Mathematics

# Evaluate $\displaystyle\int_{2}^{3}{\dfrac{\sqrt{x}dx}{\sqrt{5-x}+\sqrt{x}}}$.

##### SOLUTION
Let, $I=\displaystyle\int_{2}^{3}{\dfrac{\sqrt{x}dx}{\sqrt{5-x}+\sqrt{x}}}$        .......$(1)$

We have $\displaystyle\int_{a}^{b}{f\left(x\right)dx}=\displaystyle\int_{a}^{b}{f\left(a+b-x\right)dx}$

$\Rightarrow I=\displaystyle\int_{2}^{3}{\dfrac{\sqrt{x}dx}{\sqrt{5-x}+\sqrt{x}}}$

Replace $x\rightarrow 3+2-x=5-x$

$I=\displaystyle\int_{2}^{3}{\dfrac{\sqrt{5-x}dx}{\sqrt{5-\left(5-x\right)}+\sqrt{5-x}}}$

$I=\displaystyle\int_{2}^{3}{\dfrac{\sqrt{5-x}dx}{\sqrt{x}+\sqrt{5-x}}}$     .......$(2)$

Adding $(1)$ and $(2)$ we get

$2I=\displaystyle\int_{2}^{3}{\dfrac{\sqrt{x}dx}{\sqrt{5-x}+\sqrt{x}}}+\displaystyle\int_{2}^{3}{\dfrac{\sqrt{5-x}dx}{\sqrt{x}+\sqrt{5-x}}}$

$2I=\displaystyle\int_{2}^{3}{\dfrac{\left(\sqrt{x}+\sqrt{5-x}\right)dx}{\sqrt{5-x}+\sqrt{x}}}$

$\Rightarrow 2I=\displaystyle\int_{2}^{3}{dx}$

$\Rightarrow 2I=\left[x\right]_{2}^{3}$

$\Rightarrow 2I=3-2=1$

$\therefore I=\dfrac{1}{2}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
If ${ I }_{ n }=\int { \tan ^{ n }{ x } } dx$ then ${ I }_{ 0 }+{ I }_{ 1 }+2\left( { I }_{ 2 }+{ I }_{ 3 }+....+{ I }_{ 8 } \right) +{ I }_{ 9 }+{ I }_{ 10 }$ equals $\left( n\in N \right)$
• A. $\displaystyle 1+\sum _{ n=1 }^{ 8 }{ \dfrac { \tan ^{ n }{ x } }{ n } }$
• B. $\displaystyle \sum _{ n=2 }^{ 8 }{ \dfrac { \tan ^{ n }{ x } }{ n+1 } }$
• C. $\displaystyle \sum _{ n=1 }^{ 9 }{ \dfrac { \tan ^{ n }{ x } }{ n+1 } }$
• D. $\displaystyle \sum _{ n=1 }^{ 9 }{ \dfrac { \tan ^{ n }{ x } }{ n } }$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Evaluate the following integrals:

$\displaystyle \int \dfrac{x^3}{2x + 1}dx$
• A. $\dfrac{x^3}{3}+\dfrac{x^2}{8}+\dfrac{1}{8x}-\dfrac{1}{16}log(2x+1)+C$
• B. $\dfrac{x^3}{6}-\dfrac{x^2}{6}-\dfrac{1}{8x}-\dfrac{1}{16}log(2x+1)+C$
• C. $\dfrac{x^3}{3}-\dfrac{x^2}{8}-\dfrac{1}{8x}-\dfrac{1}{16}log(2x+1)+C$
• D. $\dfrac{x^3}{6}-\dfrac{x^2}{8}+\dfrac{1}{8x}-\dfrac{1}{16}log(2x+1)+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
Evaluate: $\int \displaystyle \frac{\sin x}{\sin\, x + \cos\, x}dx$
• A. $\displaystyle \frac{1}{2} (x - { \ln |\sin\, x + \cos\, x|)} + c$
• B. $\displaystyle \frac{1}{2} x + \frac{1}{2 \ln |\sin\, x + \cos\, x|} + c$
• C. $\displaystyle \frac{1}{2} x - \frac{1}{ \ln |\sin\, x + \cos\, x|} + c$
• D. $\displaystyle \frac{1}{2} (x - { \ln |\sin\, x - \cos\, x|)} + c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate the integrals:
$\displaystyle \int \dfrac{1}{\sqrt{4x+3}}dx$

The value of $\int _{ 0 }^{ { \pi }/{ 2 } }{ \cfrac { dx }{ 1+\tan ^{ 3 }{ x } } }$ is