2.1 Übungen
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
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|width="50%"| <math>\displaystyle\int_{-1}^{2}|x| \, dx</math> | |width="50%"| <math>\displaystyle\int_{-1}^{2}|x| \, dx</math> | ||
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| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 2.1:1|Solution a|Solution 2.1:1a|Solution b|Solution 2.1:1b|Solution c|Solution 2.1:1c|Solution d|Solution 2.1:1d}} |
===Übung 2.1:2=== | ===Übung 2.1:2=== | ||
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|width="50%"| <math>\displaystyle\int_{1}^{4} \displaystyle\frac{\sqrt{x}}{x^2}\, dx</math> | |width="50%"| <math>\displaystyle\int_{1}^{4} \displaystyle\frac{\sqrt{x}}{x^2}\, dx</math> | ||
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| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 2.1:2|Solution a|Solution 2.1:2a|Solution b|Solution 2.1:2b|Solution c|Solution 2.1:2c|Solution d|Solution 2.1:2d}} |
===Übung 2.1:3=== | ===Übung 2.1:3=== | ||
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|width="50%"| <math>\displaystyle\int \displaystyle\frac{x^2+1}{x}\, dx</math> | |width="50%"| <math>\displaystyle\int \displaystyle\frac{x^2+1}{x}\, dx</math> | ||
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| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 2.1:3|Solution a|Solution 2.1:3a|Solution b|Solution 2.1:3b|Solution c|Solution 2.1:3c|Solution d|Solution 2.1:3d}} |
===Übung 2.1:4=== | ===Übung 2.1:4=== | ||
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|width="100%"| Calculate the area of the region given by the inequality, <math>x^2\le y\le x+2</math>. | |width="100%"| Calculate the area of the region given by the inequality, <math>x^2\le y\le x+2</math>. | ||
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| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 2.1:4|Solution a|Solution 2.1:4a|Solution b|Solution 2.1:4b|Solution c|Solution 2.1:4c|Solution d|Solution 2.1:4d|Solution e|Solution 2.1:4e}} |
===Übung 2.1:5=== | ===Übung 2.1:5=== | ||
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|width="100%"| <math>\displaystyle \int \sin^2 x\ dx\quad</math> (Hint: rewrite the integrand using a trigonometric formula) | |width="100%"| <math>\displaystyle \int \sin^2 x\ dx\quad</math> (Hint: rewrite the integrand using a trigonometric formula) | ||
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| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 2.1:5|Solution a|Solution 2.1:5a|Solution b|Solution 2.1:5b}} |
Version vom 13:29, 10. Mär. 2009
| Theorie | Übungen |
Übung 2.1:1
Interpret each integral as an area, and determine its value.
| a) | \displaystyle \displaystyle\int_{-1}^{2} 2\, dx | b) | \displaystyle \displaystyle\int_{0}^{1} (2x+1)\, dx |
| c) | \displaystyle \displaystyle \int_{0}^{2} (3-2x)\, dx | d) | \displaystyle \displaystyle\int_{-1}^{2}|x| \, dx |
Antwort
Laden...
Solution a
Solution b
Solution c
Solution d
Übung 2.1:2
Calculate the integrals
| a) | \displaystyle \displaystyle\int_{0}^{2} (x^2+3x^3)\, dx | b) | \displaystyle \displaystyle\int_{-1}^{2} (x-2)(x+1)\, dx |
| c) | \displaystyle \displaystyle\int_{4}^{9} \left(\sqrt{x} - \displaystyle\frac{1}{\sqrt{x}}\right)\, dx | d) | \displaystyle \displaystyle\int_{1}^{4} \displaystyle\frac{\sqrt{x}}{x^2}\, dx |
Antwort
Solution a
Solution b
Solution c
Solution d
Übung 2.1:3
Calculate the integrals
| a) | \displaystyle \displaystyle\int \sin x\, dx | b) | \displaystyle \displaystyle\int 2\sin x \cos x\, dx |
| c) | \displaystyle \displaystyle\int e^{2x}(e^x+1)\, dx | d) | \displaystyle \displaystyle\int \displaystyle\frac{x^2+1}{x}\, dx |
Antwort
Solution a
Solution b
Solution c
Solution d
Übung 2.1:4
| a) | Calculate the area between the curve \displaystyle y=\sin x and the \displaystyle x-axis when \displaystyle 0\le x \le \frac{5\pi}{4}. |
| b) | Calculate the area under the curve \displaystyle y=-x^2+2x+2 and above the \displaystyle x-axis. |
| c) | Calculate the area of the finite region between the curves \displaystyle y=\frac{1}{4}x^2+2 and \displaystyle y=8-\frac{1}{8}x^2 (Swedish A-level 1965). |
| d) | Calculate the area of the finite region enclosed by the curves \displaystyle y=x+2, y=1 and \displaystyle y=\frac{1}{x}. |
| e) | Calculate the area of the region given by the inequality, \displaystyle x^2\le y\le x+2. |
Antwort
Solution a
Solution b
Solution c
Solution d
Solution e
Übung 2.1:5
Calculate the integral
| a) | \displaystyle \displaystyle \int \displaystyle\frac{dx}{\sqrt{x+9}-\sqrt{x}}\quad (Hint: multiply the top and bottom by the conjugate of the denominator) |
| b) | \displaystyle \displaystyle \int \sin^2 x\ dx\quad (Hint: rewrite the integrand using a trigonometric formula) |
