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Calculus Practice Set 2

2 Hours

13 challenging problems

1Problem 1

Let y=y(x)y = y(x) be determined by x5−y=tan⁡(x−y)x^5 - y = \tan(x-y), find y(0)y(0) and y′(0)y'(0).

2Problem 2

Let function y=y(x)y=y(x) be determined by parametric equations:

{x=∫0t2e−u duy=∫0tcos⁡s2 ds\begin{cases} x = \int_0^t 2e^{-u}\,du \\ y = \int_0^t \cos s^2\,ds \end{cases}

Find d2ydx2∣x=2\frac{d^2y}{dx^2}\bigg|_{x=\sqrt{2}}.

3Problem 3

Find the limit:

lim⁡x→01+x2−cos⁡2xx2\lim_{x \to 0}\frac{\sqrt{1+x^2} - \cos 2x}{x^2}
4Problem 4

Find the limit:

lim⁡x→0(ex−1x)1x2\lim_{x \to 0}\left(\frac{e^x-1}{x}\right)^{\frac{1}{x^2}}
5Problem 5

Find the limit:

lim⁡x→0(1sin⁡2x−1x2)\lim_{x \to 0}\left(\frac{1}{\sin^2 x} - \frac{1}{x^2}\right)
6Problem 6

Calculate the integral:

∫1+∞ln⁡(1+x)x2 dx\int_1^{+\infty} \frac{\ln(1+x)}{x^2}\,dx
7Problem 7

Calculate the integral:

∫1e(2+x)2(1−x3)43 dx\int_1^e (2+x)^2(1-x^3)^{\frac{4}{3}}\,dx
8Problem 8

Prove: When 0≤x<+∞0 \leq x < +\infty, arctan⁡3x≤ln⁡(1+4x)\arctan 3x \leq \ln(1+4x). When is equality achieved (i.e., when x=0x=0)?

9Problem 9

Find ∑n=0∞(−1)2n(2n+1)(2n−2)x2n+2\sum_{n=0}^{\infty}\frac{(-1)^{2n}}{(2n+1)(2n-2)}x^{2n+2}, determine its radius of convergence, interval of convergence, and sum function.

10Problem 10

Let constant a>0a > 0, f(x)=13ax3−xf(x) = \frac{1}{3}ax^3 - x. Determine the maximum and minimum values of f(x)f(x) on interval [0,1a]\left[0, \frac{1}{a}\right].

11Problem 11

Find the curve y2=x+2y^2 = x+2 and the line l:y=xl: y = x such that the region bounded by the curve and the line equals x=2x=2, find the solid of revolution's volume when rotated around the line.

12Problem 12

Prove: if the continuous function f(x)f(x) satisfies the improper integral's Cauchy convergence criterion, show:

(1) lim⁡x→∞f(x)=lim⁡x→∞g(x)=0\lim_{x \to \infty}f(x) = \lim_{x \to \infty}g(x) = 0;

(2) f(x),g(x)f(x), g(x) are small order positive infinitesimals of (x0)n(x_0)^n as nn increases, and g′(x)≠0g'(x) \neq 0;

(3) lim⁡x→∞f(x)g(x)=A(∞∞)\lim_{x \to \infty}\frac{f(x)}{g(x)} = A(\infty\infty), then lim⁡x→∞f(x)g(x)=lim⁡x→∞f(x)g′(x)=A\lim_{x \to \infty}\frac{f(x)}{g(x)} = \lim_{x \to \infty}\frac{f(x)}{g'(x)} = A.

Please verify that given condition (3) does not hold, then lim⁡x→∞f(x)g(x)\lim_{x \to \infty}\frac{f(x)}{g(x)} exists, but cannot use L'Hôpital's rule.

13Problem 13

Let f(x)=−cos⁡πx+(2x−3)3+12(x−1)f(x) = -\cos\pi x + (2x-3)^3 + \frac{1}{2}(x-1). Verify by calculating that f(x)=0f(x) = 0 has a unique zero.