Lösung 2.1:2a
Aus Online Mathematik Brückenkurs 2
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The foremost difficulty with calculating an integral is finding a primitive function of the integrand. Once we have done that, the integral is calculated as the difference between the primitive function's values in the upper and lower limits of integration. | The foremost difficulty with calculating an integral is finding a primitive function of the integrand. Once we have done that, the integral is calculated as the difference between the primitive function's values in the upper and lower limits of integration. | ||
| - | The integrand in our case consists of two terms in the form | + | The integrand in our case consists of two terms in the form <math>x^n</math>, and so we can use the rule |
| - | <math>x^ | + | |
| - | + | ||
| - | + | ||
| - | + | ||
| + | {{Displayed math||<math>\int x^n\,dx = \frac{x^{n+1}}{n+1}+C</math>}} | ||
on the terms individually to obtain that | on the terms individually to obtain that | ||
| - | + | {{Displayed math||<math>F(x) = \frac{x^{2+1}}{2+1} + 3\cdot \frac{x^{3+1}}{3+1}</math>}} | |
| - | <math>F | + | |
| - | + | ||
is a primitive function of the integrand. | is a primitive function of the integrand. | ||
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The integrand's value is thus | The integrand's value is thus | ||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \int\limits_{0}^{2} \bigl( x^2+3x^3\bigr)\,dx | ||
| + | &= \Bigl[\ \frac{x^3}{3} + 3\cdot\frac{x^4}{4}\Bigr]_0^2\\ | ||
| + | &= \frac{2^3}{3} + 3\cdot\frac{2^4}{4} - \Bigl(\frac{0^3}{3} + 3\cdot\frac{0^4}{4} \Bigr)\\[5pt] | ||
| + | &= \frac{8}{3} + \frac{3\cdot 16}{4}\\[5pt] | ||
| + | &= \frac{44}{3}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| - | <math>\begin{align} | ||
| - | & \int\limits_{0}^{2}{\left( x^{2}+3x^{3} \right)}\,dx=\left[ \frac{x^{3}}{3}+3\centerdot \frac{x^{4}}{4} \right]_{0}^{2} \\ | ||
| - | & =\frac{2^{3}}{3}+3\centerdot \frac{2^{4}}{4}-\left( \frac{0^{3}}{3}+3\centerdot \frac{0^{4}}{4} \right) \\ | ||
| - | & =\frac{8}{3}+\frac{3\centerdot 16}{4}=\frac{44}{3} \\ | ||
| - | \end{align}</math> | ||
| - | |||
| - | NOTE: One way to check that | ||
| - | <math>F\left( x \right)=\frac{1}{3}x^{3}+\frac{3}{4}x^{4}</math> | ||
| - | is a primitive function of the integral is to differentiate | ||
| - | <math>F\left( x \right)</math> | ||
| - | and to see that we obtain | ||
| + | Note: One way to check that <math>F(x) = \tfrac{1}{3}x^3 + \tfrac{3}{4}x^4</math> is a primitive function of the integral is to differentiate <math>F(x)</math> and to see that we obtain | ||
| - | <math>\begin{align} | + | {{Displayed math||<math>\begin{align} |
| - | + | F'(x) &= \tfrac{1}{3}\bigl(x^3\bigr)' + \tfrac{3}{4}\bigl(x^4\bigr)'\\[5pt] | |
| - | & =\ | + | &= \tfrac{1}{3}\cdot 3x^2 + \tfrac{3}{4}\cdot 4x^3\\[5pt] |
| - | \end{align}</math> | + | &= x^2+3x^3 |
| + | \end{align}</math>}} | ||
as the integrand. | as the integrand. | ||
Version vom 12:34, 21. Okt. 2008
The foremost difficulty with calculating an integral is finding a primitive function of the integrand. Once we have done that, the integral is calculated as the difference between the primitive function's values in the upper and lower limits of integration.
The integrand in our case consists of two terms in the form \displaystyle x^n, and so we can use the rule
| \displaystyle \int x^n\,dx = \frac{x^{n+1}}{n+1}+C |
on the terms individually to obtain that
| \displaystyle F(x) = \frac{x^{2+1}}{2+1} + 3\cdot \frac{x^{3+1}}{3+1} |
is a primitive function of the integrand.
The integrand's value is thus
| \displaystyle \begin{align}
\int\limits_{0}^{2} \bigl( x^2+3x^3\bigr)\,dx &= \Bigl[\ \frac{x^3}{3} + 3\cdot\frac{x^4}{4}\Bigr]_0^2\\ &= \frac{2^3}{3} + 3\cdot\frac{2^4}{4} - \Bigl(\frac{0^3}{3} + 3\cdot\frac{0^4}{4} \Bigr)\\[5pt] &= \frac{8}{3} + \frac{3\cdot 16}{4}\\[5pt] &= \frac{44}{3}\,\textrm{.} \end{align} |
Note: One way to check that \displaystyle F(x) = \tfrac{1}{3}x^3 + \tfrac{3}{4}x^4 is a primitive function of the integral is to differentiate \displaystyle F(x) and to see that we obtain
| \displaystyle \begin{align}
F'(x) &= \tfrac{1}{3}\bigl(x^3\bigr)' + \tfrac{3}{4}\bigl(x^4\bigr)'\\[5pt] &= \tfrac{1}{3}\cdot 3x^2 + \tfrac{3}{4}\cdot 4x^3\\[5pt] &= x^2+3x^3 \end{align} |
as the integrand.
