Question

Find the intergration of the following​

Answer 1

Given Question :-

Evaluate the following

$\: \: \: \: (a) \: \: \displaystyle\int\rm {y}^{2} \sqrt[3]{y} \: dy$

$\: \: \: \: (b) \: \: \displaystyle\int^{9}_{4}\rm ( \sqrt{x} + \frac{1}{ \sqrt[3]{x} } ) \: dx$

CALCULATIONS

FORMULA USED

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

$\red{\rm :\longmapsto\:\boxed{\tt{ {x}^{m} \times {x}^{n} = {x}^{m + n} \: }}}$

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

$\: \: \: \: (a) \: \: \displaystyle\int\rm {y}^{2} \sqrt[3]{y} \: dy$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm {y}^{2} \times {\bigg(y\bigg) }^{\dfrac{1}{3} } \: dy$

$\rm \:  =  \: \displaystyle\int\rm {\bigg(y\bigg) }^{2 + \dfrac{1}{3} } \: dy$

$\rm \:  =  \: \displaystyle\int\rm {\bigg(y\bigg) }^{\dfrac{6 + 1}{3} } \: dy$

$\rm \:  =  \: \displaystyle\int\rm {\bigg(y\bigg) }^{\dfrac{7}{3} } \: dy$

$\rm \:  =  \: \dfrac{ {\bigg(y\bigg) }^{\dfrac{7}{3} + 1} }{\dfrac{7}{3} + 1 } + c$

$\rm \:  =  \: \dfrac{ {\bigg(y\bigg) }^{\dfrac{7 + 3}{3}} }{\dfrac{7 + 3}{3}} + c$

$\rm \:  =  \: \dfrac{ {\bigg(y\bigg) }^{\dfrac{10}{3}} }{\dfrac{10}{3}} + c$

$\rm \:  =  \:\dfrac{3}{10} {\bigg(y\bigg) }^{\dfrac{10}{3}} + c$

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

$\: \: \: \: (b) \: \: \displaystyle\int^{9}_{4}\rm ( \sqrt{x} + \frac{1}{ \sqrt[3]{x} } ) \: dx$

can be rewritten as

$\rm \:  =  \: \displaystyle\int^{9}_{4} \rm \: \bigg[ {\bigg(x\bigg) }^{\dfrac{1}{2}} + {\bigg(x\bigg) }^{\dfrac{ - 1}{3} } \bigg] \: dx$

$\rm \:  =  \: \dfrac{ {\bigg(x\bigg) }^{\dfrac{1}{2} + 1} }{\dfrac{1}{2} + 1 }\bigg]^{9}_{4} \: \: + \: \: \dfrac{ {\bigg(x\bigg) }^{\dfrac{ - 1}{3} + 1} }{\dfrac{ - 1}{3} + 1 }\bigg]^{9}_{4}$

$\rm \:  =  \: \dfrac{ {\bigg(x\bigg) }^{\dfrac{3}{2}} }{\dfrac{3}{2}}\bigg]^{9}_{4} \: \: + \: \: \dfrac{ {\bigg(x\bigg) }^{\dfrac{ - 2}{3}} }{\dfrac{ - 2}{3}}\bigg]^{9}_{4}$

$\rm \:  =  \:\dfrac{2}{3} {\bigg(x\bigg) }^{\dfrac{3}{2}}\bigg]^{4}_{9} \: \: - \: \: \dfrac{3}{2} {\bigg(x\bigg) }^{\dfrac{ - 2}{3}}\bigg]^{9}_{4}$

$\rm \:  =  \:\dfrac{2}{3}\bigg[{\bigg(9\bigg) }^{\dfrac{3}{2}} - {\bigg(4\bigg) }^{\dfrac{3}{2}}\bigg] - \dfrac{3}{2}\bigg[{\bigg(9\bigg) }^{\dfrac{ - 2}{3}} - {\bigg(4\bigg) }^{\dfrac{ - 2}{3}}\bigg]$

$\rm \:  =  \:\dfrac{2}{3}\bigg[{\bigg( {3}^{2} \bigg) }^{\dfrac{3}{2}} - {\bigg( {2}^{2} \bigg) }^{\dfrac{3}{2}}\bigg] - \dfrac{3}{2}\bigg[{\bigg( {3}^{2} \bigg) }^{\dfrac{ - 2}{3}} - {\bigg( {2}^{2} \bigg) }^{\dfrac{ - 2}{3}}\bigg]$

$\rm \:  =  \:\dfrac{2}{3}\bigg[27 - 8\bigg] - \dfrac{3}{2}\bigg[{\bigg(3\bigg) }^{\dfrac{ - 4}{3}} - {\bigg(2\bigg) }^{\dfrac{ - 4}{3}}\bigg]$

$\rm \:  =  \:\dfrac{2}{3}\bigg[19\bigg] - \dfrac{3}{2}\bigg[{\bigg(3\bigg) }^{\dfrac{ - 4}{3}} - {\bigg(2\bigg) }^{\dfrac{ - 4}{3}}\bigg]$

$\rm \:  =  \:\dfrac{38}{3} - \dfrac{3}{2}\bigg[{\bigg(3\bigg) }^{\dfrac{ - 4}{3}} - {\bigg(2\bigg) }^{\dfrac{ - 4}{3}}\bigg]$

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