1.

\( \int_{1}^{e} \int_{1}^{\log y \int^{e^{x}} \log z} d z \cdot d x \cdot d y \)

Answer»

\(\int\limits_1^e\int\limits_1^{log y}\int\limits_1^{e^x}logzdz.dx.dy\)

 = \(\int\limits_1^e\int\limits_1^{log y}[zlogz-z]_1^{e^x}dx.dy\)

(\(\because\int log zdz=log z\int1.dz-\int(\frac{d}{dz}log z\int1.dz)dz\) \(=zlogz-\int\frac1z.zdz=zlogz-z\))

 = \(\int\limits_1^e\int\limits_1^{log y}[e^xloge^x-e^x+1]dx.dy\) 

 = \(\int\limits_1^e\int\limits_1^{log y}(xe^x-e^x+1)dx.dy\) 

 = \(\int\limits_1^e[x\int e^x\,dx-\int(\frac{dx}{dx}\int e^xdx)dx-e^x+x]_1^{log y}dy\) 

 = \(\int\limits_1^e(xe^x-e^x-e^x+x)_1^{log y}dy\) 

 = \(\int\limits_1^e(log y.e^{log y}-2e^{log y}+log y-e+2e-1)dy\) 

  = \(\int\limits_1^e(y log y-2y+log y+(e-1))dy\) 

 = [log y\(\int\)y dy -\(\int\)(\(\frac{d}{dy}log y\int ydy\))dy - \(\frac{2y^2}2+y log y-y+(e-1)y]_1^e\) 

 = \([\frac{y^2}2log y-\frac{y^2}4-y^2+y log y+(e-2)y]_1^e\) 

 = \(\frac{e^2}2 log 2-\frac54e^2+elog e+(e-2)e+\frac54-(e-2)\) 

(\(\because log 1 = 0\))

 = \(\frac{e^2}4-\frac54e^2+e+e^2-2e+\frac54-e+2\)

 = \(\frac{e^2}4-2e+\frac{13}4\) = \(\frac14(e^2-8e+13)\)



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