Let I 1-0-4 x 2008tan x 2008 dx - I 2-0-4 x 2009tan x 2009 dx and I 3-0-4 x 2010tan x 2010 dxA- I 3- I 2- I 1B- I 2- I 3- I 1C- I 1- I 2- I 3D- I 3- I 1- I 2

The correct option is A I_3 lt; I_2 lt; I_10 lt; x lt;π/4. x^2008(tan x)^2008 gt; x^2009 (tan x)^2009 gt; x^2010 (tan x)^2010 ∫_0^π/4 x^2008 ...

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Byju's Answer

Standard XII

Mathematics

Property 4

Let I 1=∫0π/4...

Question

Let $I_{1} = \int_{0}^{\frac{π}{4}} x^{2008} (t a n x)^{2008} d x, I_{2} = \int_{0}^{\frac{π}{4}} x^{2009} (t a n x)^{2009} d x a n d I_{3} = \int_{0}^{\frac{π}{4}} x^{2010} (t a n x)^{2010} d x$

I_{2} < I_{3} < I_{1}

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I_{1} < I_{2} < I_{3}

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I_{3} < I_{1} < I_{2}

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I_{3} < I_{2} < I_{1}

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Solution

The correct option is D $I_{3} < I_{2} < I_{1}$
$0 < x < \frac{π}{4} . x^{2008} (t a n x)^{2008} > x^{2009} (t a n x)^{2009} > x^{2010} (t a n x)^{2010} \int_{0}^{\frac{π}{4}} x^{2008} (t a n x)^{2008} d x > \int_{0}^{\frac{π}{4}} x^{2009} > \int_{0}^{\frac{π}{4}} x^{2010} (t a n x)^{2010} d x . I_{1} > I_{2} > I_{3}$

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Similar questions

Q. If

I_{1} = π / 2 \int 0 cos (sin x) d x; I_{2} = π / 2 \int 0 sin (cos x) d x

and

I_{3} = π / 2 \int 0 cos x d x,

then

Q. Suppose

I_{1} = \int_{0}^{\frac{π}{2}} c o s (π s i n^{2} x) d x, I_{2} = \int_{0}^{\frac{π}{2}} c o s (2 π s i n^{2} x) d x a n d I_{3} = \int_{0}^{\frac{π}{2}} c o s (π s i n x) d x,

then

Let $I_{1} = {(\frac{π}{4})}^{2} + \sqrt{2}, I_{2} = {(t a n^{- 1} (\frac{1}{e}))}^{2} + \frac{2 e}{\sqrt{e^{2} + 1}}, I_{3} = (t a n^{- 1} e)^{2} + \frac{2}{\sqrt{e^{2} + 1}}$ , then which of the following is true?

Q. If

I_{1} = \int_{0}^{π / 2} {sin}^{4} x d x

I_{2} = \int_{0}^{π / 2} {cos}^{6} x d x

I_{3} = \int_{0}^{π / 2} {sin}^{8} x d x

I_{4} = \int_{0}^{π / 2} {cos}^{2} x d x

then the increasing order of

I_{1}, I_{2}, I_{3}, I_{4}

is?

Q. Assertion :Consider

I_{1} = \int_{0}^{\frac{π}{4}} e^{x^{2}} d x, I_{2} = \int_{0}^{\frac{π}{4}} e^{x} d x, I_{3} = \int_{0}^{\frac{π}{4}} e^{x^{2}} cos x d x, I_{4} = \int_{0}^{\frac{π}{4}} e^{x^{2}} sin x d x

,
then

I_{2} > I_{1} > I_{3} > I_{4} .

Reason: For

x (0, 1), x > x^{2}

and

sin x > cos x .

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Let I1=∫π40x2008(tan x)2008dx,I2=∫π40x2009(tan x)2009dx and I3=∫π40x2010(tan x)2010dx

The correct option is D I3<I2<I10<x<π4.x2008(tan x)2008>x2009(tan x)2009>x2010(tan x)2010∫π40 x2008(tan x)2008dx>∫π40 x2009>∫π40 x2010(tan x)2010 dx.I1>I2>I3

Let $I_{1} = \int_{0}^{\frac{π}{4}} x^{2008} (t a n x)^{2008} d x, I_{2} = \int_{0}^{\frac{π}{4}} x^{2009} (t a n x)^{2009} d x a n d I_{3} = \int_{0}^{\frac{π}{4}} x^{2010} (t a n x)^{2010} d x$