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gabyem

  • 4 years ago

f(t)=t^2(1+t^3)^4 interval [0,2] what is the average value of the function?

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  1. Hunus
    • 4 years ago
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    \[f(t)=t^{2}(1+t^{3})^{4}\] \[\frac{1}{2-0}\int\limits_0^2 t^2(1+t^3)^4dt=\frac{1}{2}\int\limits_0^2 t^2(1+t^3)^4dt\] \[u=t^3\] \[du=3t^2dt\] \[\frac{1}{2}\int\ (1+t^3)^4t^2dt\] \[t^2dt=\frac{du}{3}\] \[\frac{1}{2}\int\limits \frac{(1+u)^4du}{3}=\frac{1}{6}\int\limits (1+u)^4du\] \[\frac{1}{6}\int\limits (1+u)^4du=\frac{1}{6}[\frac{(1+u)^5}{5}]\] \[\frac{1}{30}[(1+t^3)]_0^2=\frac{1}{30}[(1+8)-(1)]=\frac{4}{15}\]

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