1.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)1 dx.
f(x) = 2x·(x² + 1)1[0, 1]
I =
2.IntegralsHard
Evaluate I = ∫₀1 x²·eˣ dx.
∫₀1 x²eˣ dx
I =
3.Sequences & seriesMedium
Find the radius of convergence R of the power series Σk=1∞ (x + 2)ᵏ/(k²·5ᵏ).
Σ (x + 2)ᵏ/(k²·5ᵏ)x₀ = −2
R =
4.Sequences & seriesHard
Use the Maclaurin polynomial of degree 3 of f(x) = ln(1 + x) to approximate f(0.5). Give P₃(0.5).
f(x) = ln(1 + x)n = 3x = 0.5
P₃(0.5) =
5.Multivariable calculusMedium
z = xy² + 5x³ + y² − 4xy with x = −3t − 2 and y = 2t² + 1. Use the chain rule to find dz/dt at t = 1.
z = xy² + 5x³ + y² − 4xyx = −3t − 2y = 2t² + 1t = 1
dz/dt =
6.Multivariable calculusHard
Use polar coordinates to evaluate I = ∬R e−(x² + y²) dA, where R is the ring 1 ≤ x² + y² ≤ 2.25.
f = e−(x² + y²)1 ≤ r ≤ 1.50 ≤ θ ≤ 2π
I =
7.Differential equationsMedium
A can of soda at 3 °C is taken out of the fridge into a room at 25 °C. After 8 min its temperature is 12 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 3 °CTa = 25 °CT(8) = 12 °C
k = min⁻¹
8.Differential equationsHard
A system obeys y″ + 3y′ + 18.25y = 91.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 3y′ + 18.25y = 91.25y(0) = 0y′(0) = 0
ymax =