Question

$\large\displaystyle \sf \red{ \int _{0}^{2} \int_{0}^{ \frac{1}{2} x}(2x + 3y) \: dy \: dx}$​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle \rm { \int _{0}^{2} \int_{0}^{ \dfrac{x}{2}} \: (2x + 3y) \: dy \: dx}$

We have to integrate first with respect to y,

So, we get

$\rm \:  =  \: \displaystyle \rm { \int _{0}^{2} \bigg(\int_{0}^{ \dfrac{x}{2}} \: (2x + 3y) \: dy\bigg) \: dx}$

We know,

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

and

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

So, using this we get

$\rm \:  =  \: \displaystyle \rm{ \int _{0}^{2} \bigg[2xy + 3\dfrac{ {y}^{2} }{2} \bigg]_{0}^{ \dfrac{x}{2}} dx}$

$\rm \:  =  \: \displaystyle \rm{ \int _{0}^{2} \bigg[ {x}^{2} + \dfrac{3{x}^{2} }{8} \bigg] dx}$

$\rm \:  =  \: \displaystyle \rm{ \int _{0}^{2} \bigg[\dfrac{8 {x}^{2} + 3{x}^{2} }{8} \bigg] dx}$

$\rm \:  =  \: \displaystyle \rm{ \int _{0}^{2} \bigg[\dfrac{11{x}^{2} }{8} \bigg] dx}$

$\rm \:  =  \: \dfrac{11}{8} \displaystyle \rm{ \int _{0}^{2} {x}^{2} dx}$

$\rm \:  =  \: \dfrac{11}{8} \times \bigg[\dfrac{ {x}^{3} }{3} \bigg]_{0}^{2}$

$\rm \:  =  \: \dfrac{11}{8} \times \dfrac{8}{3}$

$\rm \:  =  \: \dfrac{11}{3}$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle \rm { \int _{0}^{2} \int_{0}^{ \dfrac{x}{2}} \: (2x + 3y) \: dy \: dx} = \frac{11}{3}}}$

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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

Given :-

$\quad \leadsto \quad \displaystyle \sf { \int _{0}^{2} \int_{0}^{ \footnotesize \frac{1}{2} x}(2x + 3y) \: dy \: dx}$

To Find :-

The value of given definite double integral

Solution :-

At first , see the given double integral here order is dydx . So , we have to first integrate w.r.t.y ( treating ' x ' as constant ) and then w.r.t.x ( treating ' y ' as constant )

Consider ;

${\bf I = \displaystyle \sf { \int _{0}^{2} \int_{0}^{ \footnotesize \frac{1}{2} x}(2x + 3y) \: dy \: dx}}$

${ : \implies \quad \bf I = \displaystyle \sf \int_{0}^{2} \bigg\{ \int_{0}^{ \footnotesize \dfrac{x}{2}} ( 2x + 3y ) dy \bigg\} dx }$

${ : \implies \quad \bf I = \displaystyle \sf \int_{0}^{2} \bigg\{ \int_{0}^{\footnotesize \dfrac{x}{2}} 2x \: dy + \int_{0}^{\footnotesize \dfrac{x}{2}} 3y dy \bigg\}dx }$

${ : \implies \quad \bf I = \displaystyle \sf \int_{0}^{2} \bigg\{ 2x \int_{0}^{\footnotesize \dfrac{x}{2}} \: dy + 3 \int_{0}^{\footnotesize \dfrac{x}{2}} y \: dy \bigg\}dx}$

We knows that ;

$\quad \qquad { \bigstar { \underline { \boxed { \red { \bf { \displaystyle \bf \int x^{n} dx = \dfrac{{x}^{n+1}}{n+1} + C }}}}}}{\bigstar}$

$\quad \qquad { \bigstar { \underline { \boxed { \red { \bf { \displaystyle \bf \int a f ( x ) \: dx = a \int f ( x ) \: dx + C }}}}}}{\bigstar}$

• Where ' a ' is constant

$\quad \qquad { \bigstar { \underline { \boxed { \red { \bf { \displaystyle \bf \int dn = n + C }}}}}}{\bigstar}$

Using these we have ;

${ : \implies \quad \bf I = \displaystyle \sf \int_{0}^{2} \bigg\{ 2x \left[ y \right]_{0}^{\footnotesize x/2 } + 3 \times \left[ \dfrac{y^{1+1}}{1+1} \right]_{0}^{\footnotesize \dfrac{x}{2}} \bigg\} dx}$

${ : \implies \quad \bf I = \displaystyle \sf \int_{0}^{2} \bigg\{ 2x \left[ y \right]_{0}^{\footnotesize x/2 } + 3 \times \left[ \dfrac{y^{2}}{2} \right]_{0}^{\footnotesize \dfrac{x}{2}} \bigg\} dx}$

${ : \implies \quad \bf I = \displaystyle \sf \int_{0}^{2} \bigg[ 2x \times \bigg\{ \dfrac{x}{2} - 0 \bigg\} + 3 \times \bigg\{ \dfrac{{\bigg(\dfrac{x}{2}\bigg)}^{2}}{2} - \dfrac{0²}{2} \bigg\} \bigg] dx }$

${ : \implies \quad \bf I = \displaystyle \sf \int_{0}^{2} \bigg\{ 2x \times \dfrac{x}{2} + 3 \bigg( \dfrac{x²}{4} ÷ 2 \bigg) \bigg\} dx }$

${ : \implies \quad \bf I = \displaystyle \sf \int_{0}^{2} \bigg\{ \cancel{2} \times x \times \dfrac{x}{\cancel{2}} + 3 \bigg( \dfrac{x²}{4 \times 2} \bigg) \bigg\} dx }$

${ : \implies \quad \bf I = \displaystyle \sf \int_{0}^{2} \bigg\{ x² + \dfrac{3x²}{8} \bigg\} dx }$

${ : \implies \quad \bf I = \displaystyle \sf \int_{0}^{2} x² \: dx + \int_{0}^{2} \dfrac{3x²}{8} \: dx }$

${ : \implies \quad \bf I = \displaystyle \sf \bigg\{ \dfrac{x^{2+1}}{2+1} \bigg\}_{0}^{2} + \dfrac{3}{8} \: \int_{0}^{2} x² \: dx }$

${ : \implies \quad \bf I = \displaystyle \sf \bigg\{ \dfrac{x^{3}}{3} \bigg\}_{0}^{2} + \dfrac{3}{8} \times \bigg\{ \dfrac{x^{2+1}}{2+1} \bigg\}_{0}^{2} }$

${ : \implies \quad \bf I = \displaystyle \sf \bigg\{ \dfrac{x^{3}}{3} \bigg\}_{0}^{2} + \dfrac{3}{8} \times \bigg\{ \dfrac{x^{3}}{3} \bigg\}_{0}^{2} }$

${ : \implies \quad \bf I = \displaystyle \sf \bigg\{ \dfrac{2^3}{3} - \dfrac{0^3}{3} \bigg\} + \dfrac{3}{8} \times \bigg\{ \dfrac{2^3}{3} - \dfrac{0^3}{3} \bigg\} }$

${ : \implies \quad \bf I = \displaystyle \sf \bigg\{ \dfrac{8}{3} - 0 \bigg\} + \dfrac{3}{8} \times \bigg\{ \dfrac{8}{3} - 0 \bigg\} }$

${ : \implies \quad \bf I = \displaystyle \sf \dfrac{8}{3} + \dfrac{3}{8} \times \dfrac{8}{3} }$

${ : \implies \quad \bf I = \displaystyle \sf \dfrac{8}{3} + \dfrac{\cancel{3}}{\cancel{8}} \times \dfrac{\cancel{8}}{\cancel{3}} }$

${ : \implies \quad \bf I = \displaystyle \sf \dfrac{8}{3} + 1}$

${ : \implies \quad \bf I = \displaystyle \sf \dfrac{8+3}{3} }$

${ : \implies \quad \bf I = \displaystyle \tt \dfrac{11}{3} }$

${ \bigstar { \underline { \boxed { \red { \underbrace { \bf { \therefore \displaystyle \bf \int _{0}^{2} \int_{0}^{ \footnotesize \dfrac{x}{2}}(2x + 3y) \: dy \: dx = \dfrac{11}{3} }}}}}}}{\bigstar}$

Henceforth , The Required Answer is $\bf \dfrac{11}{3}$ :)