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台灣大學 107 年度 微積分(A)

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100
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主要題型
Infinite Series 2The Limit of a Function 1Integrals 1Vector calculus 1

微積分(A) 其他年度考古題

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120The Limit of a Function
(a)10
Find an explicit value of ϵ>0\epsilon > 0 such that for every x[0,1]x \in [0,1]
xx+ϵ<1200.|\sqrt{x} - \sqrt{x+\epsilon}| < \frac{1}{200}.
(b)10
Find an explicit integer NN such that there exists a polynomial PP of degree at most NN such that for every x[0,1]x \in [0,1]
xP(x)<1100.|\sqrt{x} - P(x)| < \frac{1}{100}.
Hint: You can use the expansion of x+ϵ\sqrt{x+\epsilon} in power series in (x1)(x-1).
210Integrals
Let f:[0,1]Rf : [0,1] \mapsto \mathbb{R} be monotone increasing, i.e. if c<dc < d implies f(c)f(d)f(c) \leq f(d). Use the definition of the Riemann integral (comparing upper and lower sums relative to a partition of [0,1][0,1]) to prove that 01f(x)dx\int_0^1 f(x)dx exists.
320Vector calculus
(a)10
Evaluate 02y24ycos(x2)dxdy\int_0^2 \int_{y^2}^{4} y\cos(x^2)dxdy
(b)10
Let FF be a vector field F=(x3,y3,z3)F = (x^3, y^3, z^3) and Ω\Omega be the solid region in R3\mathbb{R}^3 bounded by x2+y2z2,x2+y2+z49,yxx^2 + y^2 \geq z^2, x^2 + y^2 + z^4 \leq 9, y \geq |x|. Evaluate Ω(x2+y2+z2)dV\iiint_\Omega (x^2+y^2+z^2)dV and find the flux of FF through the boundary surface of Ω\Omega, oriented outwards (i.e. ΩFndS\iint_{\partial\Omega} F \cdot n dS).
410Derivatives of multi-variable functions
Among all planes that are tangent to the surface xy2z2=1xy^2z^2 = 1, find the ones that are farthest from the origin.
520Infinite Series
Let {an}n=1\{a_n\}_{n=1}^\infty be a sequence and bn=1n3(k=1nk2ak)b_n = \frac{1}{n^3}(\sum_{k=1}^n k^2a_k).
(a)10
Prove or disprove: If {an}n=1\{a_n\}_{n=1}^\infty converges then {bn}n=1\{b_n\}_{n=1}^\infty converges.
(b)10
Prove or disprove: If {bn}n=1\{b_n\}_{n=1}^\infty converges then {an}n=1\{a_n\}_{n=1}^\infty converges.
620Infinite Series
(a)10
Find domain of convergence of n=1(n+x)nnn+x\sum_{n=1}^\infty \frac{(n+x)^n}{n^{n+x}}.
(b)10
Does the series
n=1exn(1)nn\sum_{n=1}^\infty e^{-\frac{x}{n}} \frac{(-1)^n}{n}
converge uniformly on [0,)[0,\infty)? Prove or disprove.