Lösung 2.1:2a
Aus Online Mathematik Brückenkurs 2
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The integrand in our case consists of two terms in the form <math>x^n</math>, and so we can use the rule | The integrand in our case consists of two terms in the form <math>x^n</math>, and so we can use the rule | ||
| - | {{ | + | {{Abgesetzte Formel||<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 | ||
| - | {{ | + | {{Abgesetzte Formel||<math>F(x) = \frac{x^{2+1}}{2+1} + 3\cdot \frac{x^{3+1}}{3+1}</math>}} |
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 | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\int\limits_{0}^{2} \bigl( x^2+3x^3\bigr)\,dx | \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\\ | &= \Bigl[\ \frac{x^3}{3} + 3\cdot\frac{x^4}{4}\Bigr]_0^2\\ | ||
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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 | 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 | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
F'(x) &= \tfrac{1}{3}\bigl(x^3\bigr)' + \tfrac{3}{4}\bigl(x^4\bigr)'\\[5pt] | 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] | &= \tfrac{1}{3}\cdot 3x^2 + \tfrac{3}{4}\cdot 4x^3\\[5pt] | ||
Version vom 12:58, 10. Mär. 2009
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.
