1.IntegralsMedium
Evaluate I = ∫₀2 x·sin x dx (x in radians).
∫₀2 x·sin x dx
I =
2.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = 4x.
y = x2y = 4x
A =
3.Sequences & seriesMedium
The power series Σk=1∞ (x − 3)ᵏ/4ᵏ converges on an interval around x₀ = 3. Find the left end x of that interval.
Σ (x − 3)ᵏ/4ᵏx₀ = 3
x =
4.Sequences & seriesHard
f(x) = eˣ 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 = 2, find the smallest degree n that guarantees an error below 0.0005.
f(x) = eˣx = 0.5M = 2ε = 0.0005
n =
5.Multivariable calculusMedium
Use the tangent plane of f(x, y) = 2 ln(x² + 2y²) at (−1, 0.5) to approximate f(−0.95, 0.3). Give the linear approximation L.
f(x, y) = 2 ln(x² + 2y²)(a, b) = (−1, 0.5)(x, y) = (−0.95, 0.3)
L =
6.Multivariable calculusHard
The marked points are the critical points of f(x, y) = x³ − 3x + y² − 4y. Click the saddle point.
f(x, y) = x³ − 3x + y² − 4y
- 1A (1, 2)
- 2B (−1, 2)
7.Differential equationsMedium
Solve y″ + 4y′ + 4y = 0 with y(0) = 3 and y′(0) = 2. Find y(1).
y″ + 4y′ + 4y = 0y(0) = 3y′(0) = 2
y(1) =
8.Differential equationsHard
Solve y″ + 2y′ + 5y = 0 with y(0) = 1 and y′(0) = 0 (released from rest). Find y(0.75); angles in radians.
y″ + 2y′ + 5y = 0y(0) = 1y′(0) = 0
y(0.75) =