Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 6

Year : MAY 2015

$$Let \space I=\int\limits_0^{\log 2}\int\limits_0^x \int\limits_0^{x+\log y} e^{x+y+z} dz\space dy\space dx$$ Integrating w.r.t z $$\therefore I= \int\limits_0^{\log 2}\int\limits_0^x e^xe^y[e^z]_0^{x+\log y}dy\space dx$$ $ =\int\limits_0^{\log 2}\int\limits_0^x e^xe^y[e^{x+\log y}-e^0]dy\space dx\\ =\int\limits_0^{\log 2}\int\limits_0^xe^xe^y[e^x.e^{\log y}-1]dy\space dx\\ =\int\limits_0^{\log 2}\int\limits_0^x[e^{2x}e^yy-e^xe^y]dy\space dx$

Integrating w.r.t y

$$=\int\limits_0^{\log 2}[e^{2x}(y.e^y-1.e^y)-e^xe^y]_0^xdx$$

$= \int\limits_0^{\log 2}[e^{2x}(xe^x-e^x)-e^x.e^x]-[e^{2x}(0-e^0)-e^x.e^0]dx\\ =\int\limits_0^{\log 2} (e^{3x}.x-e^{3x}-e^{2x}+e^{2x}+e^x) dx\\ =\int\limits_0^{\log 2}[e^{3x}(x-1)+e^{x}]dx\\ =\Bigg[(x-1) \dfrac {e^{3x}}3- \int\limits_0^{\log 2}1\dfrac {e^{3x}}3dx+e^x\Bigg]_0^{\log 2}\\ =\Bigg[\dfrac {(x-)}3e^{3x}- \dfrac {e^{3x}}{3\times 3}+e^{x}\Bigg]_0^{\log 2}\\ =\Bigg[\dfrac {(\log 2-1)}3e^{3\log2}-\dfrac {e^{3\log 2}}9 +e^{\log 2}\Bigg]-\Bigg[\dfrac {0-1}3e^0- \dfrac {e^0}9 + e^0\Bigg]\\ = \Bigg[ \dfrac {(\log 2-1)}3 e^{\log 8}-\dfrac {e^{\log 8}}9 +2\Bigg]-\Bigg[\dfrac {-1}3-\dfrac 19 +1\Bigg] \\ =\dfrac 83(\log 2-1)-\dfrac {-8}9 +2 -\Bigg[\dfrac 59\Bigg] \\ =\dfrac 83\log 2-\dfrac 83+\dfrac 59\\ =\dfrac 83\log 2-\dfrac {-19}9$