Monster Quest | AI tutor The No.1 Homework Finishing Free App
Functions and Analysis
Analysis - Basic Differential Equations | AI tutor The No.1 Homework Finishing Free App
Q.03
'Proof of the Properties of Definite Integrals (A), (B), (C)'
A. ...
Q.04
'(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} \\]'
A. ...
Q.06
'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'
A. ...
Q.08
'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'
A. ...
Q.09
'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'
A. ...
Q.10
"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."
A. ...
Q.11
'Proof of inequality using the mean value theorem for integrals'
A. ...
Q.12
"Second derivative and extrema: When , if exists, it can be used for determining the extrema."
A. ...
Q.13
'Prove using mathematical induction that equation (1) holds for all natural numbers n.'
A. ...
Q.14
'The derivative at x=a of the function y=f(x) is defined as follows.'
A. ...
Q.15
'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]'
A. ...
Q.16
'Substitution and Integration by Parts for Indefinite Integrals'
A. ...
Q.17
'Proof of (arithmetic mean) ≥ (geometric mean) by geometric shapes'
A. ...
Q.19
'There is a problem with the proof of the definite integral of Iₙ in mathematics (1).'
A. ...
Q.20
'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}}} \'
A. ...
Q.21
'Find the area 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 -axis, and the line .'
A. ...
Q.22
'Use integration by parts to find the following integral.'
A. ...
Q.23
"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."
A. ...
Q.24
'Assuming , from (*) we get , hence . Substituting and into (6) and simplifying, we get . Therefore, , and since , we have , which leads to . Therefore, .'
A. ...
Q.25
'(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.'
A. ...
Q.28
'Find the function f(x) that satisfies the following equation: (3) f(x)=1/2 x+∫[0, x](t-x) sin t d t'
A. ...
Q.29
'Please explain definite integrals and limits of sums.'
A. ...
Q.32
'Find the definite integral: \\\int_{0}^{1} \\frac{1}{x^{3}+8} d x\'
A. ...
Q.33
"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)."
A. ...
Updated: 12/12/2024