1.IntegralsMedium
Evaluate I = ∫₀1 x·cos x dx (x in radians).
∫₀1 x·cos 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∞ 9ᵏx²ᵏ/k.
Σ 9ᵏx²ᵏ/kx₀ = 0
R =
4.Sequences & seriesHard
f(x) = eˣ is approximated by its Maclaurin polynomial of degree 2 at x = −0.5. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = 1 to find the largest possible error |R₂|.
f(x) = eˣn = 2x = −0.5M = 1
|R₂| =
5.Multivariable calculusMedium
For f(x, y) = 3 sin(−2x + y), find the direction of steepest ascent at (2, −1) as the angle θ from the positive x-axis, in degrees between −180° and 180°.
f(x, y) = 3 sin(−2x + y)P = (2, −1)
θ = °
6.Multivariable calculusHard
The marked points are the critical points of f(x, y) = x³ + y³ + 9xy. Click the local maximum.
f(x, y) = x³ + y³ + 9xy
- 1A (0, 0)
- 2B (−3, −3)
7.Differential equationsMedium
A can of soda at 6 °C is taken out of the fridge into a room at 27 °C. After 6 min its temperature is 15 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 6 °CTa = 27 °CT(6) = 15 °C
k = min⁻¹
8.Differential equationsHard
A system obeys y″ + y′ + 16.25y = 16.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + y′ + 16.25y = 16.25y(0) = 0y′(0) = 0
ymax =