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- Thread starter dwsmith
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- Jan 26, 2012

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$$\int z e^{z^{2}}\,dz=\frac{1}{2}\int 2z e^{z^{2}}\,dz.$$How can I find the primitive of $\int_{\gamma}ze^{z^2}dz$ from $i$ to $2-i$?

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So $\left(\frac{e^{z^2}}{2}\right)'=\int ze^{z^2}dz$ Then to solve the integral I just integrate g'(z) right?$$\int z e^{z^{2}}\,dz=\frac{1}{2}\int 2z e^{z^{2}}\,dz.$$

Can you finish?

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- Jan 26, 2012

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Actually, I would have said thatSo $\left(\frac{e^{z^2}}{2}\right)'=\int ze^{z^2}dz$ Then to solve the integral I just integrate g'(z) right?

$$\left(\frac{e^{z^{2}}}{2}\right)'=ze^{z^{2}}.$$

Then just use the Fundamenal Theorem of the Calculus, which works because your function is analytic.