QBManager

Integral Calculus

86454 $\int_{0}^{10} \frac{x^{10}}{(10 - x)^{10} + x^{10}} d x$ is equal to

1 10

2 5

3 2

4

\frac{1}{2}

5 0

Explanation:

(B) : Given, $I = \int_{0}^{10} \frac{x^{10}}{(10 - x)^{10} + x^{10}} d x$
$∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
$I = \int_{0}^{10} \frac{(10 - x)^{10}}{x^{10} + (10 - x)^{10}} d x$
Adding equation (i) and (ii), we get-
$2 I = \int_{0}^{10} \frac{(10 - x)^{10} + x^{10}}{(10 - x)^{10} + x^{10}} d x$
$2 I = \int_{0}^{10} 1 d x$
$2 I = [x]_{0}^{10}$
$I = \frac{1}{2} [10 - 0] = 5$

Kerala CEE-2012

Integral Calculus

86455 $\int_{0}^{1} x^{- 5 x} d x$ is equal to

1

\frac{1}{25} - \frac{6 e^{- 5}}{25}

2

\frac{1}{25} + \frac{6 e^{- 5}}{25}

3

- \frac{1}{25} - \frac{6 e^{- 5}}{25}

4

\frac{1}{25} - \frac{1}{5} e^{- 5}

5

\frac{1}{25} + \frac{1}{5} e^{- 5}

Explanation:

(A) : Given,
$I = \int_{0}^{1} x e^{- 5 x} dx$
$∵ \int {xe}^{- 5 x} dx = - \frac{x}{5} e^{- 5 x} - \int 1 \cdot \frac{e^{- 5 x}}{- 5} dx$
$= \frac{- x}{5} e^{- 5 x} + \frac{1}{5} \int e^{- 5 x} d x$
$= - \frac{x}{5} e^{- 5 x} + \frac{1}{5} \times - \frac{1}{5} e^{- 5 x} = - \frac{e^{- 5 x}}{25} (5 x + 1)$
$So, I = {[- \frac{e^{- 5 x}}{25} (5 x + 1)]}_{0}^{1}$
$= [- \frac{e^{- 5}}{25} (5 \times 1 + 1) - (- \frac{e^{0}}{25} (0 + 1))]$
$= [- \frac{e^{- 5}}{25} (5 \times 1 + 1) - (- \frac{e^{0}}{25} (0 + 1))] = \frac{1}{25} - \frac{6 e^{- 5}}{25}$

Kerala CEE-2011

Integral Calculus

86456 If $\int_{0}^{π} x f (\sin x) d x = A \int_{0}^{π / 2} f (\sin x) d x$ , then $A$ is equal to

1 0

2

π

3

\frac{π}{4}

4

2 π

5

3 π

Explanation:

(B) : Given,
$\int_{0}^{π} x f (\sin x) d x = A \int_{0}^{π / 2} f (\sin x) d x$
Let, $I = \int_{0}^{π} x f (\sin x) dx$
$= \int_{0}^{π} (π - x) f (\sin (π - x)) dx$
$(∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x)$
$I = \int_{0}^{π} (π - x) f (\sin x) d x$
On adding equation (i )and (ii), we get -
$2 I = π \int_{0}^{π} f (\sin x) d x$
$(∵ \int_{0}^{a} f (x) d x = 2 \int_{0}^{a / 2} f (x) d x)$
$I = \frac{π}{2} \int_{0}^{π} f (\sin x) dx = π \int_{0}^{π / 2} f (\sin x) dx$
$A \int_{0}^{π / 2} f (\sin x) d x = π \int_{0}^{π / 2} f (\sin x) d x$
$A = π$

Kerala CEE-2008

Integral Calculus

86457 $\int_{- 1}^{1} \frac{17 x^{5} - x^{4} + 29 x^{3} - 31 x + 1}{x^{2} + 1} d x$ is

1

4 / 5

2

5 / 4

3

4 / 3

4

3 / 4

5 6

Explanation:

(C) : Given,
$I = \int_{- 1}^{1} \frac{17 x^{5} - x^{4} + 29 x^{3} - 31 x + 1}{x^{2} + 1} d x$
$= \int_{- 1}^{1} (\frac{17 x^{5} + 29 x^{3} - 31 x}{x^{2} + 1} - \frac{x^{4} - 1}{x^{2} + 1}) d x$
$= \int_{- 1}^{1} \frac{17 x^{5} + 29 x^{3} - 31 x}{x^{2} + 1} d x - \int_{- 1}^{1} \frac{(x^{2} + 1) (x^{2} - 1)}{x^{2} + 1} d x$
$= \int_{- 1}^{1} \frac{17 x^{5} + 29 x^{3} - 31 x}{x^{2} + 1} d x - \int_{- 1}^{1} (x^{2} - 1) d x$
$= 0 - 2 \int_{0}^{1} \frac{(x^{2} - 1) (x^{2} + 1)}{(x^{2} + 1)} dx$
$= - 2 {[(\frac{x^{3}}{3} - x)]}_{0}^{1} = - 2 [\frac{1}{3} - 1] = \frac{4}{3}$

Kerala CEE-2007

Integral Calculus

86454 $\int_{0}^{10} \frac{x^{10}}{(10 - x)^{10} + x^{10}} d x$ is equal to

1 10

2 5

3 2

4

\frac{1}{2}

5 0

Explanation:

(B) : Given, $I = \int_{0}^{10} \frac{x^{10}}{(10 - x)^{10} + x^{10}} d x$
$∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
$I = \int_{0}^{10} \frac{(10 - x)^{10}}{x^{10} + (10 - x)^{10}} d x$
Adding equation (i) and (ii), we get-
$2 I = \int_{0}^{10} \frac{(10 - x)^{10} + x^{10}}{(10 - x)^{10} + x^{10}} d x$
$2 I = \int_{0}^{10} 1 d x$
$2 I = [x]_{0}^{10}$
$I = \frac{1}{2} [10 - 0] = 5$

Kerala CEE-2012

Integral Calculus

86455 $\int_{0}^{1} x^{- 5 x} d x$ is equal to

1

\frac{1}{25} - \frac{6 e^{- 5}}{25}

2

\frac{1}{25} + \frac{6 e^{- 5}}{25}

3

- \frac{1}{25} - \frac{6 e^{- 5}}{25}

4

\frac{1}{25} - \frac{1}{5} e^{- 5}

5

\frac{1}{25} + \frac{1}{5} e^{- 5}

Explanation:

(A) : Given,
$I = \int_{0}^{1} x e^{- 5 x} dx$
$∵ \int {xe}^{- 5 x} dx = - \frac{x}{5} e^{- 5 x} - \int 1 \cdot \frac{e^{- 5 x}}{- 5} dx$
$= \frac{- x}{5} e^{- 5 x} + \frac{1}{5} \int e^{- 5 x} d x$
$= - \frac{x}{5} e^{- 5 x} + \frac{1}{5} \times - \frac{1}{5} e^{- 5 x} = - \frac{e^{- 5 x}}{25} (5 x + 1)$
$So, I = {[- \frac{e^{- 5 x}}{25} (5 x + 1)]}_{0}^{1}$
$= [- \frac{e^{- 5}}{25} (5 \times 1 + 1) - (- \frac{e^{0}}{25} (0 + 1))]$
$= [- \frac{e^{- 5}}{25} (5 \times 1 + 1) - (- \frac{e^{0}}{25} (0 + 1))] = \frac{1}{25} - \frac{6 e^{- 5}}{25}$

Kerala CEE-2011

Integral Calculus

86456 If $\int_{0}^{π} x f (\sin x) d x = A \int_{0}^{π / 2} f (\sin x) d x$ , then $A$ is equal to

1 0

2

π

3

\frac{π}{4}

4

2 π

5

3 π

Explanation:

(B) : Given,
$\int_{0}^{π} x f (\sin x) d x = A \int_{0}^{π / 2} f (\sin x) d x$
Let, $I = \int_{0}^{π} x f (\sin x) dx$
$= \int_{0}^{π} (π - x) f (\sin (π - x)) dx$
$(∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x)$
$I = \int_{0}^{π} (π - x) f (\sin x) d x$
On adding equation (i )and (ii), we get -
$2 I = π \int_{0}^{π} f (\sin x) d x$
$(∵ \int_{0}^{a} f (x) d x = 2 \int_{0}^{a / 2} f (x) d x)$
$I = \frac{π}{2} \int_{0}^{π} f (\sin x) dx = π \int_{0}^{π / 2} f (\sin x) dx$
$A \int_{0}^{π / 2} f (\sin x) d x = π \int_{0}^{π / 2} f (\sin x) d x$
$A = π$

Kerala CEE-2008

Integral Calculus

86457 $\int_{- 1}^{1} \frac{17 x^{5} - x^{4} + 29 x^{3} - 31 x + 1}{x^{2} + 1} d x$ is

1

4 / 5

2

5 / 4

3

4 / 3

4

3 / 4

5 6

Explanation:

(C) : Given,
$I = \int_{- 1}^{1} \frac{17 x^{5} - x^{4} + 29 x^{3} - 31 x + 1}{x^{2} + 1} d x$
$= \int_{- 1}^{1} (\frac{17 x^{5} + 29 x^{3} - 31 x}{x^{2} + 1} - \frac{x^{4} - 1}{x^{2} + 1}) d x$
$= \int_{- 1}^{1} \frac{17 x^{5} + 29 x^{3} - 31 x}{x^{2} + 1} d x - \int_{- 1}^{1} \frac{(x^{2} + 1) (x^{2} - 1)}{x^{2} + 1} d x$
$= \int_{- 1}^{1} \frac{17 x^{5} + 29 x^{3} - 31 x}{x^{2} + 1} d x - \int_{- 1}^{1} (x^{2} - 1) d x$
$= 0 - 2 \int_{0}^{1} \frac{(x^{2} - 1) (x^{2} + 1)}{(x^{2} + 1)} dx$
$= - 2 {[(\frac{x^{3}}{3} - x)]}_{0}^{1} = - 2 [\frac{1}{3} - 1] = \frac{4}{3}$

Kerala CEE-2007

Integral Calculus

86454 $\int_{0}^{10} \frac{x^{10}}{(10 - x)^{10} + x^{10}} d x$ is equal to

1 10

2 5

3 2

4

\frac{1}{2}

5 0

Explanation:

(B) : Given, $I = \int_{0}^{10} \frac{x^{10}}{(10 - x)^{10} + x^{10}} d x$
$∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
$I = \int_{0}^{10} \frac{(10 - x)^{10}}{x^{10} + (10 - x)^{10}} d x$
Adding equation (i) and (ii), we get-
$2 I = \int_{0}^{10} \frac{(10 - x)^{10} + x^{10}}{(10 - x)^{10} + x^{10}} d x$
$2 I = \int_{0}^{10} 1 d x$
$2 I = [x]_{0}^{10}$
$I = \frac{1}{2} [10 - 0] = 5$

Kerala CEE-2012

Integral Calculus

86455 $\int_{0}^{1} x^{- 5 x} d x$ is equal to

1

\frac{1}{25} - \frac{6 e^{- 5}}{25}

2

\frac{1}{25} + \frac{6 e^{- 5}}{25}

3

- \frac{1}{25} - \frac{6 e^{- 5}}{25}

4

\frac{1}{25} - \frac{1}{5} e^{- 5}

5

\frac{1}{25} + \frac{1}{5} e^{- 5}

Explanation:

(A) : Given,
$I = \int_{0}^{1} x e^{- 5 x} dx$
$∵ \int {xe}^{- 5 x} dx = - \frac{x}{5} e^{- 5 x} - \int 1 \cdot \frac{e^{- 5 x}}{- 5} dx$
$= \frac{- x}{5} e^{- 5 x} + \frac{1}{5} \int e^{- 5 x} d x$
$= - \frac{x}{5} e^{- 5 x} + \frac{1}{5} \times - \frac{1}{5} e^{- 5 x} = - \frac{e^{- 5 x}}{25} (5 x + 1)$
$So, I = {[- \frac{e^{- 5 x}}{25} (5 x + 1)]}_{0}^{1}$
$= [- \frac{e^{- 5}}{25} (5 \times 1 + 1) - (- \frac{e^{0}}{25} (0 + 1))]$
$= [- \frac{e^{- 5}}{25} (5 \times 1 + 1) - (- \frac{e^{0}}{25} (0 + 1))] = \frac{1}{25} - \frac{6 e^{- 5}}{25}$

Kerala CEE-2011

Integral Calculus

86456 If $\int_{0}^{π} x f (\sin x) d x = A \int_{0}^{π / 2} f (\sin x) d x$ , then $A$ is equal to

1 0

2

π

3

\frac{π}{4}

4

2 π

5

3 π

Explanation:

(B) : Given,
$\int_{0}^{π} x f (\sin x) d x = A \int_{0}^{π / 2} f (\sin x) d x$
Let, $I = \int_{0}^{π} x f (\sin x) dx$
$= \int_{0}^{π} (π - x) f (\sin (π - x)) dx$
$(∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x)$
$I = \int_{0}^{π} (π - x) f (\sin x) d x$
On adding equation (i )and (ii), we get -
$2 I = π \int_{0}^{π} f (\sin x) d x$
$(∵ \int_{0}^{a} f (x) d x = 2 \int_{0}^{a / 2} f (x) d x)$
$I = \frac{π}{2} \int_{0}^{π} f (\sin x) dx = π \int_{0}^{π / 2} f (\sin x) dx$
$A \int_{0}^{π / 2} f (\sin x) d x = π \int_{0}^{π / 2} f (\sin x) d x$
$A = π$

Kerala CEE-2008

Integral Calculus

86457 $\int_{- 1}^{1} \frac{17 x^{5} - x^{4} + 29 x^{3} - 31 x + 1}{x^{2} + 1} d x$ is

1

4 / 5

2

5 / 4

3

4 / 3

4

3 / 4

5 6

Explanation:

(C) : Given,
$I = \int_{- 1}^{1} \frac{17 x^{5} - x^{4} + 29 x^{3} - 31 x + 1}{x^{2} + 1} d x$
$= \int_{- 1}^{1} (\frac{17 x^{5} + 29 x^{3} - 31 x}{x^{2} + 1} - \frac{x^{4} - 1}{x^{2} + 1}) d x$
$= \int_{- 1}^{1} \frac{17 x^{5} + 29 x^{3} - 31 x}{x^{2} + 1} d x - \int_{- 1}^{1} \frac{(x^{2} + 1) (x^{2} - 1)}{x^{2} + 1} d x$
$= \int_{- 1}^{1} \frac{17 x^{5} + 29 x^{3} - 31 x}{x^{2} + 1} d x - \int_{- 1}^{1} (x^{2} - 1) d x$
$= 0 - 2 \int_{0}^{1} \frac{(x^{2} - 1) (x^{2} + 1)}{(x^{2} + 1)} dx$
$= - 2 {[(\frac{x^{3}}{3} - x)]}_{0}^{1} = - 2 [\frac{1}{3} - 1] = \frac{4}{3}$

Kerala CEE-2007

Integral Calculus

86454 $\int_{0}^{10} \frac{x^{10}}{(10 - x)^{10} + x^{10}} d x$ is equal to

1 10

2 5

3 2

4

\frac{1}{2}

5 0

Explanation:

(B) : Given, $I = \int_{0}^{10} \frac{x^{10}}{(10 - x)^{10} + x^{10}} d x$
$∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
$I = \int_{0}^{10} \frac{(10 - x)^{10}}{x^{10} + (10 - x)^{10}} d x$
Adding equation (i) and (ii), we get-
$2 I = \int_{0}^{10} \frac{(10 - x)^{10} + x^{10}}{(10 - x)^{10} + x^{10}} d x$
$2 I = \int_{0}^{10} 1 d x$
$2 I = [x]_{0}^{10}$
$I = \frac{1}{2} [10 - 0] = 5$

Kerala CEE-2012

Integral Calculus

86455 $\int_{0}^{1} x^{- 5 x} d x$ is equal to

1

\frac{1}{25} - \frac{6 e^{- 5}}{25}

2

\frac{1}{25} + \frac{6 e^{- 5}}{25}

3

- \frac{1}{25} - \frac{6 e^{- 5}}{25}

4

\frac{1}{25} - \frac{1}{5} e^{- 5}

5

\frac{1}{25} + \frac{1}{5} e^{- 5}

Explanation:

(A) : Given,
$I = \int_{0}^{1} x e^{- 5 x} dx$
$∵ \int {xe}^{- 5 x} dx = - \frac{x}{5} e^{- 5 x} - \int 1 \cdot \frac{e^{- 5 x}}{- 5} dx$
$= \frac{- x}{5} e^{- 5 x} + \frac{1}{5} \int e^{- 5 x} d x$
$= - \frac{x}{5} e^{- 5 x} + \frac{1}{5} \times - \frac{1}{5} e^{- 5 x} = - \frac{e^{- 5 x}}{25} (5 x + 1)$
$So, I = {[- \frac{e^{- 5 x}}{25} (5 x + 1)]}_{0}^{1}$
$= [- \frac{e^{- 5}}{25} (5 \times 1 + 1) - (- \frac{e^{0}}{25} (0 + 1))]$
$= [- \frac{e^{- 5}}{25} (5 \times 1 + 1) - (- \frac{e^{0}}{25} (0 + 1))] = \frac{1}{25} - \frac{6 e^{- 5}}{25}$

Kerala CEE-2011

Integral Calculus

86456 If $\int_{0}^{π} x f (\sin x) d x = A \int_{0}^{π / 2} f (\sin x) d x$ , then $A$ is equal to

1 0

2

π

3

\frac{π}{4}

4

2 π

5

3 π

Explanation:

(B) : Given,
$\int_{0}^{π} x f (\sin x) d x = A \int_{0}^{π / 2} f (\sin x) d x$
Let, $I = \int_{0}^{π} x f (\sin x) dx$
$= \int_{0}^{π} (π - x) f (\sin (π - x)) dx$
$(∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x)$
$I = \int_{0}^{π} (π - x) f (\sin x) d x$
On adding equation (i )and (ii), we get -
$2 I = π \int_{0}^{π} f (\sin x) d x$
$(∵ \int_{0}^{a} f (x) d x = 2 \int_{0}^{a / 2} f (x) d x)$
$I = \frac{π}{2} \int_{0}^{π} f (\sin x) dx = π \int_{0}^{π / 2} f (\sin x) dx$
$A \int_{0}^{π / 2} f (\sin x) d x = π \int_{0}^{π / 2} f (\sin x) d x$
$A = π$

Kerala CEE-2008

Integral Calculus

86457 $\int_{- 1}^{1} \frac{17 x^{5} - x^{4} + 29 x^{3} - 31 x + 1}{x^{2} + 1} d x$ is

1

4 / 5

2

5 / 4

3

4 / 3

4

3 / 4

5 6

Explanation:

(C) : Given,
$I = \int_{- 1}^{1} \frac{17 x^{5} - x^{4} + 29 x^{3} - 31 x + 1}{x^{2} + 1} d x$
$= \int_{- 1}^{1} (\frac{17 x^{5} + 29 x^{3} - 31 x}{x^{2} + 1} - \frac{x^{4} - 1}{x^{2} + 1}) d x$
$= \int_{- 1}^{1} \frac{17 x^{5} + 29 x^{3} - 31 x}{x^{2} + 1} d x - \int_{- 1}^{1} \frac{(x^{2} + 1) (x^{2} - 1)}{x^{2} + 1} d x$
$= \int_{- 1}^{1} \frac{17 x^{5} + 29 x^{3} - 31 x}{x^{2} + 1} d x - \int_{- 1}^{1} (x^{2} - 1) d x$
$= 0 - 2 \int_{0}^{1} \frac{(x^{2} - 1) (x^{2} + 1)}{(x^{2} + 1)} dx$
$= - 2 {[(\frac{x^{3}}{3} - x)]}_{0}^{1} = - 2 [\frac{1}{3} - 1] = \frac{4}{3}$

Kerala CEE-2007