Mathematics

# $\displaystyle\int \limits_{ -1 }^{ 1 }|x| dx = a$ then -

$a=1$

##### SOLUTION
Given,
$\displaystyle\int \limits_{ -1 }^{ 1 }|x| dx$
$=2\displaystyle\int \limits_{ 0 }^{ 1 }x dx$ [ Since $|x|$ is an even function]
$=2\times \left[\dfrac{x^2}{2}\right]_{0}^{1}$
$=2\times \dfrac{1}{2}$
$=1$.

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

#### Realted Questions

Q1 Single Correct Medium
$\int_{0}^{a} \dfrac{d x}{x+\sqrt{a^{2}-x^{2}}}$ is
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• B. $\dfrac{\pi}{2}$
• C. $\pi$
• D. $\dfrac{\pi}{4}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\int \dfrac {x+2}{(x^2 + 3x +3) \sqrt{x+1} } dx$ is equal to :
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• B. $\dfrac {1}{ \sqrt3} \tan^{-1} \left( \dfrac {x} { \sqrt{x+1} } \right) + C$
• C. None of these
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1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $\int { \cfrac { \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+4 \right) }{ \left( { x }^{ 2 }+3 \right) \left( { x }^{ 2 }-5 \right) } } dx=\int { \left\{ 1+\cfrac { f(x) }{ \left( { x }^{ 2 }+3 \right) \left( { x }^{ 2 }-5 \right) } \right\} } dx$
$x+A\tan ^{ -1 }{ \left( \cfrac { x }{ A' } \right) } +B\log { \left( \cfrac { x-l }{ x+m } \right) } +K\quad$ then which of the following is correct
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1 Verified Answer | Published on 17th 09, 2020

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1 Verified Answer | Published on 17th 09, 2020

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Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
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