QBManager

Integral Calculus

86707 $\int_{0}^{1} | 5 x - 3 | d x$ is equal to

1

\frac{10}{13}

2

\frac{31}{10}

3

\frac{13}{10}

4 None of these

Explanation:

(C) : $\int_{0}^{1} | 5 x - 3 | d x$
$| 5 x - 3 | = {\begin{cases} 5 x - 3 x \geq \frac{3}{5} \\ - (5 x - 3) x \leq \frac{3}{5} \end{cases}$
$I = \int_{0}^{1} | 5 x - 3 | d x = \int_{0}^{3 / 5} - (5 x - 3) d x + \int_{3 / 5}^{1} (5 x - 3) d x$
$= - {[\frac{5}{2} x^{2} - 3 x]}_{0}^{3 / 5} + {[\frac{5 x^{2}}{2} - 3 x]}_{3 / 5}^{1}$
$= - [(\frac{5}{2} \times {(\frac{3}{5})}^{2} - 3 \times \frac{3}{5}) - (0 - 0)] + [(\frac{5}{2} \times 1 - 3 \times 1) - (\frac{5}{2} \times \frac{9}{25} - \frac{9}{5})]$
$= - [(\frac{5}{2} \times {(\frac{3}{5})}^{2} - 3 \times \frac{3}{5})] + [(\frac{5}{2} - 3) - (\frac{45}{50} - \frac{9}{5})]$
$= - \frac{45}{50} + \frac{9}{5} - \frac{1}{2} - \frac{45}{50} + \frac{9}{5} = \frac{13}{10}$

UPSEE-2010

Integral Calculus

86728 $\int_{0}^{1} \log (\frac{1}{x} - 1) d x$ is equal to

1 1

2 0

3 2

4 None of the above

Explanation:

(B) : Given,
$\int_{0}^{1} \log (\frac{1}{x} - 1) d x$
Let, $I = \int_{0}^{1} \log (\frac{1}{x} - 1)$
$= \int_{0}^{1} \log (\frac{1 - x}{x}) d x$
$I = \int_{0}^{1} \log (\frac{1 - (1 - x)}{1 - x}) d x$
$I = \int_{0}^{1} \log (\frac{x}{1 - x}) d x$
Adding equation (i) and (ii)
$2 I = \int_{0}^{1} \log (\frac{1 - x}{x}) d x + \int_{0}^{1} \log (\frac{x}{1 - x}) d x$
$= \int_{0}^{1} \log (\frac{1 - x}{x} \cdot \frac{x}{1 - x}) d x = \int_{0}^{1} \log 1 d x = 0$
$∴ I = 0$

WB JEE-2019

Integral Calculus

86699 $\int_{2}^{1} | [x - 1] | d x =$
(where, [.] is greater integer function)

1 -3

2 6

3 -6

4 3

Explanation:

(B) : $\int_{- 2}^{1} | [x - 1] | d x =$
$- 2 < x < 1$
$- 2 - 1 < x - 1 < 1 - 1$
$- 3 < x - 1 < 0$
$∴ [x - 1] = - 3, - 2, - 1$
$| [x - 1] | = 3, 2, 1$
$\int_{- 2}^{1} | [x - 1] | d x = \int_{- 2}^{+ 1} 3 d x + \int_{- 1}^{0} 2 d x + \int_{0}^{1} 1 d x$
$= 3 [x]_{- 2}^{1} + 2 [x]_{- 1}^{0} + [x]_{0}^{1}$
$= 3 (1) + 2 (1) + 1 = 6$

MHT CET-2019

Integral Calculus

86700 $\int_{0}^{3} [x] d x =$
where $[x]$ is greatest integer function.

1 3

2 0

3 2

4 1

Explanation:

(A) : Given,
$\int_{0}^{1} 0 d x + \int_{1}^{2} 1 d x + \int_{2}^{3} 2 d x$
$= [x]_{1}^{2} + 2 [x]_{2}^{3}$
$= (2 - 1) + 2 (3 - 2)$
$= 1 + 2$
$= 3$

MHT CET-2017

Integral Calculus

86701 Evaluate $\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} d x$

1 1

2 0

3

1 / 2

4 2

Explanation:

(C) : Let $I = \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} d x$
Then, $I = \int_{1}^{2} \frac{\sqrt{3 - x}}{\sqrt{3 - (3 - x)} + \sqrt{3 - x}} d x$
$I = \int_{1}^{2} \frac{\sqrt{3 - x}}{\sqrt{x} + \sqrt{3 - x}} d x$
Adding equation (i) and (ii), we get-
$2 I = \int_{1}^{2} 1 . dx = [x]_{1}^{2} = 2 - 1 = 1$
$I = 1 / 2$

BITSAT-2005

Integral Calculus

86707 $\int_{0}^{1} | 5 x - 3 | d x$ is equal to

1

\frac{10}{13}

2

\frac{31}{10}

3

\frac{13}{10}

4 None of these

Explanation:

(C) : $\int_{0}^{1} | 5 x - 3 | d x$
$| 5 x - 3 | = {\begin{cases} 5 x - 3 x \geq \frac{3}{5} \\ - (5 x - 3) x \leq \frac{3}{5} \end{cases}$
$I = \int_{0}^{1} | 5 x - 3 | d x = \int_{0}^{3 / 5} - (5 x - 3) d x + \int_{3 / 5}^{1} (5 x - 3) d x$
$= - {[\frac{5}{2} x^{2} - 3 x]}_{0}^{3 / 5} + {[\frac{5 x^{2}}{2} - 3 x]}_{3 / 5}^{1}$
$= - [(\frac{5}{2} \times {(\frac{3}{5})}^{2} - 3 \times \frac{3}{5}) - (0 - 0)] + [(\frac{5}{2} \times 1 - 3 \times 1) - (\frac{5}{2} \times \frac{9}{25} - \frac{9}{5})]$
$= - [(\frac{5}{2} \times {(\frac{3}{5})}^{2} - 3 \times \frac{3}{5})] + [(\frac{5}{2} - 3) - (\frac{45}{50} - \frac{9}{5})]$
$= - \frac{45}{50} + \frac{9}{5} - \frac{1}{2} - \frac{45}{50} + \frac{9}{5} = \frac{13}{10}$

UPSEE-2010

Integral Calculus

86728 $\int_{0}^{1} \log (\frac{1}{x} - 1) d x$ is equal to

1 1

2 0

3 2

4 None of the above

Explanation:

(B) : Given,
$\int_{0}^{1} \log (\frac{1}{x} - 1) d x$
Let, $I = \int_{0}^{1} \log (\frac{1}{x} - 1)$
$= \int_{0}^{1} \log (\frac{1 - x}{x}) d x$
$I = \int_{0}^{1} \log (\frac{1 - (1 - x)}{1 - x}) d x$
$I = \int_{0}^{1} \log (\frac{x}{1 - x}) d x$
Adding equation (i) and (ii)
$2 I = \int_{0}^{1} \log (\frac{1 - x}{x}) d x + \int_{0}^{1} \log (\frac{x}{1 - x}) d x$
$= \int_{0}^{1} \log (\frac{1 - x}{x} \cdot \frac{x}{1 - x}) d x = \int_{0}^{1} \log 1 d x = 0$
$∴ I = 0$

WB JEE-2019

Integral Calculus

86699 $\int_{2}^{1} | [x - 1] | d x =$
(where, [.] is greater integer function)

1 -3

2 6

3 -6

4 3

Explanation:

(B) : $\int_{- 2}^{1} | [x - 1] | d x =$
$- 2 < x < 1$
$- 2 - 1 < x - 1 < 1 - 1$
$- 3 < x - 1 < 0$
$∴ [x - 1] = - 3, - 2, - 1$
$| [x - 1] | = 3, 2, 1$
$\int_{- 2}^{1} | [x - 1] | d x = \int_{- 2}^{+ 1} 3 d x + \int_{- 1}^{0} 2 d x + \int_{0}^{1} 1 d x$
$= 3 [x]_{- 2}^{1} + 2 [x]_{- 1}^{0} + [x]_{0}^{1}$
$= 3 (1) + 2 (1) + 1 = 6$

MHT CET-2019

Integral Calculus

86700 $\int_{0}^{3} [x] d x =$
where $[x]$ is greatest integer function.

1 3

2 0

3 2

4 1

Explanation:

(A) : Given,
$\int_{0}^{1} 0 d x + \int_{1}^{2} 1 d x + \int_{2}^{3} 2 d x$
$= [x]_{1}^{2} + 2 [x]_{2}^{3}$
$= (2 - 1) + 2 (3 - 2)$
$= 1 + 2$
$= 3$

MHT CET-2017

Integral Calculus

86701 Evaluate $\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} d x$

1 1

2 0

3

1 / 2

4 2

Explanation:

(C) : Let $I = \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} d x$
Then, $I = \int_{1}^{2} \frac{\sqrt{3 - x}}{\sqrt{3 - (3 - x)} + \sqrt{3 - x}} d x$
$I = \int_{1}^{2} \frac{\sqrt{3 - x}}{\sqrt{x} + \sqrt{3 - x}} d x$
Adding equation (i) and (ii), we get-
$2 I = \int_{1}^{2} 1 . dx = [x]_{1}^{2} = 2 - 1 = 1$
$I = 1 / 2$

BITSAT-2005

Integral Calculus

86707 $\int_{0}^{1} | 5 x - 3 | d x$ is equal to

1

\frac{10}{13}

2

\frac{31}{10}

3

\frac{13}{10}

4 None of these

Explanation:

(C) : $\int_{0}^{1} | 5 x - 3 | d x$
$| 5 x - 3 | = {\begin{cases} 5 x - 3 x \geq \frac{3}{5} \\ - (5 x - 3) x \leq \frac{3}{5} \end{cases}$
$I = \int_{0}^{1} | 5 x - 3 | d x = \int_{0}^{3 / 5} - (5 x - 3) d x + \int_{3 / 5}^{1} (5 x - 3) d x$
$= - {[\frac{5}{2} x^{2} - 3 x]}_{0}^{3 / 5} + {[\frac{5 x^{2}}{2} - 3 x]}_{3 / 5}^{1}$
$= - [(\frac{5}{2} \times {(\frac{3}{5})}^{2} - 3 \times \frac{3}{5}) - (0 - 0)] + [(\frac{5}{2} \times 1 - 3 \times 1) - (\frac{5}{2} \times \frac{9}{25} - \frac{9}{5})]$
$= - [(\frac{5}{2} \times {(\frac{3}{5})}^{2} - 3 \times \frac{3}{5})] + [(\frac{5}{2} - 3) - (\frac{45}{50} - \frac{9}{5})]$
$= - \frac{45}{50} + \frac{9}{5} - \frac{1}{2} - \frac{45}{50} + \frac{9}{5} = \frac{13}{10}$

UPSEE-2010

Integral Calculus

86728 $\int_{0}^{1} \log (\frac{1}{x} - 1) d x$ is equal to

1 1

2 0

3 2

4 None of the above

Explanation:

(B) : Given,
$\int_{0}^{1} \log (\frac{1}{x} - 1) d x$
Let, $I = \int_{0}^{1} \log (\frac{1}{x} - 1)$
$= \int_{0}^{1} \log (\frac{1 - x}{x}) d x$
$I = \int_{0}^{1} \log (\frac{1 - (1 - x)}{1 - x}) d x$
$I = \int_{0}^{1} \log (\frac{x}{1 - x}) d x$
Adding equation (i) and (ii)
$2 I = \int_{0}^{1} \log (\frac{1 - x}{x}) d x + \int_{0}^{1} \log (\frac{x}{1 - x}) d x$
$= \int_{0}^{1} \log (\frac{1 - x}{x} \cdot \frac{x}{1 - x}) d x = \int_{0}^{1} \log 1 d x = 0$
$∴ I = 0$

WB JEE-2019

Integral Calculus

86699 $\int_{2}^{1} | [x - 1] | d x =$
(where, [.] is greater integer function)

1 -3

2 6

3 -6

4 3

Explanation:

(B) : $\int_{- 2}^{1} | [x - 1] | d x =$
$- 2 < x < 1$
$- 2 - 1 < x - 1 < 1 - 1$
$- 3 < x - 1 < 0$
$∴ [x - 1] = - 3, - 2, - 1$
$| [x - 1] | = 3, 2, 1$
$\int_{- 2}^{1} | [x - 1] | d x = \int_{- 2}^{+ 1} 3 d x + \int_{- 1}^{0} 2 d x + \int_{0}^{1} 1 d x$
$= 3 [x]_{- 2}^{1} + 2 [x]_{- 1}^{0} + [x]_{0}^{1}$
$= 3 (1) + 2 (1) + 1 = 6$

MHT CET-2019

Integral Calculus

86700 $\int_{0}^{3} [x] d x =$
where $[x]$ is greatest integer function.

1 3

2 0

3 2

4 1

Explanation:

(A) : Given,
$\int_{0}^{1} 0 d x + \int_{1}^{2} 1 d x + \int_{2}^{3} 2 d x$
$= [x]_{1}^{2} + 2 [x]_{2}^{3}$
$= (2 - 1) + 2 (3 - 2)$
$= 1 + 2$
$= 3$

MHT CET-2017

Integral Calculus

86701 Evaluate $\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} d x$

1 1

2 0

3

1 / 2

4 2

Explanation:

(C) : Let $I = \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} d x$
Then, $I = \int_{1}^{2} \frac{\sqrt{3 - x}}{\sqrt{3 - (3 - x)} + \sqrt{3 - x}} d x$
$I = \int_{1}^{2} \frac{\sqrt{3 - x}}{\sqrt{x} + \sqrt{3 - x}} d x$
Adding equation (i) and (ii), we get-
$2 I = \int_{1}^{2} 1 . dx = [x]_{1}^{2} = 2 - 1 = 1$
$I = 1 / 2$

BITSAT-2005

Integral Calculus

86707 $\int_{0}^{1} | 5 x - 3 | d x$ is equal to

1

\frac{10}{13}

2

\frac{31}{10}

3

\frac{13}{10}

4 None of these

Explanation:

(C) : $\int_{0}^{1} | 5 x - 3 | d x$
$| 5 x - 3 | = {\begin{cases} 5 x - 3 x \geq \frac{3}{5} \\ - (5 x - 3) x \leq \frac{3}{5} \end{cases}$
$I = \int_{0}^{1} | 5 x - 3 | d x = \int_{0}^{3 / 5} - (5 x - 3) d x + \int_{3 / 5}^{1} (5 x - 3) d x$
$= - {[\frac{5}{2} x^{2} - 3 x]}_{0}^{3 / 5} + {[\frac{5 x^{2}}{2} - 3 x]}_{3 / 5}^{1}$
$= - [(\frac{5}{2} \times {(\frac{3}{5})}^{2} - 3 \times \frac{3}{5}) - (0 - 0)] + [(\frac{5}{2} \times 1 - 3 \times 1) - (\frac{5}{2} \times \frac{9}{25} - \frac{9}{5})]$
$= - [(\frac{5}{2} \times {(\frac{3}{5})}^{2} - 3 \times \frac{3}{5})] + [(\frac{5}{2} - 3) - (\frac{45}{50} - \frac{9}{5})]$
$= - \frac{45}{50} + \frac{9}{5} - \frac{1}{2} - \frac{45}{50} + \frac{9}{5} = \frac{13}{10}$

UPSEE-2010

Integral Calculus

86728 $\int_{0}^{1} \log (\frac{1}{x} - 1) d x$ is equal to

1 1

2 0

3 2

4 None of the above

Explanation:

(B) : Given,
$\int_{0}^{1} \log (\frac{1}{x} - 1) d x$
Let, $I = \int_{0}^{1} \log (\frac{1}{x} - 1)$
$= \int_{0}^{1} \log (\frac{1 - x}{x}) d x$
$I = \int_{0}^{1} \log (\frac{1 - (1 - x)}{1 - x}) d x$
$I = \int_{0}^{1} \log (\frac{x}{1 - x}) d x$
Adding equation (i) and (ii)
$2 I = \int_{0}^{1} \log (\frac{1 - x}{x}) d x + \int_{0}^{1} \log (\frac{x}{1 - x}) d x$
$= \int_{0}^{1} \log (\frac{1 - x}{x} \cdot \frac{x}{1 - x}) d x = \int_{0}^{1} \log 1 d x = 0$
$∴ I = 0$

WB JEE-2019

Integral Calculus

86699 $\int_{2}^{1} | [x - 1] | d x =$
(where, [.] is greater integer function)

1 -3

2 6

3 -6

4 3

Explanation:

(B) : $\int_{- 2}^{1} | [x - 1] | d x =$
$- 2 < x < 1$
$- 2 - 1 < x - 1 < 1 - 1$
$- 3 < x - 1 < 0$
$∴ [x - 1] = - 3, - 2, - 1$
$| [x - 1] | = 3, 2, 1$
$\int_{- 2}^{1} | [x - 1] | d x = \int_{- 2}^{+ 1} 3 d x + \int_{- 1}^{0} 2 d x + \int_{0}^{1} 1 d x$
$= 3 [x]_{- 2}^{1} + 2 [x]_{- 1}^{0} + [x]_{0}^{1}$
$= 3 (1) + 2 (1) + 1 = 6$

MHT CET-2019

Integral Calculus

86700 $\int_{0}^{3} [x] d x =$
where $[x]$ is greatest integer function.

1 3

2 0

3 2

4 1

Explanation:

(A) : Given,
$\int_{0}^{1} 0 d x + \int_{1}^{2} 1 d x + \int_{2}^{3} 2 d x$
$= [x]_{1}^{2} + 2 [x]_{2}^{3}$
$= (2 - 1) + 2 (3 - 2)$
$= 1 + 2$
$= 3$

MHT CET-2017

Integral Calculus

86701 Evaluate $\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} d x$

1 1

2 0

3

1 / 2

4 2

Explanation:

(C) : Let $I = \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} d x$
Then, $I = \int_{1}^{2} \frac{\sqrt{3 - x}}{\sqrt{3 - (3 - x)} + \sqrt{3 - x}} d x$
$I = \int_{1}^{2} \frac{\sqrt{3 - x}}{\sqrt{x} + \sqrt{3 - x}} d x$
Adding equation (i) and (ii), we get-
$2 I = \int_{1}^{2} 1 . dx = [x]_{1}^{2} = 2 - 1 = 1$
$I = 1 / 2$

BITSAT-2005

Integral Calculus

86707 $\int_{0}^{1} | 5 x - 3 | d x$ is equal to

1

\frac{10}{13}

2

\frac{31}{10}

3

\frac{13}{10}

4 None of these

Explanation:

(C) : $\int_{0}^{1} | 5 x - 3 | d x$
$| 5 x - 3 | = {\begin{cases} 5 x - 3 x \geq \frac{3}{5} \\ - (5 x - 3) x \leq \frac{3}{5} \end{cases}$
$I = \int_{0}^{1} | 5 x - 3 | d x = \int_{0}^{3 / 5} - (5 x - 3) d x + \int_{3 / 5}^{1} (5 x - 3) d x$
$= - {[\frac{5}{2} x^{2} - 3 x]}_{0}^{3 / 5} + {[\frac{5 x^{2}}{2} - 3 x]}_{3 / 5}^{1}$
$= - [(\frac{5}{2} \times {(\frac{3}{5})}^{2} - 3 \times \frac{3}{5}) - (0 - 0)] + [(\frac{5}{2} \times 1 - 3 \times 1) - (\frac{5}{2} \times \frac{9}{25} - \frac{9}{5})]$
$= - [(\frac{5}{2} \times {(\frac{3}{5})}^{2} - 3 \times \frac{3}{5})] + [(\frac{5}{2} - 3) - (\frac{45}{50} - \frac{9}{5})]$
$= - \frac{45}{50} + \frac{9}{5} - \frac{1}{2} - \frac{45}{50} + \frac{9}{5} = \frac{13}{10}$

UPSEE-2010

Integral Calculus

86728 $\int_{0}^{1} \log (\frac{1}{x} - 1) d x$ is equal to

1 1

2 0

3 2

4 None of the above

Explanation:

(B) : Given,
$\int_{0}^{1} \log (\frac{1}{x} - 1) d x$
Let, $I = \int_{0}^{1} \log (\frac{1}{x} - 1)$
$= \int_{0}^{1} \log (\frac{1 - x}{x}) d x$
$I = \int_{0}^{1} \log (\frac{1 - (1 - x)}{1 - x}) d x$
$I = \int_{0}^{1} \log (\frac{x}{1 - x}) d x$
Adding equation (i) and (ii)
$2 I = \int_{0}^{1} \log (\frac{1 - x}{x}) d x + \int_{0}^{1} \log (\frac{x}{1 - x}) d x$
$= \int_{0}^{1} \log (\frac{1 - x}{x} \cdot \frac{x}{1 - x}) d x = \int_{0}^{1} \log 1 d x = 0$
$∴ I = 0$

WB JEE-2019

Integral Calculus

86699 $\int_{2}^{1} | [x - 1] | d x =$
(where, [.] is greater integer function)

1 -3

2 6

3 -6

4 3

Explanation:

(B) : $\int_{- 2}^{1} | [x - 1] | d x =$
$- 2 < x < 1$
$- 2 - 1 < x - 1 < 1 - 1$
$- 3 < x - 1 < 0$
$∴ [x - 1] = - 3, - 2, - 1$
$| [x - 1] | = 3, 2, 1$
$\int_{- 2}^{1} | [x - 1] | d x = \int_{- 2}^{+ 1} 3 d x + \int_{- 1}^{0} 2 d x + \int_{0}^{1} 1 d x$
$= 3 [x]_{- 2}^{1} + 2 [x]_{- 1}^{0} + [x]_{0}^{1}$
$= 3 (1) + 2 (1) + 1 = 6$

MHT CET-2019

Integral Calculus

86700 $\int_{0}^{3} [x] d x =$
where $[x]$ is greatest integer function.

1 3

2 0

3 2

4 1

Explanation:

(A) : Given,
$\int_{0}^{1} 0 d x + \int_{1}^{2} 1 d x + \int_{2}^{3} 2 d x$
$= [x]_{1}^{2} + 2 [x]_{2}^{3}$
$= (2 - 1) + 2 (3 - 2)$
$= 1 + 2$
$= 3$

MHT CET-2017

Integral Calculus

86701 Evaluate $\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} d x$

1 1

2 0

3

1 / 2

4 2

Explanation:

(C) : Let $I = \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} d x$
Then, $I = \int_{1}^{2} \frac{\sqrt{3 - x}}{\sqrt{3 - (3 - x)} + \sqrt{3 - x}} d x$
$I = \int_{1}^{2} \frac{\sqrt{3 - x}}{\sqrt{x} + \sqrt{3 - x}} d x$
Adding equation (i) and (ii), we get-
$2 I = \int_{1}^{2} 1 . dx = [x]_{1}^{2} = 2 - 1 = 1$
$I = 1 / 2$

BITSAT-2005

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