Let gi- -8- 3-8 - i-1- 2 and f- - -8- 3-8- - be functions such that g1x-1- g2x-4x- and fx-sin2x- for all x- - -8-3-8-Define Si-83-8fx- gix dx- i-1- 2The value of 48S2-2 is

S_2 = ∫^3 π/8_π/8sin^2 x 4x π dx nbsp;= ∫^3 π/8_π/84 sin^2 x. x π4 dx Let x π4 = t ⇒ dx = dt S_2 = ∫_ π/8^π/84 sin^2 ( π4 + t ) t dt =∫_ π/8^π/82[1 co ...

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Byju's Answer

Standard XII

Mathematics

First Derivative Test for Local Maximum

Let gi: [π8, ...

Question

Let $g_{i} : [\frac{π}{8}, \frac{3 π}{8}] \to R, i = 1, 2$ and $f : [\frac{π}{8}, \frac{3 π}{8}] \to R$ be functions such that $g_{1} (x) = 1, g_{2} (x) = | 4 x - π |$ and $f (x) = {sin}^{2} x,$ for all $x \in [\frac{π}{8}, \frac{3 π}{8}] .$
Define $S_{i} = \int_{\frac{π}{8}}^{\frac{3 π}{8}} f (x) \cdot g_{i} (x) d x, i = 1, 2$

The value of $\frac{48 S_{2}}{π^{2}}$ is

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Solution

$S_{2} = \int_{\frac{π}{8}}^{\frac{3 π}{8}} {sin}^{2} x | \begin{matrix} 4 x - π \end{matrix} | d x$
$= \int_{\frac{π}{8}}^{\frac{3 π}{8}} 4 {sin}^{2} x . ∣ ∣ \begin{matrix} x - \frac{π}{4} \end{matrix} ∣ ∣ d x$
Let $x - \frac{π}{4} = t \Rightarrow d x = d t$
$S_{2} = \int_{- \frac{π}{8}}^{\frac{π}{8}} 4 {sin}^{2} (\frac{π}{4} + t) | \begin{matrix} t \end{matrix} | d t = \int_{- \frac{π}{8}}^{\frac{π}{8}} 2 [1 - cos (2 (\frac{π}{4} + t))] | \begin{matrix} t \end{matrix} | d t = \int_{- \frac{π}{8}}^{\frac{π}{8}} (2 + 2 sin 2 t) | \begin{matrix} t \end{matrix} | d t = \int_{- \frac{π}{8}}^{\frac{π}{8}} 2 | \begin{matrix} t \end{matrix} | d t + 2 \int_{- \frac{π}{8}}^{\frac{π}{8}} | \begin{matrix} t \end{matrix} | sin (2 t) d t$
$= 4 \int_{0}^{\frac{π}{8}} t d t + 0$
$S_{2} = {\begin{matrix} 2 t^{2} \end{matrix} |}_{0}^{π / 8} = \frac{π^{2}}{32}$
$\Rightarrow \frac{48 S_{2}}{π^{2}} = \frac{3}{2}$

Alternate Method
$S_{2} = \int_{\frac{π}{8}}^{\frac{3 π}{8}} {sin}^{2} x | \begin{matrix} 4 x - π \end{matrix} | d x \dots (1)$

$\Rightarrow S_{2} = \int_{\frac{π}{8}}^{\frac{3 π}{8}} {sin}^{2} (\frac{π}{8} + \frac{3 π}{8} - x) ∣ ∣ ∣ \begin{matrix} 4 (\frac{π}{8} + \frac{3 π}{8} - x) - π \end{matrix} ∣ ∣ ∣ d x$

$\Rightarrow S_{2} = \int_{\frac{π}{8}}^{\frac{3 π}{8}} {cos}^{2} x | \begin{matrix} 4 x - π \end{matrix} | d x \dots (2)$

From $(1) + (2)$
$2 S_{2} = \int_{\frac{π}{8}}^{\frac{3 π}{8}} | \begin{matrix} 4 x - π \end{matrix} | d x$

$\Rightarrow 2 S_{2} = \int_{\frac{π}{8}}^{\frac{π}{4}} (π - 4 x) d x + \int_{\frac{π}{4}}^{\frac{3 π}{8}} (4 x - π) d x$

$\Rightarrow 2 S_{2} = {[π x - 2 x^{2}]}_{π / 8}^{2} + {[2 x^{2} - π x]}_{π / 4}^{2}$

$\Rightarrow 2 S_{2} = \frac{π^{2}}{16}$

Now
$\Rightarrow \frac{48 S_{2}}{π^{2}} = \frac{3}{2}$

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Similar questions

Q. Let

g_{i} : [\frac{π}{8}, \frac{3 π}{8}] \to R, i = 1, 2

and

f : [\frac{π}{8}, \frac{3 π}{8}] \to R

be functions such that

g_{1} (x) = 1, g_{2} (x) = | 4 x - π |

and

f (x) = {sin}^{2} x,

for all

x \in [\frac{π}{8}, \frac{3 π}{8}] .

Define

S_{i} = \int_{\frac{π}{8}}^{\frac{3 π}{8}} f (x) \cdot g_{i} (x) d x, i = 1, 2

The value of

\frac{16 S_{1}}{π}

Q. Let

f (x) = {sin}^{4} x + {cos}^{4} x

. Then

f

is an increasing function in the interval :

Q. Let

f : (0, π) \to R

be a twice differentiable function such that

lim t \to x \frac{f (x) sin t - f (t) sin x}{t - x} = {sin}^{2} x

for all

x ϵ (0, π)

f (\frac{π}{6}) = - \frac{π}{12}

, then which of the following statement(s) is (are) TRUE?

Q. Let

f : (0, π) \to R

be a twice differentiable function such that

lim t \to x \frac{f (x) sin t - f (t) sin x}{t - x} = {sin}^{2} x

for all

x \in (0, π)

.
If

f (\frac{π}{6}) = - \frac{π}{12}

, then which of the following statement(s) is (are) TRUE?

Q. Find

α

and

β

, so that the function

f (x)

defined by

f (x) = - 2 sin x,

for

- π \leq x \leq - \frac{π}{2}

= α sin x + β,

for

- \frac{π}{2} < x < \frac{π}{2}

= cos x,

for

\frac{π}{2} \leq x \leq π

is continuous on

[- π, π]

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Let gi:[π8,3π8]→R,i=1,2 and f:[π8,3π8]→R be functions such that g1(x)=1,g2(x)=|4x−π| and f(x)=sin2x, for all x∈[π8,3π8]. Define Si=∫3π8π8f(x)⋅gi(x)dx, i=1,2 The value of 48S2π2 is