Question

please solve this questions please ​

Answer 1

Question :

Find the value of

$\sf\int\limits_{0}^1 \dfrac{x^4(1-x)^4}{1+x^2}dx$

Solution :

We have to evaluate:

I=$\sf\int\limits_{0}^1 \dfrac{x^4(1-x)^4}{1+x^2}dx$

First Solve :

$\sf\int\dfrac{x^4(1-x)^4}{1+x^2}dx$

$\implies\sf\int\dfrac{x^4[(1-x)^2]^2}{1+x^2}dx$

$\implies\sf\int\dfrac{x^4[1+x^2-2x]^2}{1+x^2}dx$

$\implies\sf\int\dfrac{x^4[1+x^4+4x^2+2x^2-4x^3-4x]}{1+x^2}dx$

$\implies\sf\int\dfrac{x^4[1+x^4+6x^2-4x^3-4x]}{1+x^2}dx$

$\implies\sf\int\dfrac{x^8-4x^7+6x^6-4x^5+x^4}{1+x^2}dx$

By long - Division :

$\begin{array}{c | c c c c c c c} \cline{2-8} & x^6 & -4x^5 & +5x^4&-4x^2&+4&& \\ \cline{1-8} (x^2 +1) & x^8&-4x^7&+6x^6&-4x^5&+x^4&&\\ & x^8& & +x^6 \\ &(-)&&(-) \\ \cline{2-8} &&-4x^7&+5x^6&-4x^5&+x^4&& \\ &&-4x^7&&-4x^5 \\ &&(+) &&(+)\\ \cline{2-8}&&&+5x^6&&+x^4&& \\ &&&+5x^6&&+5x^4\\ &&&(-)&&(-)\\ \cline{2-8}&&&&&-4x^4&&\\ &&&&&-4x^4&-4x^2\\ &&&&&(+)&(+)\\ \cline{2-8}&&&&&&+4x^2&\\ &&&&&&+4x^2&+4\\&&&&&&(-)&(-)\\ \cline{2-8}&&&&&&&-4\\ \cline{8-8} \end{array}$

We know that

Dividend = Divisor × Quotient+Remainder

$\sf\:x^8-4x^7+6x^6-4x^5+x^4=(1+x^2)\times(x^6-4x^5+5x^4-4x^2+4)+(-4)$

Now divide both sides by $\sf\:1+x^2$

$\implies\sf\dfrac{x^8-4x^7+6x^6-4x^5+x^4}{1+x^2}=(x^6-4x^5+5x^4-4x^2+4)-\dfrac{4}{1+x^2}$

Thus ,

$\sf\int\dfrac{x^8-4x^7+6x^6-4x^5+x^4}{1+x^2}dx$=$\sf\int(x^6-4x^5+5x^4-4x^2-\dfrac{4}{x^2+1}+4)dx$

Now , Apply limits of Integration and Integrate Directly .

$\sf\int\limits_{0}^1(x^6-4x^5+5x^4-4x^2-\dfrac{4}{x^2+1}+4)dx$

$I=\sf[\frac{1}{7}x^7-\frac{2}{3}x^6+x^5-\frac{4}{3}x^3-4\tan^-1(x)+4x]^1_{0}$

I =$\sf[\frac{1}{7}+5-2-\pi]$

I =$\sf[\frac{21+1}{7}-\pi]$

I =$\sf[\frac{22}{7}-\pi]$

Therefore ,

$\sf\int\limits_{0}^1 \dfrac{x^4(1-x)^4}{1+x^2}dx$=$\sf\dfrac{22}{7}-\pi$

Correct option : a)

Answer 2

Answer:

Find the value of

\sf\int\limits_{0}^1 \dfrac{x^4(1-x)^4}{1+x^2}dx

0

∫

1

1+x

2

x

4

(1−x)

4

dx

Solution :

We have to evaluate:

I=\sf\int\limits_{0}^1 \dfrac{x^4(1-x)^4}{1+x^2}dx

0

∫

1

1+x

2

x

4

(1−x)

4

dx

First Solve :

\sf\int\dfrac{x^4(1-x)^4}{1+x^2}dx∫

1+x

2

x

4

(1−x)

4

dx

\implies\sf\int\dfrac{x^4[(1-x)^2]^2}{1+x^2}dx⟹∫

1+x

2

x

4

[(1−x)

2

]

2

dx

\implies\sf\int\dfrac{x^4[1+x^2-2x]^2}{1+x^2}dx⟹∫

1+x

2

x

4

[1+x

2

−2x]

2

dx

\implies\sf\int\dfrac{x^4[1+x^4+4x^2+2x^2-4x^3-4x]}{1+x^2}dx⟹∫

1+x

2

x

4

[1+x

4

+4x

2

+2x

2

−4x

3

−4x]

dx

\implies\sf\int\dfrac{x^4[1+x^4+6x^2-4x^3-4x]}{1+x^2}dx⟹∫

1+x

2

x

4

[1+x

4

+6x

2

−4x

3

−4x]

dx

\implies\sf\int\dfrac{x^8-4x^7+6x^6-4x^5+x^4}{1+x^2}dx⟹∫

1+x

2

x

8

−4x

7

+6x

6

−4x

5

+x

4

dx

By long - Division :

\begin{gathered}\begin{array}{c | c c c c c c c} \cline{2-8} & x^6 & -4x^5 & +5x^4&-4x^2&+4&& \\ \cline{1-8} (x^2 +1) & x^8&-4x^7&+6x^6&-4x^5&+x^4&&\\ & x^8& & +x^6 \\ &(-)&&(-) \\ \cline{2-8} &&-4x^7&+5x^6&-4x^5&+x^4&& \\ &&-4x^7&&-4x^5 \\ &&(+) &&(+)\\ \cline{2-8}&&&+5x^6&&+x^4&& \\ &&&+5x^6&&+5x^4\\ &&&(-)&&(-)\\ \cline{2-8}&&&&&-4x^4&&\\ &&&&&-4x^4&-4x^2\\ &&&&&(+)&(+)\\ \cline{2-8}&&&&&&+4x^2&\\ &&&&&&+4x^2&+4\\&&&&&&(-)&(-)\\ \cline{2-8}&&&&&&&-4\\ \cline{8-8} \end{array}\end{gathered}

\cline2−8

\cline1−8(x

2

+1)

\cline2−8

\cline2−8

\cline2−8

\cline2−8

\cline2−8

\cline8−8

x

6

x

8

x

8

(−)

−4x

5

−4x

7

−4x

7

−4x

7

(+)

+5x

4

+6x

6

+x

6

(−)

+5x

6

+5x

6

+5x

6

(−)

−4x

2

−4x

5

−4x

5

−4x

5

(+)

+4

+x

4

+x

4

+x

4

+5x

4

(−)

−4x

4

−4x

4

(+)

−4x

2

(+)

+4x

2

+4x

2

(−)

+4

(−)

−4

We know that

Dividend = Divisor × Quotient+Remainder

\sf\:x^8-4x^7+6x^6-4x^5+x^4=(1+x^2)\times(x^6-4x^5+5x^4-4x^2+4)+(-4)x

8

−4x

7

+6x

6

−4x

5

+x

4

=(1+x

2

)×(x

6

−4x

5

+5x

4

−4x

2

+4)+(−4)

Now divide both sides by \sf\:1+x^21+x

2

\implies\sf\dfrac{x^8-4x^7+6x^6-4x^5+x^4}{1+x^2}=(x^6-4x^5+5x^4-4x^2+4)-\dfrac{4}{1+x^2}⟹

1+x

2

x

8

−4x

7

+6x

6

−4x

5

+x

4

=(x

6

−4x

5

+5x

4

−4x

2

+4)−

1+x

2

4

Thus ,

\sf\int\dfrac{x^8-4x^7+6x^6-4x^5+x^4}{1+x^2}dx∫

1+x

2

x

8

−4x

7

+6x

6

−4x

5

+x

4

dx =\sf\int(x^6-4x^5+5x^4-4x^2-\dfrac{4}{x^2+1}+4)dx∫(x

6

−4x

5

+5x

4

−4x

2

−

x

2

+1

4

+4)dx

Now , Apply limits of Integration and Integrate Directly .

\sf\int\limits_{0}^1(x^6-4x^5+5x^4-4x^2-\dfrac{4}{x^2+1}+4)dx

0

∫

1

(x

6

−4x

5

+5x

4

−4x

2

−

x

2

+1

4

+4)dx

I=\sf[\frac{1}{7}x^7-\frac{2}{3}x^6+x^5-\frac{4}{3}x^3-4\tan^-1(x)+4x]^1_{0}I=[

7

1

x

7

−

3

2

x

6

+x

5

−

3

4

x

3

−4tan

−

1(x)+4x]

0

1

I =\sf[\frac{1}{7}+5-2-\pi][

7

1

+5−2−π]

I =\sf[\frac{21+1}{7}-\pi][

7

21+1

−π]

I =\sf[\frac{22}{7}-\pi][

7

22

−π]

Therefore ,

\sf\int\limits_{0}^1 \dfrac{x^4(1-x)^4}{1+x^2}dx

0

∫

1

1+x

2

x

4

(1−x)

4

dx =\sf\dfrac{22}{7}-\pi

7

22

−π

Correct option : a