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## Functions and Analysis

### Analysis - Basic Differential Equations

#### Q.03

'Proof of the Properties of Definite Integrals (A), (B), (C)'

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'(2) \ 2 a x-a^{2}=2 b x-b^{2} \ Let us denote that \\( 2(a-b) x=a^{2}-b^{2} \\) Therefore, the \ x \ coordinate of the intersection of \ \\ell_{1} \ and \ \\ell_{2} \ is\n\n\ x=\\frac{a+b}{2} \\n\nTherefore, the area \\( S(a) \\) enclosed by \ \\ell_{1}, \\ell_{2} \ and the parabola C is, from the right figure\n\n\\[\n\egin{aligned}\nS(a)= & \\int_{b}^{\\frac{a+b}{2}}\\left\\{x^{2}-\\left(2 b x-b^{2}\\right)\\right\\} d x \\\\\n& +\\int_{\\frac{a+b}{2}}^{a}\\left\\{x^{2}-\\left(2 a x-a^{2}\\right)\\right\\} d x \\\\\n= & \\int_{b}^{\\frac{a+b}{2}}(x-b)^{2} d x+\\int_{\\frac{a+b}{2}}^{a}(x-a)^{2} d x \\\\\n= & {\\left[\\frac{(x-b)^{3}}{3}\\right]_{b}^{\\frac{a+b}{2}}+\\left[\\frac{(x-a)^{3}}{3}\\right]_{\\frac{a+b}{2}}^{a} } \\\\\n= & \\frac{1}{3}\\left(\\frac{a-b}{2}\\right)^{3}-\\frac{1}{3}\\left(\\frac{b-a}{2}\\right)^{3} \\\\\n= & \\frac{1}{12}(a-b)^{3}=\\frac{1}{12}\\left(a+\\frac{1}{4 a}\\right)^{3}\n\\end{aligned} \\]'

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'The x-coordinate of the intersection of the two parabolas is the solution to the equation x^{2}+x+2=x^{2}-7x+10. Therefore, x=1'

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'Prove that the differential equation can be transformed into the following form and find the solution. By letting y = uv, we have du/dx * v + u * dv/dx - uv/x = x'

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'Given f (x) = ∫_0^x t cos t dt - x ∫_0^x cos t dt, hence f’(x) = d/dx ∫_0^x t cos t dt - { (x)’ ∫_0^x cos t dt + x ( d/dx ∫_0^x cos t dt ) } = x cos x - { [sin t ]_0^x + x cos x } = -sin x'

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"Prove the following theorem: For all natural numbers n, if the nth derivative functions f^{(n)}(x) and g^{(n)}(x) of f(x) and g(x) exist, then the nth derivative function of the product f(x) g(x) can be expressed as stated in Leibniz's theorem."

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'Proof of inequality using the mean value theorem for integrals'

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"Second derivative and extrema: When $f'(a)=0$, if $f''(a)$ exists, it can be used for determining the extrema."

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'Prove using mathematical induction that equation (1) holds for all natural numbers n.'

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'The derivative at x=a of the function y=f(x) is defined as follows.'

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'Example 122 Maximum and Minimum of a Function Represented by a Definite Integral (2)\nWhen a real number t moves in the range 1 ≤ t ≤ e, find the maximum and minimum values of S(t) = ∫_{0}^{1} |e^{x} - t|dx.\n[Nagaoka Institute of Technology]'

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'Substitution and Integration by Parts for Indefinite Integrals'

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'Proof of (arithmetic mean) ≥ (geometric mean) by geometric shapes'

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'There is a problem with the proof of the definite integral of Iₙ in mathematics (1).'

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'Find the following definite integrals. In (1), let a be a positive constant.\n(1) \ \\int_{0}^{\\frac{a}{2}} \\sqrt{a^{2}-x^{2}} d x \\n(2) \ \\int_{0}^{\\sqrt{2}} \\frac{d x}{\\sqrt{4-x^{2}}} \'

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'Find the area $S$ enclosed by the curve \\left\\{\egin{\overlineray}{l} x=t-\\sin t \\\\ y=1-\\cos t \\end{\overlineray} (0 \\leq t \\leq \\pi)\\right., the $x$-axis, and the line $x=\\pi$.'

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'Use integration by parts to find the following integral.'

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"Find f(x) when the differentiable function f(x) (x>0) satisfies the equation f(x)=x log x + ∫1^e t f'(t) dt."

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'Assuming $\\frac{dy}{dx}=-2$, from (*) we get $-\\frac{8x+y}{x-3y}=-2$, hence $6x+7y=0$. Substituting $x=\\frac{e^{t}+3e^{-t}}{2}$ and $y=e^{t}-2e^{-t}$ into (6) and simplifying, we get $2e^{t}-e^{-t}=0$. Therefore, $e^{-t}(2e^{2t}-1)=0$, and since $e^{-t}>0$, we have $e^{2t}=\\frac{1}{2}$, which leads to $2t=-\\log 2$. Therefore, $t=-\\frac{1}{2}\\log 2$.'

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'(1) In the important example 220, let Jₙ=∫0π/2 cosⁿxdx (where n is a non-negative integer), prove that Iₙ=Jₙ when n is greater than or equal to 0.'

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'Find the function f(x) that satisfies the following equation: (3) f(x)=1/2 x+∫[0, x](t-x) sin t d t'

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'Please explain definite integrals and limits of sums.'

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'Find the definite integral: \\\int_{0}^{1} \\frac{1}{x^{3}+8} d x\'

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"Solve the following differential equation and find the maximum value. Let f'(x)=0, then x / sqrt(4-x²)=1. Therefore, sqrt(4-x²)=x. Squaring both sides gives 4-x²=x². Thus x²=2, hence x=±sqrt(2). Among these, x>0 so x=sqrt(2). Next, f''(sqrt(2))=-4/(2*sqrt(2))=-sqrt(2)<0 so x=sqrt(2) has a maximum value. Therefore, the maximum value is f(sqrt(2))=sqrt(2)-2+sqrt(4-2)=2(sqrt(2)-1)."

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Verify the following equation for subsets $A, B$ of the universal set $U$ using a diagram:
\( \overline{(ar{A} \cap B)}=A \cup ar{B} \)