Answer to Question #207583 in Calculus for AGRAHARAPU SAMPATH

Question #207583

State Bonnet’s mean value theorem for integrals. Apply it to show that:

|35 cosxdx/x|≤ 2/3



1
Expert's answer
2021-06-16T17:25:52-0400

Bonnet Mean Value Theorem .

Suppose "f" is Lebesgue integrable on "[a, b]" and "g:[a,b]\\to\\R" is monotone. 

i) If "g" is non-negative, decreasing and greater than or equal to "0," for "A\\in \\R," "A\\geq \\lim\\limits_{x\\to a^{+}}g(x)" there exists "C" such that "a\\leq C\\leq b" and


"\\displaystyle\\int_{a}^{b}f(x)g(x)dx=A\\displaystyle\\int_{a}^{C}f(x)dx"

ii) If "g" is non-negative, increasing and greater than or equal to "0," for "B\\in \\R," "A\\geq \\lim\\limits_{x\\to b^{-}}g(x)" there exists "C" such that "a\\leq C\\leq b" and


"\\displaystyle\\int_{a}^{b}f(x)g(x)dx=B\\displaystyle\\int_{C}^{b}f(x)dx"

Consider


"\\displaystyle\\int_{3}^{5}\\dfrac{1}{x}\\cos xdx"

The function "f(x)=\\cos x" is integrable on "[3, 5]."

The function "g(x)=\\dfrac{1}{x}" is non-negative, monotone decreasing on "[3, 5]."

Then by the Bonnet Mean Value Theorem, for "A\\geq \\lim\\limits_{x\\to 3^{+}}\\dfrac{1}{x}" there exists "C" such that "3\\leq C\\leq 5" and


"\\displaystyle\\int_{3}^{5}\\dfrac{1}{x}\\cos xdx=A\\displaystyle\\int_{3}^{C}\\cos xdx"

Let "A=\\dfrac{1}{3}"


"\\bigg|\\displaystyle\\int_{3}^{5}\\dfrac{1}{x}\\cos xdx\\bigg|=\\dfrac{1}{3}\\bigg|\\displaystyle\\int_{3}^{C}\\cos xdx\\bigg|"

"=\\dfrac{1}{3}\\big|[\\sin x]\\big|\\begin{matrix}\n 5 \\\\\n 3\n\\end{matrix}\\leq\\dfrac{1}{3}(2)=\\dfrac{2}{3}"

Therefore


"\\bigg|\\displaystyle\\int_{3}^{5}\\dfrac{1}{x}\\cos xdx\\bigg|\\leq\\dfrac{2}{3}"


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