1.IntegralsMedium
Evaluate I = ∫₁5 x·ln x dx.
∫₁5 x ln x dx
I =
2.IntegralsHard
Find the area A of the region between the parabola y = 1 − x² and the x-axis.
y = 1 − x²
A =
3.Sequences & seriesMedium
Find the radius of convergence R of the power series Σk=1∞ (2k)!·xᵏ/(k!)².
Σ (2k)!·xᵏ/(k!)²x₀ = 0
R =
4.Sequences & seriesHard
Use the Maclaurin polynomial of degree 3 of f(x) = sin(−x) to approximate f(−1.5). Give P₃(−1.5).
f(x) = sin(−x)n = 3x = −1.5
P₃(−1.5) =
5.Multivariable calculusMedium
For f(x, y) = −xy² + 5x³ + 4y² + 3xy, find the direction of steepest ascent at (1, 1) as the angle θ from the positive x-axis, in degrees between −180° and 180°.
f(x, y) = −xy² + 5x³ + 4y² + 3xyP = (1, 1)
θ = °
6.Multivariable calculusHard
The marked points are the critical points of f(x, y) = x³ − 27x + y². Click the local minimum.
f(x, y) = x³ − 27x + y²
- 1A (−3, 0)
- 2B (3, 0)
7.Differential equationsMedium
A can of soda at 4 °C is taken out of the fridge into a room at 29 °C. After 28 min its temperature is 24 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 4 °CTa = 29 °CT(28) = 24 °C
k = min⁻¹
8.Differential equationsHard
A system obeys y″ + 4y′ + 8y = 80 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 4y′ + 8y = 80y(0) = 0y′(0) = 0
ymax =