\left(g\right)}\cdot g^x\cdot ln\left(g\right)=g^x}$.
   2. Als ${\displaystyle f\left(x\right)={\text{e}}^x}$ dan is ${\displaystyle F\left(x\right)=\frac{1}{ln\left(\text{e}\right)}\cdot {\text{e}}^x+c={\text{e}}^x+c}$.
   3. ${\displaystyle {\int }_0^{2}h\left(x\right)\text{d}x=[2x+\mathrm{0,5}{\text{e}}^{2x}\cdot 2]\begin{array}{l}2\\ 0\end{array}=[2x+{\text{e}}^{2x}]\begin{array}{l}2\\ 0\end{array}=3+{\text{e}}^4}$.
   4. De oppervlakte van het gebied begrensd door de grafiek van ${\displaystyle h}$, de ${\displaystyle x}$-as en de lijnen ${\displaystyle x=0}$ en ${\displaystyle x=2}$.
2. 1. ${\displaystyle F\text{'}\left(x\right)=\frac{1}{x}}$ als ${\displaystyle x>0}$ en ${\displaystyle F\text{'}\left(x\right)=\frac{1}{x}}$ als .
   2. ${\displaystyle G\left(x\right)=4ln|x+2|+c}$.
   3. ${\displaystyle H\left(x\right)=4ln|2x+4|\cdot \frac{1}{2}+c=2ln|2x+4|+c}$.
3. 1. ${\displaystyle F\text{'}\left(x\right)=1\cdot ln\left(x\right)+x\cdot \frac{1}{x}-1=ln\left(x\right)}$.
   2. ${\displaystyle F\left(x\right)=\frac{1}{ln\left(g\right)}\cdot (xln\left(x\right)-x)+c}$.
4. 1. ${\displaystyle F\left(x\right)={\text{e}}^{x+1}+c}$
   2. ${\displaystyle F\left(x\right)=5x+3ln\left(2\right)\cdot 2^{4x}\cdot \frac{1}{4}=5x+\frac{3}{4}ln\left(2\right)\cdot 2^{4x}}$
   3. ${\displaystyle F\left(x\right)=x^2+ln\mid x\}+c}$
   4. ${\displaystyle F\left(x\right)=\mathrm{1,5}ln(2x+1)+c}$
   5. ${\displaystyle f\left(x\right)=(3xln\left(3x\right)-3x)\cdot \frac{1}{3}+c=xln\left(3x\right)-x+c}$
   6. ${\displaystyle f\left(x\right)=2{\text{e}}^{-4x}\cdot -\frac{1}{4}+c=-\frac{1}{2{\text{e}}^{-4x}}+c}$
5. 1. ${\displaystyle {\int }_0^2\frac{3}{3x+4}\text{d}x=[ln|3x+4|]\begin{array}{l}2\\ 0\end{array}=ln\left(10\right)-ln\left(4\right)}$.
   2. ${\displaystyle {\int }_0^4{\mathrm{0,5}}^{2x-1}\text{d}x=\left[\frac{1}{2}ln\left(\mathrm{0,5}\right){\mathrm{0,5}}^{2x-1}\right]\begin{array}{l}4\\ 0\end{array}=\frac{1}{2}ln\left(\mathrm{0,5}\right){\mathrm{0,5}}^7-\frac{1}{2}ln\left(\mathrm{0,5}\right){\mathrm{0,5}}^{-1}}$.
   3. ${\displaystyle {\int }_1^2\frac{x^4+5x^2}{x^3}\text{d}x={\int }_1^2(x+\frac{5}{x})\text{d}x=[\mathrm{0,5}{x}^2+5ln\left|x\right|]\begin{array}{l}2\\ 1\end{array}=\mathrm{1,5}+5ln\left(2\right)}$.
   4. ${\displaystyle {\int }_{\mathrm{0,25}}^{\text{e}}ln\left(4x\right)\text{d}x=[xln\left(4x\right)-x]\begin{array}{l}\text{e}\\ \mathrm{0,25}\end{array}=\text{e}ln\left(4\text{e}\right)-\text{e}+\mathrm{0,25}}$.
6. 1. ${\displaystyle {\int }_0^2{\text{e}}^x\text{d}x=\left[{\text{e}}^x\right]\begin{array}{l}2\\ 0\end{array}={\text{e}}^2-1}$.
   2. ${\displaystyle {\int }_0^2\pi {\text{e}}^{2x}\text{d}x=\left[\frac{1}{2}\pi {\text{e}}^{2x}\right]\begin{array}{l}2\\ 0\end{array}=\frac{1}{2}\pi ({\text{e}}^4-{\text{e}}^2)}$.
   3. ${\displaystyle \pi \cdot 2^2\cdot 1+{\int }_1^{{\text{e}}^2}\pi {ln}^2\left(y\right)\text{d}y\approx \mathrm{52,71}}$.
   4. ${\displaystyle 1+2+{\text{e}}^2+{\int }_0^2\sqrt{1+{\text{e}}^{2x}}\text{d}x\approx \mathrm{17,18}}$.
7. 1. ${\displaystyle f\left(x\right)=2\frac{1}{2}}$ geeft ${\displaystyle x=\frac{1}{2}\vee x=2}$.
      De oppervlakte is ${\displaystyle 2\frac{1}{2}-{\int }_{\mathrm{0,5}}^2(x+\frac{1}{x})\text{d}x=2\frac{1}{2}-[\frac{1}{2}{x}^2+ln\left(x\right)]\begin{array}{l}2\\ \mathrm{0,5}\end{array}=\frac{5}{8}-2ln\left(2\right)}$.
   2. De inhoud is ${\displaystyle {\int }_{\mathrm{0,5}}^2\pi \cdot {\left(\mathrm{2,5}\right)}^2\text{d}x-{\int }_{\mathrm{0,5}}^2\pi {(x+\frac{1}{x})}^2\text{d}x=\left[\mathrm{6,25}\pi x\right]\begin{array}{l}2\\ \mathrm{0,5}\end{array}-\left[\pi (\frac{1}{3}{x}^3+2ln\left(x\right)-\frac{1}{x})\right]\begin{array}{l}2\\ \mathrm{0,5}\end{array}=\mathrm{7,25}\pi -4ln\left(2\right)-\mathrm{1,5}}$.
   3. De omtrek is ${\displaystyle 1\frac{1}{2}+{\int }_{\mathrm{0,5}}^2\sqrt{1+{(1-\frac{1}{x^2})}^2}\text{d}x\approx \mathrm{3,40}}$.
8. 1. ${\displaystyle [\frac{1}{2}ln|2x+1|]\begin{array}{l}1\\ 0\end{array}=\frac{1}{2}ln\left(3\right)}$.
   2. ${\displaystyle [\frac{1}{2\text{e}+1}\cdot x^{2\text{e}+1}-\frac{1}{2}{\text{e}}^{2x}]\begin{array}{l}1\\ 0\end{array}=\frac{1}{2\text{e}+1}-\frac{1}{2}{\text{e}}^2+\frac{1}{2}}$.
   3. ${\displaystyle [xln\left|4x\right|-x]\begin{array}{l}1\\ \mathrm{0,25}\end{array}=ln\left(4\right)-\mathrm{0,75}}$.
   4. ${\displaystyle [\frac{1}{2}x+2ln\left|x\right|]\begin{array}{l}4\\ 1\end{array}=\mathrm{1,5}+2ln\left(4\right)}$.
   5. ${\displaystyle [2x+\frac{10}{ln\left(10\right)}\cdot {10}^{\mathrm{0,5}x}]\begin{array}{l}1\\ 0\end{array}=2+\frac{10}{ln\left(10\right)}(\sqrt{1}-1)}$.
   6. ${\displaystyle [\frac{1}{ln\left(10\right)\cdot (xln\left|3x\right|-x)}\begin{array}{l}2\\ 1\end{array}=\frac{1}{ln\left(10\right)\cdot (2ln\left(6\right)-1)}}$.
9. 1. ${\displaystyle F\left(x\right)=\mathrm{0,5}{\text{e}}^{-\frac{1}{2}x}+\text{e}x-\mathrm{0,5}}$.
   2. ${\displaystyle F\left(x\right)=x-\frac{1}{ln\left(10\right)\cdot (xln\left|x\right|-x)-1-\frac{1}{ln\left(10\right)}}}$.
10. 1. ${\displaystyle opp\left(V\right)={\int }_2^4\frac{1}{2}x-\frac{2}{x}\text{d}x=[\frac{1}{4}{x}^2-2ln\left|x\right|]\begin{array}{l}4\\ 2\end{array}=3-2ln\left(2\right)}$.
    2. ${\displaystyle l\left(V\right)=2+\mathrm{1,5}+{\int }_2^4\sqrt{1+{(\frac{1}{2}+\frac{1}{x^2})}^2}\text{d}x\approx \mathrm{5,96}}$.
    3. ${\displaystyle I\left(V\right)={\int }_2^4\pi {(\frac{1}{2}x-\frac{2}{x})}^2\text{d}x=\pi {\int }_2^4\frac{1}{4}{x}^2-2+\frac{4}{x^2}\text{d}x=\pi [\frac{1}{12}{x}^3-2x-\frac{4}{x}]\begin{array}{l}4\\ 2\end{array}=1\frac{2}{3}\pi }$.
11. 1. Doen, gebruik de quotiëntregel.
    2. ${\displaystyle opp\left(V\right)={\int }_0^{2}g\left(x\right)\text{d}x=\left[\frac{2{\text{e}}^x}{{\text{e}}^x+1}\right]\begin{array}{l}2\\ 0\end{array}=\frac{2{\text{e}}^2}{{\text{e}}^2+1}-1}$.
    3. ${\displaystyle \left[\frac{2{\text{e}}^x}{{\text{e}}^x+1}\right]\begin{array}{l}a\\ -a\end{array}=\frac{2{\text{e}}^a}{{\text{e}}^a+1}-\frac{2{\text{e}}^{-a}}{{\text{e}}^{-a}+1}}$.
       Herleiden: ${\displaystyle \frac{2{\text{e}}^a}{{\text{e}}^a+1}-\frac{2{\text{e}}^{-a}}{{\text{e}}^{-a}+1}\cdot \frac{{\text{e}}^a}{{\text{e}}^a}=\frac{2{\text{e}}^a-2}{{\text{e}}^a+1}=1}$ geeft ${\displaystyle {\text{e}}^x=3}$ en dus ${\displaystyle x=ln\left(3\right)}$.
12. 1. Differentieer ${\displaystyle F\left(x\right)}$ en laat zien dat ${\displaystyle F\text{'}\left(x\right)=f\left(x\right)}$.
    2. ${\displaystyle opp\left(V_1\right)=[-\mathrm{0,5}{\text{e}}^{-x^2}]\begin{array}{l}2\\ 0\end{array}=\frac{1}{2}-\frac{1}{2{\text{e}}^4}}$.
    3. ${\displaystyle opp\left(V_1\right)=[-\mathrm{0,5}p{\text{e}}^{-x^2}]\begin{array}{l}2\\ 0\end{array}=(\frac{1}{2}-\frac{1}{2{\text{e}}^4})\cdot p=10}$ geeft ${\displaystyle p=\frac{20{\text{e}}^4}{{\text{e}}^4-1}}$.
13. 1. ${\displaystyle [ln\left|x\right|+{\text{e}}^x]\begin{array}{l}2\\ 1\end{array}=2ln\left(2\right)+{\text{e}}^2-\text{e}}$.
    2. ${\displaystyle [x+{\text{e}}^x]\begin{array}{l}1\\ 0\end{array}=\text{e}}$.
    3. ${\displaystyle [2ln|2x+3|]\begin{array}{l}2\\ 1\end{array}=2ln\left(\mathrm{1,4}\right)}$.
    4. ${\displaystyle [x-6ln\left|x\right|-\frac{9}{x}]\begin{array}{l}2\\ 1\end{array}=\mathrm{5,5}-6ln\left(2\right)}$.
    5. ${\displaystyle [(x-1)ln(x-1)-(x-1)]\begin{array}{l}4\\ 2\end{array}=3ln\left(3\right)-2}$.
    6. ${\displaystyle [\frac{1}{3ln\left(2\right)}\cdot 2^{3x}]\begin{array}{l}2\\ 0\end{array}=\frac{2^6-1}{3ln\left(2\right)}}$.
14. ${\displaystyle {\int }_0^6{\text{e}}^{\frac{x}{3}}-2x+5\text{d}x-\frac{1}{2}\cdot 5\cdot \mathrm{2,5}=[3{\text{e}}^{\frac{x}{3}}-x^2+5x]\begin{array}{l}6\\ 0\end{array}-\mathrm{6,25}=3{\text{e}}^2-\mathrm{15,25}}$.
    
15. 1. ${\displaystyle f\text{'}\left(x\right)={\text{e}}^x-{\text{e}}^{-x}=0}$ geeft ${\displaystyle {\text{e}}^{2x}=1}$ en dus ${\displaystyle x=0}$.
       Min.${\displaystyle f\left(0\right)=2}$.
    2. ${\displaystyle f\left(x\right)=4}$ geeft ${\displaystyle {\text{e}}^x+{\text{e}}^{-x}=4}$ en dus ${\displaystyle {\text{e}}^{2x}-4{\text{e}}^x+1=0}$ en ${\displaystyle {\text{e}}^x=\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}}$. Oplossing: .
    3. ${\displaystyle opp\left(V\right)=4\cdot 2ln(2+\sqrt{3})-{\int }_{ln(2-\sqrt{3})}^{ln(2+\sqrt{3})}{\text{e}}^x+{\text{e}}^{-x}\text{d}x=8ln(2+\sqrt{3})-[{\text{e}}^x-{\text{e}}^{-x}]\begin{array}{l}ln(2+\sqrt{3})\\ ln(2-\sqrt{3})\end{array}=8ln(2+\sqrt{3})-4\sqrt{3}}$.