Let I 1-3-6sin x - x dx - I 2-3-6sin sin x -sin x dx - I 3-3-6sin tan x -tan x d x- then- I 2- I 1- I 3- I 1- I 2- I 3- I 3- I 1- I 2D- I 3- I 2- I 1

The correct option is C I_3 lt; I_1 lt; I_2f(x) =sin x/x is a decreasing function and sin x/x gt; 0 for all x in (0, π) Since, sin x lt; x lt; ta ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Second Fundamental Theorem of Calculus

Let I 1=∫π/3π...

Question

Let $I_{1} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin x}{x} d x, I_{2} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin (sin x)}{sin x} d x, I_{3} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin (tan x)}{tan x} d x$ , then

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $I_{3} < I_{1} < I_{2}$
$f (x) = \frac{sin x}{x}$ is a decreasing function and $\frac{sin x}{x} > 0$ for all $x$ in $(0, π)$
Since, $sin x < x < tan x$
$\Rightarrow \frac{sin (sin x)}{sin x} > \frac{sin x}{x} > \frac{sin (tan x)}{tan x} for \frac{π}{6} < x < \frac{π}{3}$
$∴ I_{2} > I_{1} > I_{3}$

Suggest Corrections

Similar questions

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?

I f I_{1} = \int_{0}^{1} 2^{x^{2}} d x, I_{2} = \int_{0}^{1} 2^{x^{3}} d x, I_{3} = \int_{1}^{2} 2^{x^{2}} d x a n d I_{4} = \int_{1}^{2} 2 x^{3} d x t h e n

Q. 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

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

Join BYJU'S Learning Program

Let I1=∫π6π3sinxxdx,I2=∫π6π3sin(sinx)sinxdx,I3=∫π6π3sin(tanx)tanxdx, then

The correct option is C I3<I1<I2f(x)=sinxx is a decreasing function and sinxx>0 for all x in (0,π) Since, sinx<x<tanx ⇒sin(sinx)sinx>sinxx>sin(tanx)tanx for π6<x<π3 ∴I2>I1>I3

Let $I_{1} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin x}{x} d x, I_{2} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin (sin x)}{sin x} d x, I_{3} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin (tan x)}{tan x} d x$ , then