Mathematics

# If  $\mathrm { I } _ { 1 } = \int _ { 1 } ^ { 2 } x [ \sqrt { x } + \sqrt { 3 - x } ] d x$  and  $\mathrm { I } _ { 2 } = \int _ { 1 } ^ { 2 } ( \sqrt { x } + \sqrt { 3 - x } ) d x$  then  $\dfrac { I _ { 1 } } { I _ { 2 } } =$

$3/2$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
If $\displaystyle \int_{0}^{b} \displaystyle \frac{dx}{1\, +\, x^2}\, =\, \displaystyle \int_{b}^{\infty} \displaystyle \frac{dx}{1\, +\, x^2}$, then $b =$
• A. $\tan ^{-1}\, \left (\displaystyle \frac{1}{3} \right )$
• B. $\displaystyle \frac{\sqrt 3}{2}$
• C. $\sqrt 3$
• D. $1$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
If $f(x)=\left\{\begin{matrix} 0 , & where\ x=\displaystyle\frac{n}{n+1}, n=1, 2, 3....., \\ 1, & elsewhere \end{matrix}\right\}$ then the value of $\displaystyle\int^2_0 f(x)dx$
• A. $0$
• B. $2$
• C. $\infty$
• D. $1$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
$\int_{}^{} {\frac{1}{{\sqrt {1 + x.} }}\,dx}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
$\int {\frac{{2x - 1}}{{\sqrt {9{x^2} - 4} }}dx}$

$\int { \cfrac { f'(x) }{ f(x) } dx } =\log { [f(x)] } +c$