Question

☆BRAINLIEAST question ☆ answer the attachment question.

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Answer 1

Answer:-

$⇒dx = 2(b - a) \sin∅ \cos∅ \: d∅$

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

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

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

$\text{Let } \: \sin∅ = t$

$⇒ \cos∅ \: d∅ = dt$

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

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

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

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

Answer 2

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

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