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Question Number 69027 by rajesh4661kumar@gmail.com last updated on 18/Sep/19

Commented by mathmax by abdo last updated on 18/Sep/19
$I =∫_0 ^(2π) sinxdx −∫_0 ^(2π) ∣sinx∣dx but ∫_0 ^(2π) sinx dx =_(x=π+t) ∫_(−π) ^π sin(π+t)dt =−∫_(−π) ^π sint dt =0(odd function) ∫_0 ^(2π) ∣sinx∣dx =_(x=π+t) ∫_(−π) ^π ∣sint∣dt =2 ∫_0 ^π sint dt =2[−cost]_0 ^π =2{1−(−1)} =4$
$I = \int_{0}^{2 π} s i n x d x - \int_{0}^{2 π} ∣ s i n x ∣ d x b u t$ $\int_{0}^{2 π} s i n x d x =_{x = π + t} \int_{- π}^{π} s i n (π + t) d t = - \int_{- π}^{π} s i n t d t = 0 (o d d f u n c t i o n)$ $\int_{0}^{2 π} ∣ s i n x ∣ d x =_{x = π + t} \int_{- π}^{π} ∣ s i n t ∣ d t = 2 \int_{0}^{π} s i n t d t = 2 {[- c o s t]}_{0}^{π}$ $= 2 {1 - (- 1)} = 4$
Commented by mathmax by abdo last updated on 18/Sep/19

$\Rightarrow I = 0 - 4 = - 4 .$
	Answered by MJS last updated on 18/Sep/19
	$sin x −∣sin x∣= { ((0; 0≤x<π)),((2sin x; π≤x≤2π)) :} ⇒ ∫_0 ^(2π) sin x −∣sin x∣ dx=2∫_π ^(2π) sin x dx=−2∫_0 ^π sin x dx= =2[cos x]_0 ^π =−4$
	$\sin x - ∣ \sin x ∣= {\begin{cases} 0; 0 ⩽ x < π \\ 2 \sin x; π ⩽ x ⩽ 2 π \end{cases}$ $\Rightarrow \underset{0}{\int^{2 π}} \sin x - ∣ \sin x ∣ d x = 2 \underset{π}{\int^{2 π}} \sin x d x = - 2 \underset{0}{\int^{π}} \sin x d x =$ $= 2 {[\cos x]}_{0}^{π} = - 4$
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