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Determine a antiderivada: \[\int (6e^{3r} +12r^3 -5)\; dr\]
xmhintEsta é uma pergunta sobre antiderivadas. Ela pede a função mais geral \(f(r)\) tal
que \(f'(r) = 6e^{3r}+12r^3-5\). Para resolvê-la, você precisa usar tanto a regra da exponencial quanto a regra
da potência para antiderivadas.
xmhintA regra da exponencial para antiderivadas diz que \(\int e^u \; du = e^u+C\) e, neste caso, com \(u= 3r\), \(\frac {du}{dr} = 3\), de
modo que \(du = 3dr\) e \[\int e^{u}\; du = \int e^{3r} 3\; dr\] e, portanto, \[\int 6e^{3r}\; dr = 2 \int \left (3e^{3r}\right )\; dr = 2\int e^u\; du = 2e^u +C = 2e^{3r}+C.\]
A regra da potência para antiderivadas diz que \[\int r^m \; dr = \frac {r^{m+1}}{m+1}+C\] sempre que \(m\neq -1\). Neste problema, temos
um termo com \(m=3\) e um termo com \(m=0\), de modo que a regra da potência se aplica e \[\int 12r^3-5\; dr = 12\left (\frac {r^4}{4}\right )-5r+C = 3r^4-5r+C.\]
Juntando tudo, \[\int 6e^{3r}+12r^3-5\; dr = 2e^{3r}+3r^4-5r+C.\]
\(18e^{3r}+36r^2\) \(3e^{3r}+4r^4-5r+C\) \(3e^{3r}+4r^4-5r\) \(2e^{3r}+3r^4-5r\) \(2e^{3r}+3r^4-5r+C\)