Math, asked by rohithkrhoypuc1, 2 months ago

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Answered by itsPapaKaHelicopter
1

Answer:-

⇒dx = 2(b - a) \sin∅  \cos∅  \: d∅

∴I = \int\limits_{}^{b}( x - a {)}^{3} (b - x {)}^{4}  \: dx = 2(b - a {)}^{8}

⇒\int\limits_{0}^{\pi/2}  { \sin}^{7} ∅ \:  { \cos }^{9}  \: ∅ \: d∅ = 2( b - a {)}^{8}

⇒\int\limits_{0}^{\pi/2}  { \sin}^{7} ∅(1 -  { \sin}^{2} ∅ {)}^{4}  \:  \cos \: ∅ \: d∅

\text{Let }  \:  \sin∅ = t

⇒ \cos∅ \: d∅ = dt

⇒I = 2(b - a {)}^{8} \int\limits_{0}^{1}  {t}^{7} ( 1 -  {t}^{2}  {)}^{4} dt

⇒2(b - a {)}^{8} \int\limits_{0}^{1}  {t}^{7} (I - 4 {t}^{2}  + 6 {t}^{4}  - 4 {t}^{6}  +  {t}^{8} dt

 =   \frac{(b - a {)}^{8} }{280}

 \\  \\  \\  \\ \sf \colorbox{lightgreen} {\red★ANSWER ᵇʸɴᴀᴡᴀʙﷻ}

Answered by mathdude500
4

Appropriate Question

The value of

\red{\rm :\longmapsto\:\displaystyle\int_b^a \tt  {(x - a)}^{3} {(b - x)}^{4}dx \: is \: \dfrac{ {(b - a)}^{m} }{n} }

then ( m, n ) is

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

Given integral is

\red{\rm :\longmapsto\:\displaystyle\int_b^a \tt  {(x - a)}^{3} {(b - x)}^{4}dx}

Put x = a + y

So, dx = dy

Since, in definite integrals, when we use substitution method, we have to change the limits too.

So,

When x = a, y = 0

and

When x = b, y = b - a

So, given integral reduced to

\rm \:  =  \:  \: \:\displaystyle\int_{b - a}^0 \tt  {(a + y - a)}^{3} {(b - a - y)}^{4}dy

\rm \:  =  \:  \: \:\displaystyle\int_{b - a}^0 \tt  {( y )}^{3} {(b - a - y)}^{4}dy

Now,

Again we use method of Substitution,

Put y = (b - a)x

So that dy = (b - a) dx

Now, again we have to change the limits.

When y = 0, x = 0

When y = b - a, x = 1

So, above integral reduced to

\rm \:  =  \:  \: \:\displaystyle\int_{1}^0 \tt  {(b - a)}^{3}  {x}^{3} {(b - a - (b - a)x)}^{4}(b - a)dy

\rm \:  =  \:  \:  {(b - a)}^{8}\displaystyle\int_1^0 \tt  {x}^{3} {(1 - x)}^{4}dx

\rm \:  =  \:  \: -   {(b - a)}^{8}\displaystyle\int_0^1 \tt  {x}^{3} {(1 - x)}^{4}dx

We know,

\rm :\longmapsto\:\rm \: B(m, n)  =  \: \displaystyle\int_0^1 \tt  {x}^{m - 1} {(1 - x)}^{n - 1}dx

So, above integral can be rewritten as

\rm \:  =  \:  \: -   {(b - a)}^{8}B(4, 5)

We know,

\boxed{ \rm{ B(m, n)  =  \frac{\Gamma m \: \Gamma n}{\Gamma (m + n)} }}

So, above integral reduced to

\rm \:  =  \:  \:  -  {(b - a)}^{8} \:  \:  \dfrac{\Gamma 4 \: \Gamma 5}{\Gamma (4 + 5)}

\rm \:  =  \:  \:  -  {(b - a)}^{8} \:  \:  \dfrac{\Gamma 4 \: \Gamma 5}{\Gamma 9}

We know,

\boxed{ \rm{ \Gamma (n) = (n - 1)!}}

So, using this identity, we have

\rm \:  =  \:   - \:  {(b - a)}^{8} \times \dfrac{3! \times 4!}{8!}

\rm \:  =  \:  -  \:  {(b - a)}^{8} \times \dfrac{3 \times 2 \times 1 \times 4!}{8 \times 7 \times 6 \times 5 \times 4!}

\rm \:  =  \:  -  \:  {(b - a)}^{8} \times \dfrac{1}{280}

So, it implies,

\red{\rm :\longmapsto\:\displaystyle\int_b^a \tt  {(x - a)}^{3} {(b - x)}^{4}dx = \dfrac{ {(b - a)}^{8} }{ -  \: 280} }

But it is given that,

\red{\rm :\longmapsto\:\displaystyle\int_b^a \tt  {(x - a)}^{3} {(b - x)}^{4}dx \: is \: \dfrac{ {(b - a)}^{m} }{n} }

So, on comparing we get

 \:  \:  \:  \:  \: \bold{\underbrace{\boxed{ \bf{ m \:  =  \: 8 \:  \:  \: and \:  \:  \: n \:  =  \:  -  \: 280 \: }}}}

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