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1

WB JEE 2009

MCQ (Single Correct Answer)

$$\int\limits_{ - 1}^4 {f(x)dx = 4} $$ and $$\int\limits_2^4 {\{ 3 - f(x)\} dx = 7} $$, then the value of $$\int\limits_{ - 1}^2 {f(x)dx} $$ is

A
$$-$$2
B
3
C
4
D
5

Explanation

Given $$\int\limits_{ - 1}^4 {f(x)dx = 4} $$ ..... (i)

$$\int\limits_2^4 {\{ 3 - f(x)\} dx = 7 \Rightarrow 3\,.\,2 - \int\limits_2^4 {f(x)dx = 7} } $$

$$ \Rightarrow \int\limits_2^4 {f(x)dx = - 1} $$

$$ \Rightarrow - \int\limits_4^2 {f(x)dx = - 1 \Rightarrow \int\limits_4^2 {f(x)dx = 1} } $$ .... (ii)

($$\because$$ $$\int\limits_a^b {f(x)dx = - \int\limits_b^a {f(x)dx} } $$)

Adding (i) and (ii)

$$\int\limits_{ - 1}^4 {f(x)dx + \int\limits_4^2 {f(x)dx = 4 + 1 \Rightarrow \int\limits_{ - 1}^2 {f(x)dx = 5} } } $$ [$$\because$$ $$\int\limits_a^b {f(x)dx + \int\limits_b^c {f(x)dx = \int\limits_a^c {f(x)dx} } } $$]

2

WB JEE 2009

MCQ (Single Correct Answer)

If $${I_1} = \int\limits_0^{\pi /4} {{{\sin }^2}xdx} $$ and $${I_2} = \int\limits_0^{\pi /4} {{{\cos }^2}xdx} $$, then

A
$${I_1} = {I_2}$$
B
$${I_1} < {I_2}$$
C
$${I_1} > {I_2}$$
D
$${I_2} = {I_1} + \pi /4$$

Explanation

$${I_1} = \int\limits_0^{\pi /4} {{{\sin }^2}xdx} $$ is the area of curve bounded between

$$y = {\sin ^2}x$$, x-axis

x = 0 and x = $$\pi$$/4

$${I_2} = \int\limits_0^{\pi /4} {{{\cos }^2}xdx} $$ is the area of curve bounded between

$$y = {\cos ^2}x$$, x-axis, x = 0 and x = $$\pi$$/4

From figure, area under I2 is greater than area under I1.

3

WB JEE 2009

MCQ (Single Correct Answer)

The value of $$\int\limits_0^\infty {{{dx} \over {({x^2} + 4)({x^2} + 9)}}} $$ is

A
$${\pi \over {60}}$$
B
$${\pi \over {20}}$$
C
$${\pi \over {40}}$$
D
$${\pi \over {80}}$$

Explanation

$$\int\limits_0^\infty {{{dx} \over {({x^2} + 4)({x^2} + 9)}} = {1 \over 5}\int\limits_0^\infty {\left[ {{1 \over {{x^2} + 4}} - {1 \over {{x^2} + 9}}} \right]dx} } $$

$$ = {1 \over 5}\left[ {\left[ {{1 \over 2}{{\tan }^{ - 1}}{x \over 2}} \right]_0^\infty - \left[ {{1 \over 3}{{\tan }^{ - 1}}{x \over 3}} \right]_0^\infty } \right]$$

$$ = {1 \over 5}\left[ {{1 \over 2}{{\tan }^{ - 1}}\infty - {1 \over 3}{{\tan }^{ - 1}}\infty } \right] = {1 \over 5}\left[ {{\pi \over 4} - {\pi \over 6}} \right] = {\pi \over {60}}$$

4

WB JEE 2009

MCQ (Single Correct Answer)

If $$f(x) = f(a - x)$$, then $$\int\limits_0^a {xf(x)dx} $$ is equal to

A
$$\int\limits_0^a {f(x)dx} $$
B
$${{{a^2}} \over 2}\int\limits_0^a {f(x)dx} $$
C
$${a \over 2}\int\limits_0^a {f(x)dx} $$
D
$$ - {a \over 2}\int\limits_0^a {f(x)dx} $$

Explanation

Let $$I = \int\limits_0^a {xf(x)dx = \int\limits_0^a {(a - x)f(a - x)dx} } $$ ($$\because$$ $$\int\limits_0^a {f(x)dx = \int\limits_0^a {f(a - x)dx} } $$)

$$ = a\int\limits_0^a {f(a - x)dx - \int\limits_0^a {xf(a - x)dx} } $$

$$ = a\int\limits_0^a {f(x)dx - \int\limits_0^a {xf(x)dx} } $$ ($$\because$$ $$f(a - x) = f(x)$$)

$$ = a\int\limits_0^a {f(x)dx - I \Rightarrow 2I = a\int\limits_0^a {f(x)dx \Rightarrow I = {a \over 2}\int\limits_0^a {f(x)dx} } } $$

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