JEE Main 2023 (Online) 30th January Evening Shift | Definite Integration Question 70 | Mathematics

$\lim\limits_{n \rightarrow \infty} \frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+\ldots+\left(3-\frac{1}{n}\right)^2\right\}$ is equal to :

$\frac{19}{3}$

JEE Main 2023 (Online) 30th January Morning Shift

MCQ (Single Correct Answer)

-1

If [t] denotes the greatest integer $$\le \mathrm{t}$$, then the value of $${{3(e - 1)} \over e}\int\limits_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx} $$ is :

$$\mathrm{e^8-e}$$

$$\mathrm{e^7-1}$$

$$\mathrm{e^9-e}$$

$$\mathrm{e^8-1}$$

JEE Main 2023 (Online) 29th January Evening Shift

MCQ (Single Correct Answer)

-1

The value of the integral $$\int_1^2 {\left( {{{{t^4} + 1} \over {{t^6} + 1}}} \right)dt} $$ is

$${\tan ^{ - 1}}{1 \over 2} - {1 \over 3}{\tan ^{ - 1}}8 + {\pi \over 3}$$

$${\tan ^{ - 1}}2 - {1 \over 3}{\tan ^{ - 1}}8 + {\pi \over 3}$$

$${\tan ^{ - 1}}2 + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}$$

$${\tan ^{ - 1}}{1 \over 2} + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}$$

JEE Main 2023 (Online) 29th January Evening Shift

MCQ (Single Correct Answer)

-1

The value of the integral $$\int\limits_{1/2}^2 {{{{{\tan }^{ - 1}}x} \over x}dx} $$ is equal to :

$${\pi \over 2}{\log _e}2$$

$${\pi \over 4}{\log _e}2$$

$${1 \over 2}{\log _e}2$$

$$\pi {\log _e}2$$

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