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Evaluate the integral $\int_0^{1+i}(x^2 - iy)dz$ along the path y = x.

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Evaluate the integral $\int_0^{1+i}(x^2 - iy)dz$ along the path y = x.

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written 3.8 years ago by | modified 3.6 years ago by |

If the curve is not a closed curve then to evaluate the value of integral we use Cartesian form of z i.e.

$z = x + iy i.e. dz = dx + idy$ , the path of the integration is the line y = x

Putting above values in the given integral ,value of the integral along the line y=x i.e. dy =dx is given by

$I = ∫_0^1(x^2-ix) \ (dx + idx)$

= $ (1+i) ∫_0^1(x^2-ix)dx$

$= (1+i) [ \frac{x^3}{3}-i \frac{x^2}{2}]_0^1 = (1+i) [ \frac{1}{3}-\frac{i}{2}] $

= $ \frac{(1+i)(2-3i)}{6} = \frac{(2-3i+2i+3)}{6} = \frac{(5-i)}{6}$

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