Lös I = ∫ (from 0 to 1) of (4z+2)e^2t wrt x

Lös ut I

I=2\left(2z+1\right)e^{2t}

Lös ut t

\left\{\begin{matrix}t=\frac{\ln(\frac{I}{2z+1})-\ln(2)}{2}\text{, }&\left(I<0\text{ and }z<-\frac{1}{2}\right)\text{ or }\left(I>0\text{ and }z>-\frac{1}{2}\right)\\t\in \mathrm{R}\text{, }&I=0\text{ and }z=-\frac{1}{2}\end{matrix}\right,

Frågesport

Integration

5 problem som liknar:

I = \int _ { 0 } ^ { 1 } ( 4 z + 2 ) e ^ { 2 t } d x

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