Question

$\begin{array}{cc} \sf \: find \: the \: area \: of \: the \: region \: enclosed \: \: \: \\ \\ \sf \: by \: the \: following \: curves \: or \: equations: \\ \\ \\ \boxed{\bf \: y = x + 6} \\ \\ \boxed{ \bf \: y = {x}^{2} } \\ \\ \boxed{ \bf \: x = 5} \\ \\ \boxed{ \bf \: x = 0} \end{array}$
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Answer 1

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

Given curves are

$\rm :\longmapsto\:y = {x}^{2}$

represents the upper parabola with vertex (0, 0).

$\rm :\longmapsto\:y = x + 6$

represents a line.

So, point of intersection of above two curves,

$\rm :\longmapsto\: {x}^{2} = x + 6$

$\rm :\longmapsto\: {x}^{2} - x - 6 = 0$

$\rm :\longmapsto\: {x}^{2} - 3x + 2x - 6 = 0$

$\rm :\longmapsto\:x(x - 3) + 2(x - 3) = 0$

$\rm :\longmapsto\:(x - 3)(x + 2) = 0$

$\bf\implies \:x = 3 \: \: or \: \: x = - 2$

So, point of intersection are

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 3 & \sf 9 \\ \\ \sf - 2 & \sf 4 \end{array}} \\ \end{gathered}$

Now, Required area bounded between the curves

$\rm :\longmapsto\:y = {x}^{2}, \: y = x + 6, \: x = 0 \: and \: x = 5 \:$

is

$\rm \:  =  \:\displaystyle\int_0^3 {x}^{2} \: dx \: + \: \displaystyle\int_3^5 \: (x + 6) \: dx$

We know,

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

$\rm \:  =  \:\bigg[\dfrac{ {x}^{3} }{3} \bigg]_0^3 + \bigg[\dfrac{ {x}^{2} }{2} + 6x\bigg]_3^5$

$\rm \:  =  \:\bigg[\dfrac{ 27 }{3} \bigg] + \bigg[\dfrac{25 - 9 }{2} + 6(5 - 3)\bigg]$

$\rm \:  =  \:9 + 8 + 12$

$\rm \:  =  \:29 \: square \: units$

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:

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