QBManager

Integral Calculus

86418 If $f : R \to R$ is a differentiable function and $f (2)$ $= 6$ , then $lim_{x \to 2} \int_{6}^{f (x)} \frac{2 tdt}{(x - 2)}$ is

1

12 f^{'} (2)

2 0

3

24 f^{'} (2)

4

2 f^{'} (2)

Explanation:

(A) : Given,
$lim_{x \to 2} \int_{6}^{f (x)} \frac{2 t d t}{(x - 2)} = lim_{x \to 2} \frac{1}{(x - 2)} {[t^{2}]}_{6}^{f (x)}$
$= lim_{x \to 2} \frac{[f (x)]^{2} - 6^{2}}{(x - 2)}$
$[\frac{0}{0}$ form $]$
By L - Hospital Rule ,
$= lim_{x \to 2} \frac{2 f (x) f^{'} (x)}{1} = 2 f (2) f^{'} (2) = 12 f^{'} (2)$

JEE Main-2019-09.04.2019

Integral Calculus

86420 The value of $lim_{n \to \infty} \frac{1}{r} \sum_{r = 0}^{2 n - 1} \frac{n^{2}}{n^{2} + 4 r^{2}}$ is

1

\frac{1}{2} \tan^{- 1} (2)

2

\frac{1}{2} \tan^{- 1} (4)

3

\tan^{- 1} (4)

4

\frac{1}{4} \tan^{- 1} (4)

Explanation:

(B)Given,
$lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{n^{2}}{n^{2} + 4 r^{2}}$
$y = lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{n^{2}}{n^{2} (1 + 4 \frac{r^{2}}{n^{2}})} = lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{1}{1 + 4 {(\frac{r}{n})}^{2}}$
$Let, \frac{r}{x} = x$
$When, r = 0, x = 0$
$r = 2 n - 1, x = 2 - \frac{1}{n} = 2 (∵ n = \infty)$
$y = \int_{0}^{2} \frac{1}{1 + 4 x^{2}} \cdot d x$
$y = \int_{0}^{2} \frac{1}{1 + (2 x)^{2}} \cdot d x$
$y = {[\frac{1}{1} \frac{\tan^{- 1} (\frac{2 x}{1})}{2}]}_{0}^{2}$
$y = \frac{1}{2} {[\tan^{- 1} (2 x)]}_{0}^{2} = \frac{1}{2} [\tan^{1} (2 \times 2) - \tan^{- 1} (0)]$
$y = \frac{1}{2} \tan^{- 1} (4)$

JEE Main-2021-26.08.2021-Shift I

Integral Calculus

86421 $lim_{x \to 0} \frac{\int_{0}^{x^{2}} (\sin \sqrt{t}) d t}{x^{3}}$ is equal $t$

1

\frac{2}{3}

2

\frac{3}{2}

3

\frac{1}{15}

4 0

Explanation:

(A) : Given $lim_{x \to 0} \frac{\int_{0}^{x^{2}} (\sin \sqrt{t}) d t}{x^{3}}$
By, L' Hospital's rule:
$lim_{x \to 0} \frac{\int_{0}^{x^{2}} (\sin \sqrt{t}) d t}{x^{3}}$
$= lim_{x \to 0} \frac{\frac{d}{d x} \int_{0}^{x^{2}} \sin \sqrt{t} d t}{\frac{d}{d x} x^{3}} [\begin{array}{l} ∵ t = x^{2} \\ d t = 2 x d x \end{array}]$
$= lim_{x \to 0} \frac{\sin (\sqrt{x^{2}}) (2 x)}{3 x^{2}} = lim_{x \to 0} \frac{2 \sin (| x |)}{3 x}$
$= lim_{x \to 0} \frac{2 \sin (x)}{3 x}$ $(x > 0)$
$= \frac{2}{3}$

JEE Main-2021-24.02.2021

Integral Calculus

86422 If $\int_{1 / 2}^{2} \frac{1}{x} {cosec}^{101} (x - \frac{1}{x}) d x = k$ then the value of $k$ is

1 1

2

\frac{1}{2}

3 0

4

\frac{1}{101}

Explanation:

(C) : Let, $I = \int_{1 / 2}^{2} \frac{1}{x} {cosec}^{101} (x - \frac{1}{x}) dx$
Let, $x = \frac{1}{t}$
$dx = \frac{1}{t^{2}} dt$
When, $x = \frac{1}{2}, t = 2$
$x = 2, t = \frac{1}{2}$
$I = \int_{2}^{1 / 2} t {coses}^{101} (\frac{1}{t} - t) (\frac{1}{t^{2}}) dt$
$= - \int_{2}^{1 / 2} \frac{1}{t} {coses}^{101} (t - \frac{1}{t}) dt$
$I = - I$
$2 I = 0$
$I = 0$

Jamia Millia Islamia-2013

Integral Calculus

86423 $\int_{0}^{π / 4} \tan^{2} (x) d x =$

1

\frac{π}{4}

2

\frac{π}{4} - 1

3

1 - \frac{π}{4}

4 0

Explanation:

(C) : Given,
$\int_{0}^{π / 4} \tan^{2} (x) dx$
$= \int_{0}^{π / 4} (\sec^{2} x - 1) d x = \int_{0}^{π / 4} \sec^{2} x \cdot d x - \int_{0}^{π / 4} 1 . d x$
$= [\tan x]_{0}^{π / 4} - [x]_{0}^{π / 4} = [\tan \frac{π}{4} - \tan 0] - [\frac{π}{4} - 0]$
$= 1 - 0 - \frac{π}{4} = 1 - \frac{π}{4}$

AP EAMCET-2020-21.09.2020

Integral Calculus

86418 If $f : R \to R$ is a differentiable function and $f (2)$ $= 6$ , then $lim_{x \to 2} \int_{6}^{f (x)} \frac{2 tdt}{(x - 2)}$ is

1

12 f^{'} (2)

2 0

3

24 f^{'} (2)

4

2 f^{'} (2)

Explanation:

(A) : Given,
$lim_{x \to 2} \int_{6}^{f (x)} \frac{2 t d t}{(x - 2)} = lim_{x \to 2} \frac{1}{(x - 2)} {[t^{2}]}_{6}^{f (x)}$
$= lim_{x \to 2} \frac{[f (x)]^{2} - 6^{2}}{(x - 2)}$
$[\frac{0}{0}$ form $]$
By L - Hospital Rule ,
$= lim_{x \to 2} \frac{2 f (x) f^{'} (x)}{1} = 2 f (2) f^{'} (2) = 12 f^{'} (2)$

JEE Main-2019-09.04.2019

Integral Calculus

86420 The value of $lim_{n \to \infty} \frac{1}{r} \sum_{r = 0}^{2 n - 1} \frac{n^{2}}{n^{2} + 4 r^{2}}$ is

1

\frac{1}{2} \tan^{- 1} (2)

2

\frac{1}{2} \tan^{- 1} (4)

3

\tan^{- 1} (4)

4

\frac{1}{4} \tan^{- 1} (4)

Explanation:

(B)Given,
$lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{n^{2}}{n^{2} + 4 r^{2}}$
$y = lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{n^{2}}{n^{2} (1 + 4 \frac{r^{2}}{n^{2}})} = lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{1}{1 + 4 {(\frac{r}{n})}^{2}}$
$Let, \frac{r}{x} = x$
$When, r = 0, x = 0$
$r = 2 n - 1, x = 2 - \frac{1}{n} = 2 (∵ n = \infty)$
$y = \int_{0}^{2} \frac{1}{1 + 4 x^{2}} \cdot d x$
$y = \int_{0}^{2} \frac{1}{1 + (2 x)^{2}} \cdot d x$
$y = {[\frac{1}{1} \frac{\tan^{- 1} (\frac{2 x}{1})}{2}]}_{0}^{2}$
$y = \frac{1}{2} {[\tan^{- 1} (2 x)]}_{0}^{2} = \frac{1}{2} [\tan^{1} (2 \times 2) - \tan^{- 1} (0)]$
$y = \frac{1}{2} \tan^{- 1} (4)$

JEE Main-2021-26.08.2021-Shift I

Integral Calculus

86421 $lim_{x \to 0} \frac{\int_{0}^{x^{2}} (\sin \sqrt{t}) d t}{x^{3}}$ is equal $t$

1

\frac{2}{3}

2

\frac{3}{2}

3

\frac{1}{15}

4 0

Explanation:

(A) : Given $lim_{x \to 0} \frac{\int_{0}^{x^{2}} (\sin \sqrt{t}) d t}{x^{3}}$
By, L' Hospital's rule:
$lim_{x \to 0} \frac{\int_{0}^{x^{2}} (\sin \sqrt{t}) d t}{x^{3}}$
$= lim_{x \to 0} \frac{\frac{d}{d x} \int_{0}^{x^{2}} \sin \sqrt{t} d t}{\frac{d}{d x} x^{3}} [\begin{array}{l} ∵ t = x^{2} \\ d t = 2 x d x \end{array}]$
$= lim_{x \to 0} \frac{\sin (\sqrt{x^{2}}) (2 x)}{3 x^{2}} = lim_{x \to 0} \frac{2 \sin (| x |)}{3 x}$
$= lim_{x \to 0} \frac{2 \sin (x)}{3 x}$ $(x > 0)$
$= \frac{2}{3}$

JEE Main-2021-24.02.2021

Integral Calculus

86422 If $\int_{1 / 2}^{2} \frac{1}{x} {cosec}^{101} (x - \frac{1}{x}) d x = k$ then the value of $k$ is

1 1

2

\frac{1}{2}

3 0

4

\frac{1}{101}

Explanation:

(C) : Let, $I = \int_{1 / 2}^{2} \frac{1}{x} {cosec}^{101} (x - \frac{1}{x}) dx$
Let, $x = \frac{1}{t}$
$dx = \frac{1}{t^{2}} dt$
When, $x = \frac{1}{2}, t = 2$
$x = 2, t = \frac{1}{2}$
$I = \int_{2}^{1 / 2} t {coses}^{101} (\frac{1}{t} - t) (\frac{1}{t^{2}}) dt$
$= - \int_{2}^{1 / 2} \frac{1}{t} {coses}^{101} (t - \frac{1}{t}) dt$
$I = - I$
$2 I = 0$
$I = 0$

Jamia Millia Islamia-2013

Integral Calculus

86423 $\int_{0}^{π / 4} \tan^{2} (x) d x =$

1

\frac{π}{4}

2

\frac{π}{4} - 1

3

1 - \frac{π}{4}

4 0

Explanation:

(C) : Given,
$\int_{0}^{π / 4} \tan^{2} (x) dx$
$= \int_{0}^{π / 4} (\sec^{2} x - 1) d x = \int_{0}^{π / 4} \sec^{2} x \cdot d x - \int_{0}^{π / 4} 1 . d x$
$= [\tan x]_{0}^{π / 4} - [x]_{0}^{π / 4} = [\tan \frac{π}{4} - \tan 0] - [\frac{π}{4} - 0]$
$= 1 - 0 - \frac{π}{4} = 1 - \frac{π}{4}$

AP EAMCET-2020-21.09.2020

Integral Calculus

86418 If $f : R \to R$ is a differentiable function and $f (2)$ $= 6$ , then $lim_{x \to 2} \int_{6}^{f (x)} \frac{2 tdt}{(x - 2)}$ is

1

12 f^{'} (2)

2 0

3

24 f^{'} (2)

4

2 f^{'} (2)

Explanation:

(A) : Given,
$lim_{x \to 2} \int_{6}^{f (x)} \frac{2 t d t}{(x - 2)} = lim_{x \to 2} \frac{1}{(x - 2)} {[t^{2}]}_{6}^{f (x)}$
$= lim_{x \to 2} \frac{[f (x)]^{2} - 6^{2}}{(x - 2)}$
$[\frac{0}{0}$ form $]$
By L - Hospital Rule ,
$= lim_{x \to 2} \frac{2 f (x) f^{'} (x)}{1} = 2 f (2) f^{'} (2) = 12 f^{'} (2)$

JEE Main-2019-09.04.2019

Integral Calculus

86420 The value of $lim_{n \to \infty} \frac{1}{r} \sum_{r = 0}^{2 n - 1} \frac{n^{2}}{n^{2} + 4 r^{2}}$ is

1

\frac{1}{2} \tan^{- 1} (2)

2

\frac{1}{2} \tan^{- 1} (4)

3

\tan^{- 1} (4)

4

\frac{1}{4} \tan^{- 1} (4)

Explanation:

(B)Given,
$lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{n^{2}}{n^{2} + 4 r^{2}}$
$y = lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{n^{2}}{n^{2} (1 + 4 \frac{r^{2}}{n^{2}})} = lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{1}{1 + 4 {(\frac{r}{n})}^{2}}$
$Let, \frac{r}{x} = x$
$When, r = 0, x = 0$
$r = 2 n - 1, x = 2 - \frac{1}{n} = 2 (∵ n = \infty)$
$y = \int_{0}^{2} \frac{1}{1 + 4 x^{2}} \cdot d x$
$y = \int_{0}^{2} \frac{1}{1 + (2 x)^{2}} \cdot d x$
$y = {[\frac{1}{1} \frac{\tan^{- 1} (\frac{2 x}{1})}{2}]}_{0}^{2}$
$y = \frac{1}{2} {[\tan^{- 1} (2 x)]}_{0}^{2} = \frac{1}{2} [\tan^{1} (2 \times 2) - \tan^{- 1} (0)]$
$y = \frac{1}{2} \tan^{- 1} (4)$

JEE Main-2021-26.08.2021-Shift I

Integral Calculus

86421 $lim_{x \to 0} \frac{\int_{0}^{x^{2}} (\sin \sqrt{t}) d t}{x^{3}}$ is equal $t$

1

\frac{2}{3}

2

\frac{3}{2}

3

\frac{1}{15}

4 0

Explanation:

(A) : Given $lim_{x \to 0} \frac{\int_{0}^{x^{2}} (\sin \sqrt{t}) d t}{x^{3}}$
By, L' Hospital's rule:
$lim_{x \to 0} \frac{\int_{0}^{x^{2}} (\sin \sqrt{t}) d t}{x^{3}}$
$= lim_{x \to 0} \frac{\frac{d}{d x} \int_{0}^{x^{2}} \sin \sqrt{t} d t}{\frac{d}{d x} x^{3}} [\begin{array}{l} ∵ t = x^{2} \\ d t = 2 x d x \end{array}]$
$= lim_{x \to 0} \frac{\sin (\sqrt{x^{2}}) (2 x)}{3 x^{2}} = lim_{x \to 0} \frac{2 \sin (| x |)}{3 x}$
$= lim_{x \to 0} \frac{2 \sin (x)}{3 x}$ $(x > 0)$
$= \frac{2}{3}$

JEE Main-2021-24.02.2021

Integral Calculus

86422 If $\int_{1 / 2}^{2} \frac{1}{x} {cosec}^{101} (x - \frac{1}{x}) d x = k$ then the value of $k$ is

1 1

2

\frac{1}{2}

3 0

4

\frac{1}{101}

Explanation:

(C) : Let, $I = \int_{1 / 2}^{2} \frac{1}{x} {cosec}^{101} (x - \frac{1}{x}) dx$
Let, $x = \frac{1}{t}$
$dx = \frac{1}{t^{2}} dt$
When, $x = \frac{1}{2}, t = 2$
$x = 2, t = \frac{1}{2}$
$I = \int_{2}^{1 / 2} t {coses}^{101} (\frac{1}{t} - t) (\frac{1}{t^{2}}) dt$
$= - \int_{2}^{1 / 2} \frac{1}{t} {coses}^{101} (t - \frac{1}{t}) dt$
$I = - I$
$2 I = 0$
$I = 0$

Jamia Millia Islamia-2013

Integral Calculus

86423 $\int_{0}^{π / 4} \tan^{2} (x) d x =$

1

\frac{π}{4}

2

\frac{π}{4} - 1

3

1 - \frac{π}{4}

4 0

Explanation:

(C) : Given,
$\int_{0}^{π / 4} \tan^{2} (x) dx$
$= \int_{0}^{π / 4} (\sec^{2} x - 1) d x = \int_{0}^{π / 4} \sec^{2} x \cdot d x - \int_{0}^{π / 4} 1 . d x$
$= [\tan x]_{0}^{π / 4} - [x]_{0}^{π / 4} = [\tan \frac{π}{4} - \tan 0] - [\frac{π}{4} - 0]$
$= 1 - 0 - \frac{π}{4} = 1 - \frac{π}{4}$

AP EAMCET-2020-21.09.2020

Integral Calculus

86418 If $f : R \to R$ is a differentiable function and $f (2)$ $= 6$ , then $lim_{x \to 2} \int_{6}^{f (x)} \frac{2 tdt}{(x - 2)}$ is

1

12 f^{'} (2)

2 0

3

24 f^{'} (2)

4

2 f^{'} (2)

Explanation:

(A) : Given,
$lim_{x \to 2} \int_{6}^{f (x)} \frac{2 t d t}{(x - 2)} = lim_{x \to 2} \frac{1}{(x - 2)} {[t^{2}]}_{6}^{f (x)}$
$= lim_{x \to 2} \frac{[f (x)]^{2} - 6^{2}}{(x - 2)}$
$[\frac{0}{0}$ form $]$
By L - Hospital Rule ,
$= lim_{x \to 2} \frac{2 f (x) f^{'} (x)}{1} = 2 f (2) f^{'} (2) = 12 f^{'} (2)$

JEE Main-2019-09.04.2019

Integral Calculus

86420 The value of $lim_{n \to \infty} \frac{1}{r} \sum_{r = 0}^{2 n - 1} \frac{n^{2}}{n^{2} + 4 r^{2}}$ is

1

\frac{1}{2} \tan^{- 1} (2)

2

\frac{1}{2} \tan^{- 1} (4)

3

\tan^{- 1} (4)

4

\frac{1}{4} \tan^{- 1} (4)

Explanation:

(B)Given,
$lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{n^{2}}{n^{2} + 4 r^{2}}$
$y = lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{n^{2}}{n^{2} (1 + 4 \frac{r^{2}}{n^{2}})} = lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{1}{1 + 4 {(\frac{r}{n})}^{2}}$
$Let, \frac{r}{x} = x$
$When, r = 0, x = 0$
$r = 2 n - 1, x = 2 - \frac{1}{n} = 2 (∵ n = \infty)$
$y = \int_{0}^{2} \frac{1}{1 + 4 x^{2}} \cdot d x$
$y = \int_{0}^{2} \frac{1}{1 + (2 x)^{2}} \cdot d x$
$y = {[\frac{1}{1} \frac{\tan^{- 1} (\frac{2 x}{1})}{2}]}_{0}^{2}$
$y = \frac{1}{2} {[\tan^{- 1} (2 x)]}_{0}^{2} = \frac{1}{2} [\tan^{1} (2 \times 2) - \tan^{- 1} (0)]$
$y = \frac{1}{2} \tan^{- 1} (4)$

JEE Main-2021-26.08.2021-Shift I

Integral Calculus

86421 $lim_{x \to 0} \frac{\int_{0}^{x^{2}} (\sin \sqrt{t}) d t}{x^{3}}$ is equal $t$

1

\frac{2}{3}

2

\frac{3}{2}

3

\frac{1}{15}

4 0

Explanation:

(A) : Given $lim_{x \to 0} \frac{\int_{0}^{x^{2}} (\sin \sqrt{t}) d t}{x^{3}}$
By, L' Hospital's rule:
$lim_{x \to 0} \frac{\int_{0}^{x^{2}} (\sin \sqrt{t}) d t}{x^{3}}$
$= lim_{x \to 0} \frac{\frac{d}{d x} \int_{0}^{x^{2}} \sin \sqrt{t} d t}{\frac{d}{d x} x^{3}} [\begin{array}{l} ∵ t = x^{2} \\ d t = 2 x d x \end{array}]$
$= lim_{x \to 0} \frac{\sin (\sqrt{x^{2}}) (2 x)}{3 x^{2}} = lim_{x \to 0} \frac{2 \sin (| x |)}{3 x}$
$= lim_{x \to 0} \frac{2 \sin (x)}{3 x}$ $(x > 0)$
$= \frac{2}{3}$

JEE Main-2021-24.02.2021

Integral Calculus

86422 If $\int_{1 / 2}^{2} \frac{1}{x} {cosec}^{101} (x - \frac{1}{x}) d x = k$ then the value of $k$ is

1 1

2

\frac{1}{2}

3 0

4

\frac{1}{101}

Explanation:

(C) : Let, $I = \int_{1 / 2}^{2} \frac{1}{x} {cosec}^{101} (x - \frac{1}{x}) dx$
Let, $x = \frac{1}{t}$
$dx = \frac{1}{t^{2}} dt$
When, $x = \frac{1}{2}, t = 2$
$x = 2, t = \frac{1}{2}$
$I = \int_{2}^{1 / 2} t {coses}^{101} (\frac{1}{t} - t) (\frac{1}{t^{2}}) dt$
$= - \int_{2}^{1 / 2} \frac{1}{t} {coses}^{101} (t - \frac{1}{t}) dt$
$I = - I$
$2 I = 0$
$I = 0$

Jamia Millia Islamia-2013

Integral Calculus

86423 $\int_{0}^{π / 4} \tan^{2} (x) d x =$

1

\frac{π}{4}

2

\frac{π}{4} - 1

3

1 - \frac{π}{4}

4 0

Explanation:

(C) : Given,
$\int_{0}^{π / 4} \tan^{2} (x) dx$
$= \int_{0}^{π / 4} (\sec^{2} x - 1) d x = \int_{0}^{π / 4} \sec^{2} x \cdot d x - \int_{0}^{π / 4} 1 . d x$
$= [\tan x]_{0}^{π / 4} - [x]_{0}^{π / 4} = [\tan \frac{π}{4} - \tan 0] - [\frac{π}{4} - 0]$
$= 1 - 0 - \frac{π}{4} = 1 - \frac{π}{4}$

AP EAMCET-2020-21.09.2020

Integral Calculus

86418 If $f : R \to R$ is a differentiable function and $f (2)$ $= 6$ , then $lim_{x \to 2} \int_{6}^{f (x)} \frac{2 tdt}{(x - 2)}$ is

1

12 f^{'} (2)

2 0

3

24 f^{'} (2)

4

2 f^{'} (2)

Explanation:

(A) : Given,
$lim_{x \to 2} \int_{6}^{f (x)} \frac{2 t d t}{(x - 2)} = lim_{x \to 2} \frac{1}{(x - 2)} {[t^{2}]}_{6}^{f (x)}$
$= lim_{x \to 2} \frac{[f (x)]^{2} - 6^{2}}{(x - 2)}$
$[\frac{0}{0}$ form $]$
By L - Hospital Rule ,
$= lim_{x \to 2} \frac{2 f (x) f^{'} (x)}{1} = 2 f (2) f^{'} (2) = 12 f^{'} (2)$

JEE Main-2019-09.04.2019

Integral Calculus

86420 The value of $lim_{n \to \infty} \frac{1}{r} \sum_{r = 0}^{2 n - 1} \frac{n^{2}}{n^{2} + 4 r^{2}}$ is

1

\frac{1}{2} \tan^{- 1} (2)

2

\frac{1}{2} \tan^{- 1} (4)

3

\tan^{- 1} (4)

4

\frac{1}{4} \tan^{- 1} (4)

Explanation:

(B)Given,
$lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{n^{2}}{n^{2} + 4 r^{2}}$
$y = lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{n^{2}}{n^{2} (1 + 4 \frac{r^{2}}{n^{2}})} = lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{1}{1 + 4 {(\frac{r}{n})}^{2}}$
$Let, \frac{r}{x} = x$
$When, r = 0, x = 0$
$r = 2 n - 1, x = 2 - \frac{1}{n} = 2 (∵ n = \infty)$
$y = \int_{0}^{2} \frac{1}{1 + 4 x^{2}} \cdot d x$
$y = \int_{0}^{2} \frac{1}{1 + (2 x)^{2}} \cdot d x$
$y = {[\frac{1}{1} \frac{\tan^{- 1} (\frac{2 x}{1})}{2}]}_{0}^{2}$
$y = \frac{1}{2} {[\tan^{- 1} (2 x)]}_{0}^{2} = \frac{1}{2} [\tan^{1} (2 \times 2) - \tan^{- 1} (0)]$
$y = \frac{1}{2} \tan^{- 1} (4)$

JEE Main-2021-26.08.2021-Shift I

Integral Calculus

86421 $lim_{x \to 0} \frac{\int_{0}^{x^{2}} (\sin \sqrt{t}) d t}{x^{3}}$ is equal $t$

1

\frac{2}{3}

2

\frac{3}{2}

3

\frac{1}{15}

4 0

Explanation:

(A) : Given $lim_{x \to 0} \frac{\int_{0}^{x^{2}} (\sin \sqrt{t}) d t}{x^{3}}$
By, L' Hospital's rule:
$lim_{x \to 0} \frac{\int_{0}^{x^{2}} (\sin \sqrt{t}) d t}{x^{3}}$
$= lim_{x \to 0} \frac{\frac{d}{d x} \int_{0}^{x^{2}} \sin \sqrt{t} d t}{\frac{d}{d x} x^{3}} [\begin{array}{l} ∵ t = x^{2} \\ d t = 2 x d x \end{array}]$
$= lim_{x \to 0} \frac{\sin (\sqrt{x^{2}}) (2 x)}{3 x^{2}} = lim_{x \to 0} \frac{2 \sin (| x |)}{3 x}$
$= lim_{x \to 0} \frac{2 \sin (x)}{3 x}$ $(x > 0)$
$= \frac{2}{3}$

JEE Main-2021-24.02.2021

Integral Calculus

86422 If $\int_{1 / 2}^{2} \frac{1}{x} {cosec}^{101} (x - \frac{1}{x}) d x = k$ then the value of $k$ is

1 1

2

\frac{1}{2}

3 0

4

\frac{1}{101}

Explanation:

(C) : Let, $I = \int_{1 / 2}^{2} \frac{1}{x} {cosec}^{101} (x - \frac{1}{x}) dx$
Let, $x = \frac{1}{t}$
$dx = \frac{1}{t^{2}} dt$
When, $x = \frac{1}{2}, t = 2$
$x = 2, t = \frac{1}{2}$
$I = \int_{2}^{1 / 2} t {coses}^{101} (\frac{1}{t} - t) (\frac{1}{t^{2}}) dt$
$= - \int_{2}^{1 / 2} \frac{1}{t} {coses}^{101} (t - \frac{1}{t}) dt$
$I = - I$
$2 I = 0$
$I = 0$

Jamia Millia Islamia-2013

Integral Calculus

86423 $\int_{0}^{π / 4} \tan^{2} (x) d x =$

1

\frac{π}{4}

2

\frac{π}{4} - 1

3

1 - \frac{π}{4}

4 0

Explanation:

(C) : Given,
$\int_{0}^{π / 4} \tan^{2} (x) dx$
$= \int_{0}^{π / 4} (\sec^{2} x - 1) d x = \int_{0}^{π / 4} \sec^{2} x \cdot d x - \int_{0}^{π / 4} 1 . d x$
$= [\tan x]_{0}^{π / 4} - [x]_{0}^{π / 4} = [\tan \frac{π}{4} - \tan 0] - [\frac{π}{4} - 0]$
$= 1 - 0 - \frac{π}{4} = 1 - \frac{π}{4}$

AP EAMCET-2020-21.09.2020

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