Math, asked by SamarendraDas143, 1 month ago

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Answers

Answered by mathdude500
5

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

Show that

\rm :\longmapsto\:\displaystyle\int_0^a\sf x {(a - x)}^{n}  =  \frac{ {a}^{n + 2} }{(n + 1)(n + 2)}

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

Consider, LHS

Let assume that,

\rm :\longmapsto\:I \:  =  \: \displaystyle\int_0^a\sf x {(a - x)}^{n}  dx

We know,

\boxed{ \bf{ \: \displaystyle\int_0^a\sf f(x) \: dx \:  =  \: \displaystyle\int_0^a\sf f(a - x) \: dx}}

So, using property of integrals, equation (1) can be rewritten as

\rm :\longmapsto\:I \:  =  \: \displaystyle\int_0^a\sf (a - x){[a - (a - x)]}^{n}  dx

\rm :\longmapsto\:I \:  =  \: \displaystyle\int_0^a\sf (a - x){[a - a  +  x]}^{n}  dx

\rm :\longmapsto\:I \:  =  \: \displaystyle\int_0^a\sf (a - x){[x]}^{n}  dx

\rm :\longmapsto\:I \:  =  \: \displaystyle\int_0^a\sf (a {x}^{n}  -  {x}^{n + 1} )  dx

We know,

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

So, using this we get

\rm :\longmapsto\:I = \bigg(\dfrac{a {x}^{n + 1} }{n + 1}  - \dfrac{ {x}^{n + 2} }{n + 2}  \bigg) _0^a

\rm :\longmapsto\:I = a \times \dfrac{ {a}^{n + 1} }{n + 1}  - \dfrac{ {a}^{n + 2} }{n + 2}

\rm :\longmapsto\:I =  \dfrac{ {a}^{n + 2} }{n + 1}  - \dfrac{ {a}^{n + 2} }{n + 2}

\rm :\longmapsto\:I =  {a}^{n + 2}\bigg(\dfrac{1}{n + 1} - \dfrac{1}{n + 2}  \bigg)

\rm :\longmapsto\:I =  {a}^{n + 2}\bigg(\dfrac{n + 2 - (n + 1)}{(n + 1)(n + 2)} \bigg)

\rm :\longmapsto\:I =  {a}^{n + 2}\bigg(\dfrac{n + 2 - n  - 1}{(n + 1)(n + 2)} \bigg)

\rm :\longmapsto\:I =  {a}^{n +  2}\bigg(\dfrac{1}{(n + 1)(n + 2)} \bigg)

\bf\implies \:I \:  =  \: \dfrac{ {a}^{n + 2} }{(n + 1)(n + 2)}

Additional Information :-

\boxed{ \bf{ \: \displaystyle\int_a^b\sf f(x)dx = \displaystyle\int_a^b\sf f(y)dy}}

\boxed{ \bf{ \: \displaystyle\int_a^b\sf f(x)dx = -  \:  \displaystyle\int_b^a\sf f(x)dx}}

\boxed{ \bf{ \: \displaystyle\int_a^b\sf f(x)dx =   \:  \displaystyle\int_a^b\sf f(a + b - x)dx}}

\boxed{ \bf{ \: \displaystyle\int_{ - a}^a\sf f(x) \: dx \:=  \: 0 \: if \: f( - x) = -  f(x) }}

\boxed{ \bf{ \: \displaystyle\int_{ - a}^a\sf f(x) \: dx \:= \displaystyle\int_0^a\sf f(x) \: dx \:  \: if \: f( - x) = f(x) }}

Answered by TrustedAnswerer19
55

Step-by-step explanation:

At first we have to know some basic formula :

\odot\sf \:  \:\displaystyle\int\sf {x}^{n}  \: dx =  \frac{ {x}^{n + 1} }{n + 1}  + c \\  \\  \pink{ \odot\:\displaystyle\int_b^a\sf \: f(x) \: dx = \:\displaystyle\int_b^a\sf \: f(a + b - x) \: dx }\\  \\ { \mathbb \red{ \sf \: now}} \\  \bf \: L.H.S = \:\displaystyle\int_0^a\bf \:  x(a - x)^{n}  \: dx \\ \\  =  \pink{ \displaystyle\int_0^a\bf \:(a + 0 - x) \{(a + 0)   - (a - x)   \}  ^{n} \: dx} \\  \\  = \displaystyle\int_0^a\bf \:(a - x). {x}^{n}  \: dx \\  \\ \displaystyle\int_0^a\bf \:(a. {x}^{n}  - x. {x}^{n} ) \: dx \\  \\  = \displaystyle\int_0^a\bf \:(a {x}^{n}  -  {x}^{n + 1} ) \: dx \\  \\  = \bf { \huge{ [ }} \frac{a {x}^{n + 1} }{n + 1}  -  \frac{ {x}^{n + 2} }{n + 2}  { \huge{ ]}} _0^a \\  \\  =  \bf \:  \frac{a. {a}^{n + 1} }{n + 1}  -  \frac{ {a}^{n + 2} }{n + 2}  - 0 + 0 \\  \\  \bf \:  =  \frac{ {a}^{n + 2} }{(n + 1)}  -  \frac{ {a}^{n + 2} }{n + 2}  \\  \\  \bf \:  =  {a}^{n + 2}  \times ( \frac{1}{n + 1}  -  \frac{1}{n + 2} ) \\  \\  \bf =  {a}^{n + 2}  \times  \{ \frac{n + 2 - n - 1}{(n + 1)(n + 2)}  \} \\  \\  \bf  =  {a}^{n + 2}  {\times  \frac{1}{(n + 1)(n + 2)} }  \\  \\  \bf  =  \frac{ {a}^{n + 2} }{(n + 1)(n +  2)}  \\  \\  \bf  = R.H.S \\  \\  \orange{ \therefore \bf \: L.H.S = R.H.S} \\  \\   \green{ \boxed{\sf \: hence \: showed}}

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