Question

$\huge \underline{ \: \: \: \: \underline{\color {red}\bf QUESTION} \: \: \: \: }$


Find the area between the curves y = x² + 5 and y = x³ and the lines x = 1 and x = 2​

Answer 1

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

The given curves are

$\rm :\longmapsto\:y = {x}^{2} + 5 - - - (1)$

$\rm :\longmapsto\:y = {x}^{3}$

and

boundary conditions

$\rm :\longmapsto\:x = 1 \: \: and \: \: x = 2$

So,

required area between the curves is

$\rm :\longmapsto\:I \: = \displaystyle\int_1^2 \tt \:(y_1 - y_2)dx$

$\rm \: \: = \: \: \displaystyle\int_1^2 \tt \:( {x}^{2} + 5 - {x}^{3})$

$\rm \: \: = \: \: \bigg(\dfrac{ {x}^{3} }{3} + 5x - \dfrac{ {x}^{4} }{4} \bigg)_1^2$

$\: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \displaystyle\int \tt \: {x}^{n}dx = \dfrac{ {x}^{n + 1} }{n + 1} + c\bigg \}}$

$\rm \: \: = \: \: \bigg(\dfrac{8}{3} + 10 - \dfrac{16}{4} \bigg) - \bigg(\dfrac{1}{3} + 5 - \dfrac{1}{4} \bigg)$

$\rm \: \: = \: \: \bigg(\dfrac{8}{3} + 10 - 4 \bigg) - \bigg(\dfrac{1}{3} + 5 - \dfrac{1}{4} \bigg)$

$\rm \: \: = \: \: \bigg(\dfrac{8}{3} +6 \bigg) - \bigg(\dfrac{1}{3} + 5 - \dfrac{1}{4} \bigg)$

$\rm \: \: = \: \: \dfrac{8}{3} - \dfrac{1}{3} + \dfrac{1}{4} + 1$

$\rm \: \: = \: \: \dfrac{8 - 1}{3} + \dfrac{1 + 4}{4}$

$\rm \: \: = \: \: \dfrac{7}{3} + \dfrac{5}{4}$

$\rm \: \: = \: \: \dfrac{28 + 15}{12}$

$\rm \: \: = \: \: \dfrac{43}{12} \: square \: units$

Answer 2

ur answer is in the attachment...

hope it helps.