Question

Evaluate the integral

$\displaystyle\sf \int_{0}^{1} \int_{ {y}^{2} }^{1 - y } \int_{0}^{1 - x} xdzdxdy$
​

Answer 1

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

$\rm :\longmapsto\:\displaystyle\sf \int_{0}^{1} \int_{ {y}^{2} }^{1 - y } \int_{0}^{1 - x} xdzdxdy$

We have integrate with respect to z

So,

$\rm \:  =  \: \displaystyle\sf \int_{0}^{1} \int_{ {y}^{2} }^{1 - y }\bigg[ \int_{0}^{1 - x} xdz\bigg]dxdy$

$\rm \:  =  \: \displaystyle\sf \int_{0}^{1} \int_{ {y}^{2} }^{1 - y }\bigg[xz\bigg]_{0}^{1 - x}dxdy$

$\rm \:  =  \: \displaystyle\sf \int_{0}^{1} \int_{ {y}^{2} }^{1 - y }\bigg[x(1 - x)\bigg]dxdy$

$\rm \:  =  \: \displaystyle\sf \int_{0}^{1} \int_{ {y}^{2} }^{1 - y }\bigg[x - {x}^{2} \bigg]dxdy$

Now we integrate with respect to x

$\rm \:  =  \: \displaystyle\sf \int_{0}^{1} \bigg[\dfrac{ {x}^{2} }{2} - \dfrac{ {x}^{3} }{3} \bigg]_{ {y}^{2} }^{1 - y}dy$

$\rm \:  =  \: \displaystyle\sf \int_{0}^{1} \bigg[\dfrac{ {(1 - y)}^{2} }{2} - \dfrac{ {(1 - y)}^{3} }{3} - \dfrac{ {y}^{4} }{2} + \dfrac{ {y}^{6} }{3} \bigg]dy$

$\rm \:  =  \: \bigg[ - \dfrac{ {(1 - y)}^{3} }{6} + \dfrac{ {(1 - y)}^{4} }{12} - \dfrac{ {y}^{5} }{10} + \dfrac{ {y}^{7} }{21} \bigg]_{0}^{1}$

$\rm \:  =  \: \bigg[ - \dfrac{1}{10} + \dfrac{1}{21} \bigg] - \bigg[ - \dfrac{1}{6} + \dfrac{1}{12} \bigg]$

$\rm \:  =  \: \bigg[ \dfrac{ - 21 + 10}{210}\bigg] - \bigg[ \dfrac{ - 2 + 1}{12} \bigg]$

$\rm \:  =  \: \bigg[ \dfrac{ - 11}{210}\bigg] - \bigg[ \dfrac{ - 1}{12} \bigg]$

$\rm \:  =  \: \dfrac{1}{12} - \dfrac{11}{210}$

$\rm \:  =  \: \dfrac{210 - 132}{12 \times 210}$

$\rm \:  =  \: \dfrac{78}{12 \times 210}$

$\rm \:  =  \: \dfrac{39}{6 \times 210}$

$\rm \:  =  \: \dfrac{13}{2\times 210}$

$\rm \:  =  \: \dfrac{13}{420}$

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More to know

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

Answer 2

Answer:

$\huge \frac{13}{420}$

Step-by-step explanation:

$\int_{0}^{1} \int_{ {y}^{2} }^{1 - y } \int_{0}^{1 - x} xdzdxdy \\ \\ = \int_{0}^{1} \int_{ {y}^{2} }^{1 - y } \int_{0}^{1 - x} \{ xz \}dxdy \\ \\ = \int_{0}^{1} \int_{ {y}^{2}} ^{1 - y} x({1 - x} )dxdy \\ \\ = \int_{0}^{1} \int_{ {y}^{2} }^{1 - y }( \frac{ {x}^{2} }{2} - \frac{ {x}^{3} }{3} )dy \\ \\ = \int_{0}^{1} (\frac{ {(1 - y)}^{2} }{2} - \frac{ {(1 - y)}^{3} }{3} - \frac{ {(y {}^{2} )}^{2} }{2} + \frac{ {( {y}^{2}) }^{3} }{3} )dy dy \\ \\using \: integral \: property \: in \: first \: two \: terms \\ \int _{0} ^{1} f(y) \: dy = \int _{0} ^{1} f(1 - y) \: dy \\ \\ = \ \int _{0} ^{1} ( \frac{ {y}^{2} }{2} - \frac{ {y}^{3} }{3} - \frac{ {y}^{4} }{2} + \frac{ {y}^{6} }{3} )dy \\ \\ = \frac{ {y}^{3} }{6} - \frac{ {y}^{4} }{12} - \frac{ {y}^{5} }{10} + \frac{ {y}^{7} }{21} \: \: \{ \: lim \: \: 0 \: to \: 1 \\ \\ = \frac{1}{6} - \frac{1}{12} - \frac{1}{10} + \frac{1}{21} \\ \\ = \frac{70 - 35 - 42 + 20}{420} \\ \\ = \frac{13}{420}$