1.IntegralsMedium
Evaluate I = ∫₀1 x·sin x dx (x in radians).
∫₀1 x·sin x dx
I =
2.IntegralsHard
Find the area A of the region between the parabola y = 4 − x² and the x-axis.
y = 4 − x²
A =
3.Sequences & seriesMedium
The power series Σk=1∞ (2k)!·(x − 2)ᵏ/(k!)² converges on an interval around x₀ = 2. Find the right end x of that interval.
Σ (2k)!·(x − 2)ᵏ/(k!)²x₀ = 2
x =
4.Sequences & seriesHard
Use the Maclaurin polynomial of degree 4 of f(x) = eˣ to approximate f(2). Give P₄(2).
f(x) = eˣn = 4x = 2
P₄(2) =
5.Multivariable calculusMedium
For f(x, y) = 4x³y² + 5x² − 5y² − xy, find the directional derivative Duf at (1, 2) in the direction of v = (−1, 3).
f(x, y) = 4x³y² + 5x² − 5y² − xyP = (1, 2)v = (−1, 3)
Duf =
6.Multivariable calculusHard
The integral I = ∫02 ∫x2 1/(1 + y²)² dy dx cannot be done in this order. Reverse the order of integration and evaluate it.
f = 1/(1 + y²)²0 ≤ x ≤ y ≤ 2
I =
7.Differential equationsMedium
A can of soda at 3 °C is taken out of the fridge into a room at 28 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.03 min⁻¹, find its temperature T after 16 min.
T0 = 3 °CTa = 28 °Ck = 0.03 min⁻¹t = 16 min
T = °C
8.Differential equationsHard
Solve y″ + 2y′ + 26y = 0 with y(0) = 5 and y′(0) = 0 (released from rest). Find y(0.75); angles in radians.
y″ + 2y′ + 26y = 0y(0) = 5y′(0) = 0
y(0.75) =