Mathematics

# $\displaystyle\int_{1}^{2} \dfrac 2x\ dx$

##### SOLUTION

$\displaystyle\int_{1}^{2} \dfrac 2x\ dx$

$\left.2\log x \right]_1^2$

$2\log 2-2\log 1$

$2\log 2$

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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 Hard
Let $f(x)$ be a positive function. Let
$I_{1} = \int_{1 - k}^{k} xf\left \{x(1 - x)\right \} dx$,
$I_{2} = \int_{1 - k}^{k} f\left \{x(1 - x) \right \} dx$,
where $2k - 1 > 0$, then $\dfrac {I_{1}}{I_{2}}$ is
• A. $2$
• B. $k$
• C. $1$
• D. $\dfrac {1}{2}$

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The value of $\displaystyle \lim_{n\rightarrow \infty } \sum_{r=1}^{4n} \dfrac{\sqrt{n}}{\sqrt{r}(3\sqrt{r}+4\sqrt{n})^{2}}$ is equal to
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