Mathematics

# The value of the define integral $\displaystyle \int^{1}_{0}e^{-x^{2}}dx+\displaystyle \int^{e}_{1}\sqrt {-ln x}dx$ is equal to

$e^{-\dfrac {1}{3}}$

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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 Hard
The value of $\int \dfrac {10^{x/2}}{\sqrt {10^{-x} - 10^{x}}}\, dx$ is
• A. $2\sqrt {10^{-x} + 10^{x}} + c$
• B. $\dfrac {1}{\log_{e}10}\sin h^{-1} (10^{x}) + c$
• C. $\dfrac {-1}{\log_{e}10}\sin h^{-1}(10^{x}) + c$
• D. $\dfrac {1}{\log_{2}10}\sin^{-1}(10^{x}) + c$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\int { \cfrac { dx }{ \left( 9+{ 4x }^{ 2 } \right) } } =$
• A. $\cfrac { 1 }{ 2 } \tan ^{ -1 }{ \cfrac { 2x }{ 3 } } +C$
• B. $\cfrac { 1 }{ 6 } \tan ^{ -1 }{ \cfrac { 3x }{ 2 } } +C$
• C. none of these
• D. $\cfrac { 1 }{ 6 } \tan ^{ -1 }{ \cfrac { 2x }{ 3 } } +C$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Prove that $\displaystyle \int_{0}^{a}f(x)dx = \int_{0}^{a} f(a - x)dx$ and hence evaluate $\displaystyle \int_{0}^{a} \dfrac {\sqrt {x}}{\sqrt {x} + \sqrt {a - x}} dx$.

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate the following integrals :
$\displaystyle\int_{0}^{\pi}x\sin x\cos^{4}x\ dx$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
Solve $\displaystyle \int \dfrac{\sqrt{1 - x^2} + \sqrt{1 + x^2}}{\sqrt{1 - x^4}}dx$;
• A. $I=\log|x-\sqrt{1+x^2}|+\sin^{-1}x+c$
• B. $I=\log|x+\sqrt{1+x^2}|-\sin^{-1}x+c$
• C. None of these
• D. $I=\log|x+\sqrt{1+x^2}|+\sin^{-1}x+c$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020