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Lösung 2.1:2a

Aus Online Mathematik Brückenkurs 2

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K (Lösning 2.1:2a moved to Solution 2.1:2a: Robot: moved page)
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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.
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The integrand in our case consists of two terms in the form
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<math>x^{n}</math>, and so we can use the rule
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<math>\int{x^{n}\,dx=\frac{x^{n+1}}{n+1}}+C</math>
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on the terms individually to obtain that
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<math>F\left( x \right)=\frac{x^{2+1}}{2+1}+3\centerdot \frac{x^{3+1}}{3+1}</math>
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is a primitive function of the integrand.
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The integrand's value is thus
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<math>\begin{align}
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& \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} \\
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& =\frac{2^{3}}{3}+3\centerdot \frac{2^{4}}{4}-\left( \frac{0^{3}}{3}+3\centerdot \frac{0^{4}}{4} \right) \\
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& =\frac{8}{3}+\frac{3\centerdot 16}{4}=\frac{44}{3} \\
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\end{align}</math>
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NOTE: One way to check that
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<math>F\left( x \right)=\frac{1}{3}x^{3}+\frac{3}{4}x^{4}</math>
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is a primitive function of the integral is to differentiate
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<math>F\left( x \right)</math>
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and to see that we obtain
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<math>\begin{align}
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& {F}'\left( x \right)=\frac{1}{3}\left( x^{3} \right)^{\prime }+\frac{3}{4}\left( x^{4} \right)^{\prime } \\
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& =\frac{1}{3}\centerdot 3x^{2}+\frac{3}{4}\centerdot 4x^{3}=x^{2}+3x^{3} \\
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\end{align}</math>
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as the integrand.

Version vom 12:44, 17. 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 xn, and so we can use the rule


xndx=xn+1n+1+C 


on the terms individually to obtain that


Fx=x2+12+1+3x3+13+1 


is a primitive function of the integrand.

The integrand's value is thus


20x2+3x3dx=3x3+34x420=323+3424−303+3404=38+4316=344

NOTE: One way to check that Fx=31x3+43x4  is a primitive function of the integral is to differentiate Fx  and to see that we obtain


Fx=31x3+43x4=313x2+434x3=x2+3x3

as the integrand.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:2a“