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Solved Subjective Question on Definite Integral Set 3

Posted on - 04-01-2017

Math

IIT JEE

Example.1

Evaluate.

Solution

I = = I1 + I2

where I1 =

Put x = – t
&⇒
dx = - dt


&⇒
I1 =

Hence I = I1 + I2 =

I =


&⇒
2I = 2p= 2p
&⇒
I = p

Example.2

If f, g, h be continuous functions on [0, a] such that f(a – x) = - f(x), g(a –x) = g(x) and 3h(x) – 4h(a –x) = 5, then prove that .

Solution

I = =

= -

7I = 3I + 4I

=

= 5(since f(a – x) g(a – x) = – f(x) g(x) ).


&⇒
I = 0
. .

Example.3

Evaluate (i)

(ii)

where [.] denotes the greatest integer function.

Solution

(i) I = 2, 0 < x < p


&⇒
0 < x2 < p2
&⇒
–3
.14 < x2 –p < 6.72.

= sin 4(p –3) + sin 3(1) + sin 2(1) + sin 1 (1) + ....+ sin (1) + .... + sin (1) + ....+ sin 6(p2 –p –6).

= 2 sin 1 + 2 sin 2 + 2 sin 3 + (p –2) sin 4 + sin 5 + (p2 –p –6) sin 6

(ii). x2 – 3x + 3 > 0, " x ∈ R (D < 0)

It attains the minimum value = at x =

And the graph is parabola

x2 –3x + 3 = 1, when x2 –3x + 2 = 0, i.e. when x = 1, 2.

x2 –3x + 3 = 2, when x2 –3x + 1 = 0, i.e. when x =

x2 –3x + 3 = 3, when x2 –3x = 0, i.e. when x = 0, 3.

Hence [x2 –3x + 3] =

I = = 0

Example.4

Evaluate a > b > 0.

Solution

I = dq …. (1)

= 2 dq.

Also, I = 2dq …. (2)

Adding (1) and (2),

2I = 2= 8a

= 8a= 8a

put a tanq = t
&⇒
I = 4dt

=

=

= = .

Example.5

Prove that .

Solution

=


&⇒
.

 
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