Question

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plz solve my question.

or any one know the ans plz ans...

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Answer 1

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

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 :-

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

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

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

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

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

Answer 2

Step-by-step explanation:

At first we have to know some basic formula :

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