f ( x ) = 1 + s i n ( x ) f(x) = 1 + sin(x) f ( x ) = 1 + s in ( x ) g ( x ) = 1 − s i n ( x ) g(x) = 1-sin(x) g ( x ) = 1 − s in ( x ) a = 0 b = π 4 a = 0 \space\space\space\space b = \large\frac{\pi}{4} a = 0 b = 4 π
We apply the formulae that the coordinates of the centroid (= center of mass assuming constant density) of the region with top f(x), bottom g(x), left-hand side x = a, and right-hand side x = b are
x = ∫ a b x ∣ f ( x ) − g ( x ) ∣ d x ∫ a b ∣ f ( x ) − g ( x ) ∣ d x , x = \large\frac{\int_a^bx|f(x)-g(x)|dx}{\int_a^b|f(x)-g(x)|dx}, x = ∫ a b ∣ f ( x ) − g ( x ) ∣ d x ∫ a b x ∣ f ( x ) − g ( x ) ∣ d x , y = ∫ a b 1 2 ∣ f ( x ) 2 − g ( x ) 2 ∣ d x ∫ a b ∣ f ( x ) − g ( x ) ∣ d x y= \large\frac{\int_a^b\frac{1}{2}|f(x)^2-g(x)^2|dx}{\int_a^b|f(x)-g(x)|dx} y = ∫ a b ∣ f ( x ) − g ( x ) ∣ d x ∫ a b 2 1 ∣ f ( x ) 2 − g ( x ) 2 ∣ d x
∫ a b ∣ f ( x ) − g ( x ) ∣ d x = ∫ 0 π 4 ∣ 1 + s i n ( x ) − 1 + s i n ( x ) ∣ = ∫ 0 π 4 2 s i n ( x ) = ∣ − 2 c o s ( x ) ∣ 0 π 4 = − 2 c o s π 4 + 2 c o s 0 = 2 − 2 \int_a^b|f(x)-g(x)|dx = \int_0^{\frac{\pi}{4}}|1+sin(x)-1+sin(x)|=\int_0^{\frac{\pi}{4}}2sin(x) = |-2cos(x)|_0^\frac{\pi}{4} = -2cos\frac{\pi}{4}+2cos0 = 2-\sqrt{2} ∫ a b ∣ f ( x ) − g ( x ) ∣ d x = ∫ 0 4 π ∣1 + s in ( x ) − 1 + s in ( x ) ∣ = ∫ 0 4 π 2 s in ( x ) = ∣ − 2 cos ( x ) ∣ 0 4 π = − 2 cos 4 π + 2 cos 0 = 2 − 2
∫ a b x ∣ f ( x ) − g ( x ) ∣ d x = ∫ 0 π 4 x ∣ 1 + s i n ( x ) − 1 + s i n ( x ) ∣ = ∫ 0 π 4 2 x s i n ( x ) = 2 ∣ s i n ( x ) − x c o s ( x ) ∣ 0 π 4 = 2 ( s i n π 4 − π 4 c o s ( π 4 ) ) − 2 ( s i n 0 − 0 ∗ c o s 0 ) = 2 ( 1 − π 4 ) \int_a^bx|f(x)-g(x)|dx = \int_0^{\frac{\pi}{4}}x|1+sin(x)-1+sin(x)|=\int_0^{\frac{\pi}{4}}2xsin(x) = 2|sin(x)-xcos(x)|_0^\frac{\pi}{4} = 2(sin\frac{\pi}{4}-\frac{\pi}{4}cos(\frac{\pi}{4}))-2(sin0-0*cos0) = \sqrt{2}(1-\frac{\pi}{4}) ∫ a b x ∣ f ( x ) − g ( x ) ∣ d x = ∫ 0 4 π x ∣1 + s in ( x ) − 1 + s in ( x ) ∣ = ∫ 0 4 π 2 x s in ( x ) = 2∣ s in ( x ) − x cos ( x ) ∣ 0 4 π = 2 ( s in 4 π − 4 π cos ( 4 π )) − 2 ( s in 0 − 0 ∗ cos 0 ) = 2 ( 1 − 4 π )
∫ a b 1 2 ∣ f ( x ) 2 − g ( x ) 2 ∣ d x = ∫ 0 π 4 ∣ 1 + 2 s i n ( x ) + s i n 2 ( x ) − 1 + 2 s i n ( x ) − s i n 2 ( x ) ∣ = ∫ 0 π 4 4 s i n ( x ) = ∣ − 4 c o s ( x ) ∣ 0 π 4 = − 4 c o s π 4 + 4 c o s 0 = 4 − 2 2 \int_a^b\frac{1}{2}|f(x)^2-g(x)^2|dx = \int_0^{\frac{\pi}{4}}|1+2sin(x)+sin^2(x)-1+2sin(x)-sin^2(x)|=\int_0^{\frac{\pi}{4}}4sin(x) = |-4cos(x)|_0^\frac{\pi}{4} = -4cos\frac{\pi}{4}+4cos0 = 4-2\sqrt{2} ∫ a b 2 1 ∣ f ( x ) 2 − g ( x ) 2 ∣ d x = ∫ 0 4 π ∣1 + 2 s in ( x ) + s i n 2 ( x ) − 1 + 2 s in ( x ) − s i n 2 ( x ) ∣ = ∫ 0 4 π 4 s in ( x ) = ∣ − 4 cos ( x ) ∣ 0 4 π = − 4 cos 4 π + 4 cos 0 = 4 − 2 2
x = ∫ a b x ∣ f ( x ) − g ( x ) ∣ d x ∫ a b ∣ f ( x ) − g ( x ) ∣ d x = 2 ( 1 − π 4 ) 2 − 2 = 4 − π 4 2 − 1 , x = \large\frac{\int_a^bx|f(x)-g(x)|dx}{\int_a^b|f(x)-g(x)|dx} = \frac{\sqrt{2}(1-\frac{\pi}{4})}{2-\sqrt{2}} = \frac{4-\pi}{4\sqrt{2}-1} , x = ∫ a b ∣ f ( x ) − g ( x ) ∣ d x ∫ a b x ∣ f ( x ) − g ( x ) ∣ d x = 2 − 2 2 ( 1 − 4 π ) = 4 2 − 1 4 − π , y = ∫ a b 1 2 ∣ f ( x ) 2 − g ( x ) 2 ∣ d x ∫ a b ∣ f ( x ) − g ( x ) ∣ d x = 4 − 2 2 2 − 2 y= \large\frac{\int_a^b\frac{1}{2}|f(x)^2-g(x)^2|dx}{\int_a^b|f(x)-g(x)|dx} = \frac{ 4-2\sqrt{2}}{2-\sqrt{2}} y = ∫ a b ∣ f ( x ) − g ( x ) ∣ d x ∫ a b 2 1 ∣ f ( x ) 2 − g ( x ) 2 ∣ d x = 2 − 2 4 − 2 2 = 2 =2 = 2
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