1.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)3 dx.
f(x) = 2x·(x² + 1)3[0, 1]
I =
2.IntegralsHard
Evaluate I = ∫₀0.5 x²·eˣ dx.
∫₀0.5 x²eˣ dx
I =
3.Sequences & seriesMedium
Find the radius of convergence R of the power series Σk=1∞ (2k)!·(x − 3)ᵏ/(k!)².
Σ (2k)!·(x − 3)ᵏ/(k!)²x₀ = 3
R =
4.Sequences & seriesHard
f(x) = sin x is approximated at x = 0.5 by its Maclaurin polynomial of degree n. With the Lagrange bound M·|x|ⁿ⁺¹/(n + 1)! and M = 1, find the smallest degree n that guarantees an error below 0.0005.
f(x) = sin xx = 0.5M = 1ε = 0.0005
n =
5.Multivariable calculusMedium
For f(x, y) = 5x²y³ + x³ − 2y³ + xy, find the directional derivative Duf at (2, 3) in the direction of v = (3, −1).
f(x, y) = 5x²y³ + x³ − 2y³ + xyP = (2, 3)v = (3, −1)
Duf =
6.Multivariable calculusHard
The marked points are the critical points of f(x, y) = x³ + y³ − 9xy. Click the saddle point.
f(x, y) = x³ + y³ − 9xy
- 1A (0, 0)
- 2B (3, 3)
7.Differential equationsMedium
Solve y″ + 7y′ + 12y = 0 with y(0) = 2 and y′(0) = -2. Find y(1).
y″ + 7y′ + 12y = 0y(0) = 2y′(0) = -2
y(1) =
8.Differential equationsHard
A system obeys y″ + 4y′ + 29y = 58 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 4y′ + 29y = 58y(0) = 0y′(0) = 0
ymax =