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 16S1- is

S_1=∫_π/8^3π/8sin^2xdx =12∫_π/8^3π/8(1 cos2x)dx =12[x sin 2x2]_π/8^3π/8 S_1=12[ (3π4 12√(2) ) ( π8 12√(2))] =π8 ⇒16S_1π=16π×π8=2 ...

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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 $\frac{16 S_{1}}{π}$ is

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Solution

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Similar questions

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

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

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

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

x \in [\frac{π}{8}, \frac{3 π}{8}] .

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

\frac{48 S_{2}}{π^{2}}

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

f

f : (0, π) \to R

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

x ϵ (0, π)

f (\frac{π}{6}) = - \frac{π}{12}

f : (0, π) \to R

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

x \in (0, π)

f (\frac{π}{6}) = - \frac{π}{12}

f (x) = sin 2 x + cos 2 x \forall x \in [0, \frac{π}{2}]

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