Question

$\displaystyle \sf\int_{0}^{2} \int_{ 0 }^{{z}^{2}} \int_{ 0 }^{{y - z}^{}} (2x - y) \: dxdydz$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle \sf\int_{0}^{2} \int_{ 0 }^{{z}^{2}} \int_{ 0 }^{{y - z}^{}} (2x - y) \: dxdydz$

Since, in the first integral, the limits are free from variable x, so we have to first integrate with respect to x.

So,

$\rm \:  =  \: \displaystyle \sf\int_{0}^{2} \int_{ 0 }^{{z}^{2}} \bigg(\int_{ 0 }^{{y - z}^{}} (2x - y) \: dx\bigg)dydz$

We know,

$\rm :\longmapsto\:\boxed{\tt{ \: \: \displaystyle \sf\int \: {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: \: }}$

and

$\rm :\longmapsto\:\boxed{\tt{ \: \: \displaystyle \sf\int \: k \: dx \: = \: kx + c \: \: }}$

So, using these Identities, we get

$\rm \:  =  \: \displaystyle \sf\int_{0}^{2} \int_{ 0 }^{{z}^{2}}\bigg[ {x}^{2} - xy \bigg]_{0}^{y - z} \:dydz$

$\rm \:  =  \: \displaystyle \sf\int_{0}^{2} \int_{ 0 }^{{z}^{2}}\bigg[ {(y - z)}^{2} - y(y - z) \bigg]\: dydz$

$\rm \:  =  \: \displaystyle \sf\int_{0}^{2} \int_{ 0 }^{{z}^{2}}\bigg[ {y}^{2} + {z}^{2} - 2yz - {y}^{2} + yz \bigg]\: dydz$

$\rm \:  =  \: \displaystyle \sf\int_{0}^{2} \int_{ 0 }^{{z}^{2}}\bigg[{z}^{2} -yz \bigg]\: dydz$

Now, the inner integral is free from variable y, so we have to integrate with respect to y, we get

$\rm \:  =  \: \displaystyle \sf\int_{0}^{2} \bigg[{z}^{2}y -z \frac{ {y}^{2} }{2} \bigg]_{0}^{ {z}^{2} }\:dz$

$\rm \:  =  \: \displaystyle \sf\int_{0}^{2} \bigg[{z}^{4} -\frac{ {z}^{5} }{2} \bigg] \:dz$

$\rm \:  =  \: \bigg[\dfrac{ {z}^{5} }{5} - \dfrac{ {z}^{6} }{12} \bigg]_{0}^{2}$

$\rm \:  =  \: \dfrac{32}{5} - \dfrac{64}{12}$

$\rm \:  =  \: \dfrac{32}{5} - \dfrac{32}{6}$

$\rm \:  =  \:32 \bigg[\dfrac{1}{5} - \dfrac{1}{6} \bigg]$

$\rm \:  =  \:32 \bigg[\dfrac{6 - 5}{30} \bigg]$

$\rm \:  =  \:16 \bigg[\dfrac{1}{15} \bigg]$

$\rm \:  =  \: \dfrac{16}{15}$

Hence,

$\purple{\rm\implies \:\boxed{\tt{ \displaystyle \sf\int_{0}^{2} \int_{ 0 }^{{z}^{2}} \int_{ 0 }^{{y - z}^{}} (2x - y) \: dxdydz = \frac{16}{15}}}}$

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ADDITIONAL INFORMATION

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle \sf\int_{0}^{2} \int_{ 0 }^{{z}^{2}} \int_{ 0 }^{{y - z}^{}} (2x - y) \: dxdydz$

Since, in the first integral, the limits are free from variable x, so we have to first integrate with respect to x.

So,

$\rm \:  =  \: \displaystyle \sf\int_{0}^{2} \int_{ 0 }^{{z}^{2}} \bigg(\int_{ 0 }^{{y - z}^{}} (2x - y) \: dx\bigg)dydz$

We know,

$\rm :\longmapsto\:\boxed{\tt{ \: \: \displaystyle \sf\int \: {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: \: }}$

and

$\rm :\longmapsto\:\boxed{\tt{ \: \: \displaystyle \sf\int \: k \: dx \: = \: kx + c \: \: }}$

So, using these Identities, we get

$\rm \:  =  \: \displaystyle \sf\int_{0}^{2} \int_{ 0 }^{{z}^{2}}\bigg[ {x}^{2} - xy \bigg]_{0}^{y - z} \:dydz$

$\rm \:  =  \: \displaystyle \sf\int_{0}^{2} \int_{ 0 }^{{z}^{2}}\bigg[ {(y - z)}^{2} - y(y - z) \bigg]\: dydz$

$\rm \:  =  \: \displaystyle \sf\int_{0}^{2} \int_{ 0 }^{{z}^{2}}\bigg[ {y}^{2} + {z}^{2} - 2yz - {y}^{2} + yz \bigg]\: dydz$

$\rm \:  =  \: \displaystyle \sf\int_{0}^{2} \int_{ 0 }^{{z}^{2}}\bigg[{z}^{2} -yz \bigg]\: dydz$

Now, the inner integral is free from variable y, so we have to integrate with respect to y, we get

$\rm \:  =  \: \displaystyle \sf\int_{0}^{2} \bigg[{z}^{2}y -z \frac{ {y}^{2} }{2} \bigg]_{0}^{ {z}^{2} }\:dz$

$\rm \:  =  \: \displaystyle \sf\int_{0}^{2} \bigg[{z}^{4} -\frac{ {z}^{5} }{2} \bigg] \:dz$

$\rm \:  =  \: \bigg[\dfrac{ {z}^{5} }{5} - \dfrac{ {z}^{6} }{12} \bigg]_{0}^{2}$

$\rm \:  =  \: \dfrac{32}{5} - \dfrac{64}{12}$

$\rm \:  =  \: \dfrac{32}{5} - \dfrac{32}{6}$

$\rm \:  =  \:32 \bigg[\dfrac{1}{5} - \dfrac{1}{6} \bigg]$

$\rm \:  =  \:32 \bigg[\dfrac{6 - 5}{30} \bigg]$

$\rm \:  =  \:16 \bigg[\dfrac{1}{15} \bigg]$

$\rm \:  =  \: \dfrac{16}{15}$

Hence,

$\pink{\rm\implies \:\boxed{\tt{ \displaystyle \sf\int_{0}^{2} \int_{ 0 }^{{z}^{2}} \int_{ 0 }^{{y - z}^{}} (2x - y) \: dxdydz = \frac{16}{15}}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

ADDITIONAL INFORMATION

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$