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1

WB JEE 2008

MCQ (Single Correct Answer)

The value of the integral $$\int\limits_{ - a}^a {{{x{e^{{x^2}}}} \over {1 + {x^2}}}dx} $$ is

A
$${{e^{{a^2}}}}$$
B
0
C
$${e^{ - {a^2}}}$$
D
a

Explanation

Let $$I = \int\limits_{ - a}^a {{{x{e^{{x^2}}}} \over {1 + {x^2}}}dx} $$

$$f(x) = {{x{e^{{x^2}}}} \over {1 + {x^2}}}$$ is an odd function

$$\therefore$$ $$I = 0$$. ($$\because$$ $$f( - x) = {{ - x{e^{{{( - x)}^2}}}} \over {(1 + {{( - x)}^2}}} = - {{x{e^{{x^2}}}} \over {1 + {x^2}}} = - f(x)$$)

2

WB JEE 2008

MCQ (Single Correct Answer)

The value of $$\int\limits_{ - 3}^3 {(a{x^5} + b{x^3} + cx + k)dx} $$, where a, b, c, k are constants, depends only on

A
a and k
B
a and b
C
a, b and c
D
k

Explanation

Let $$I = \int\limits_{ - 3}^3 {(a{x^5} + b{x^3} + cx + k)dx} $$

$$ = a\int\limits_{ - 3}^3 {{x^5}dx + b\int\limits_{ - 3}^3 {{x^3}dx + c\int\limits_{ - 3}^3 {x\,dx + k\int\limits_{ - 3}^3 {1\,dx} } } } $$

Since x5, x3, x all are odd functions

$$\therefore$$ $$I = 0 + 0 + 0 + k\left| x \right|_{ - 3}^3$$

$$\Rightarrow$$ I = 6k.

3

WB JEE 2008

MCQ (Single Correct Answer)

The value of $$\int\limits_0^\pi {|\cos x|dx} $$ is

A
2$$\pi$$
B
2
C
2/$$\pi$$
D
$$\pi$$

Explanation

Let $$I = \int\limits_0^\pi {|\cos x|dx} $$

$$I = \int\limits_0^{\pi /2} {\cos x\,dx + \int\limits_{\pi /2}^\pi { - \cos x\,dx} } $$

$$ = \left| {\sin x} \right|_0^{\pi /2} - \left| {\sin x} \right|_{\pi /2}^\pi $$

$$ = 1 - (0 - 1) = 2$$.

4

WB JEE 2008

MCQ (Single Correct Answer)

The value of the integral $$\int\limits_0^2 {|{x^2} - 1|dx} $$ is

A
0
B
2
C
$$-$$1/3
D
$$-$$2

Explanation

Let $$I = \int\limits_0^2 {|{x^2} - 1|dx = \int\limits_0^1 { - ({x^2} - 1)dx + \int\limits_1^2 {({x^2} - 1)dx} } } $$

$$ = - \left[ {{{{x^3}} \over 3} - x} \right]_0^1 + \left[ {{{{x^3}} \over 3} - x} \right]_1^2$$

$$ = - {1 \over 3} + 1 + {8 \over 3} - 2 - {1 \over 3} + 1 = {8 \over 3} - {2 \over 3} = {6 \over 3} = 2$$.

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