f(x)=∫ limits cos x sin x(1-t+2 t3) d t has in [0,2 π]

Question

f(x)=∫ limits cos x sin x(1-t+2 t3) d t has in [0,2 π] (A) a maximum at (π/4) a minimum at (3 π/4) (B) a maximum at (3 π/4) a minimum at (7 π/4) (C) a maximum at (5 π/4) a minimum at (7 π/4) (D) neither a maxima nor minima

Tardigrade Admin · Accepted Answer

f prime(x)=(1 ⋅ cos x- sin x cos x+2 sin 3 x ⋅ cos x)-(-1 ⋅ sin x+ cos x sin x-2 cos 3 x sin x) = cos x+ sin x-2 sin x ⋅ cos x+2 sin x ⋅ cos x= cos x+ sin x

Q. $f (x) = c o s x \int s i n x (1 - t + 2 t^{3}) d t$ has in $[0, 2 π]$

a maximum at $\frac{π}{4} &$ a minimum at $\frac{3 π}{4}$

a maximum at $\frac{3 π}{4}$ & a minimum at $\frac{7 π}{4}$

a maximum at $\frac{5 π}{4}$ \& a minimum at $\frac{7 π}{4}$

neither a maxima nor minima

$f^{'} (x) = (1 \cdot cos x - sin x cos x + 2 sin^{3} x \cdot cos x) - (- 1 \cdot sin x + cos x sin x - 2 cos^{3} x sin x)$
$= cos x + sin x - 2 sin x \cdot cos x + 2 sin x \cdot cos x = cos x + sin x$

Q. f(x)=cosx∫sinx​(1−t+2t3)dt has in [0,2π]

a maximum at 4π​& a minimum at 43π​

a maximum at 43π​ & a minimum at 47π​

a maximum at 45π​ \& a minimum at 47π​