Question

please answer the fast in detailed solutions no spam wrong answers will be reported
click the image for bigger view​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

Evaluate

$\displaystyle\int_{-2}^1\rm {(2 {t}^{2} - 1)}^{2} \: dt$

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

Consider,

$\: \rm :\longmapsto\:\displaystyle\int_{-2}^1\rm {(2 {t}^{2} - 1)}^{2} \: dt$

We know,

$\boxed{ \bf{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}$

So, using this identity, we get

$\rm \: = \: \displaystyle\int_{-2}^1\rm ( {4t}^{4} + 1 - {4t}^{2}) \: dt$

$\rm \: = \: \displaystyle\int_{-2}^1\rm {4t}^{4}dt \: + \: \displaystyle\int_{-2}^1\rm dt \: - \: \displaystyle\int_{-2}^1\rm {4t}^{2} \: dt$

$\rm \: = \:4 \displaystyle\int_{-2}^1\rm {t}^{4}dt \: + \: \displaystyle\int_{-2}^1\rm dt \: - \: 4\displaystyle\int_{-2}^1\rm {t}^{2} \: dt$

We know

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

So, using this result, we get

$\rm \: = \: 4\bigg(\dfrac{ {t}^{4 + 1} }{4 + 1} \bigg)_{-2}^1 + \bigg( t\bigg) _{-2}^1 - 4\bigg(\dfrac{ {t}^{2 + 1} }{2 + 1} \bigg)_{-2}^1$

$\rm \: = \:4 \bigg(\dfrac{ {t}^{5} }{5} \bigg)_{-2}^1 + \bigg( t\bigg) _{-2}^1 - 4\bigg(\dfrac{ {t}^{3} }{3} \bigg)_{-2}^1$

$\rm \: = \:4 \bigg(\dfrac{ {1}^{5} - ( - 2) {}^{5} }{5} \bigg) + \bigg( 1 - ( - 2)\bigg) - 4\bigg(\dfrac{ {1}^{3} - {( - 2)}^{3} }{3} \bigg)$

$\rm \: = \: 4\bigg(\dfrac{ 1 - ( - 32) }{5} \bigg) + \bigg( 1 + 2\bigg) - 4\bigg(\dfrac{ 1 + 8 }{3} \bigg)$

$\rm \: = \: 4\bigg(\dfrac{33}{5} \bigg) + \bigg(3\bigg) - (12)$

$\rm \: = \: \bigg(\dfrac{132}{5} \bigg) + 3 - 12$

$\rm \: = \: \bigg(\dfrac{132}{5} \bigg) - 9$

$\rm \: = \: \dfrac{132 - 45}{5}$

$\rm \: = \: \dfrac{87}{5}$

Additional Information :-

$\boxed{ \bf{ \: \displaystyle\int \: kdx \: = \: kx \: + \: c}}$

$\boxed{ \bf{ \: \displaystyle\int \: cosxdx \: = \: sinx \: + \: c}}$

$\boxed{ \bf{ \: \displaystyle\int \: sinxdx \: = - \: cosx \: + \: c}}$

$\boxed{ \bf{ \: \displaystyle\int \: tanxdx \: = \: log \: secx \: + \: c}}$

$\boxed{ \bf{ \: \displaystyle\int \: cotxdx \: = \: log \: sinx \: + \: c}}$

$\boxed{ \bf{ \: \displaystyle\int \: cosecxdx \: = \: log \: (cosecx - cotx)\: + \: c}}$

$\boxed{ \bf{ \: \displaystyle\int \: secxdx \: = \: log \: (secx + tanx)\: + \: c}}$