Calculus 2: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Integrals

Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C
Definite integral: ∫_a^b f dx = F(b) − F(a)
Common integrals: ∫eˣ dx = eˣ, ∫1/x dx = ln|x|, ∫sin x dx = −cos x
Area between curves: ∫_a^b (f − g) dx

Method

Rewrite the integrand as a sum of simple terms where possible.
Apply the power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1), and standard integrals for e^x, 1/x, sin and cos.
Use substitution when you see a function and its derivative together.
For a definite integral, evaluate F(b) − F(a).

Sequences & series

Arithmetic sequence: aₙ = a₁ + (n − 1)d, Sₙ = n(a₁ + aₙ)/2
Geometric sequence: aₙ = a₁rⁿ⁻¹, Sₙ = a₁(1 − rⁿ)/(1 − r)
Infinite geometric series: S = a₁/(1 − r), |r| < 1
p-series: Σ1/kᵖ converges only for p > 1
Ratio test: L = lim |a_k+1/a_k|: L < 1 converges, L > 1 diverges
Radius of convergence: R = lim |c_k/c_k+1|
Taylor polynomial: Pₙ(x) = Σ f⁽ᵏ⁾(a)(x − a)ᵏ/k!
Maclaurin series: eˣ = Σxᵏ/k!, sin x = x − x³/3! + …, cos x = 1 − x²/2! + …, ln(1 + x) = x − x²/2 + …
Lagrange error bound: |Rₙ| ≤ M·|x − a|ⁿ⁺¹/(n + 1)!

Method

Identify the type: arithmetic (constant difference), geometric (constant ratio), a standard series, a power series or a Taylor polynomial.
For sums use Sₙ = n(a₁ + aₙ)/2, Sₙ = a₁(1 − rⁿ)/(1 − r), or S = a₁/(1 − r) when |r| < 1, with a₁ the first term of the sum.
For convergence: n-th term test first, then compare with Σ1/kᵖ or a geometric series, and use the ratio or root test for factorials and k-th powers.
For Taylor polynomials list f(a), f′(a), f″(a), …, divide the k-th derivative by k!, and bound the error with M·|x − a|ⁿ⁺¹/(n + 1)!.

Multivariable calculus

Partial derivatives: f_x = ∂f/∂x (y constant), f_xy = f_yx
Gradient: ∇f = (f_x, f_y), largest slope |∇f|
Directional derivative: D_uf = ∇f·u, u = v/|v|
Tangent plane: z = f(a, b) + f_x(x − a) + f_y(y − b)
Chain rule: dz/dt = f_x·dx/dt + f_y·dy/dt
Second-derivative test: D = f_xxf_yy − f_xy²: D > 0, f_xx > 0 minimum; D > 0, f_xx < 0 maximum; D < 0 saddle
Double integral: ∬_R f dA = ∫_a^b∫_g(x)^h(x) f dy dx
Polar coordinates: x = r cos θ, y = r sin θ, dA = r dr dθ

Method

For a partial derivative, treat the other variable as a constant and use the one-variable rules.
For gradient questions compute ∇f = (f_x, f_y) at the point; use |∇f| for the steepest slope and ∇f·u with a unit vector u for a directional derivative.
For extrema solve f_x = 0 and f_y = 0 together, then classify every point with D = f_xx·f_yy − f_xy².
For a double integral sketch the region, choose the order (or polar coordinates for discs) so the inner limits are the bounding curves, and integrate inside first.

Differential equations

Exponential growth/decay: dy/dt = ky → y = y₀e^kt
Half-life: k = ln 2 / t_½
Newton's law of cooling: T(t) = T_a + (T₀ − T_a)e^−kt
Integrating factor: y′ + p(x)y = q(x): μ = e^{∫p dx}, (μy)′ = μq
Mixing tank: dm/dt = r·c − (r/V)·m

2Practice problems

1.IntegralsMedium

Evaluate I = ∫₀¹ 2x·(x² + 1)³ dx.

f(x) = 2x·(x² + 1)³[0, 1]

I =

2.IntegralsHard

Find the area A of the region enclosed by y = 2x² and y = 4x.

y = 2x²y = 4x

A =

3.Sequences & seriesMedium

Apply the ratio test to Σ_k=1^∞ 2ᵏ·k!/kᵏ. Find L = lim |a_k+1/a_k|.

a_k = 2ᵏ·k!/kᵏ

L =

4.Sequences & seriesHard

f(x) = eˣ is approximated by its Maclaurin polynomial of degree 5 at x = −0.5. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = 1 to find the largest possible error |R₅|.

f(x) = eˣn = 5x = −0.5M = 1

|R₅| =

5.Multivariable calculusMedium

z = −4x²y³ − 5x² with x = 2t − 2 and y = −t² + 1. Use the chain rule to find dz/dt at t = 0.

z = −4x²y³ − 5x²x = 2t − 2y = −t² + 1t = 0

dz/dt =

6.Multivariable calculusHard

Find the critical points of f(x, y) = −x³ + 27x − y² + 4y, classify them, and give the value f_max of f at its local maximum.

f(x, y) = −x³ + 27x − y² + 4y

f_max =

7.Differential equationsMedium

A forged steel part at 360 °C is left to cool in a workshop at 25 °C. After 15 min its temperature is 273 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 174 °C?

T₀ = 360 °CT_a = 25 °CT(15) = 273 °CT = 174 °C

t = min

8.Differential equationsHard

Solve y′ + 3y = 9e^−t with y(0) = 3. Find y(1).

y′ + 3y = 9e^−ty(0) = 3

y(1) =

3Solutions

1.Integrals

Answer: I = 3.75

Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.

Substitution

u = x² + 1, du = 2x dx

I = ∫₁² u³ du

Integrate and evaluate

I = [u⁴/4]₁² = (16 − 1)/4 = 3.75

Answer

I = 3.75

2.Integrals

Answer: A = 2.667

Intersection points

2x² = 4x → x = 0, x = 2

Integrate upper minus lower

A = [2x² − 0.6667x³]₀² = 2.667

Answer

A = 2.667

3.Sequences & series

Answer: L = 0.7358

Convergence: if the terms do not tend to 0 the series diverges (n-th term test). Σ1/kᵖ converges only for p > 1, and Σrᵏ only for |r| < 1. Compare a series with one of these when its terms behave like them for large k. Ratio test: L = lim |a_k+1/a_k|; root test: L = lim |a_k|^1/k. L < 1 converges, L > 1 diverges, L = 1 decides nothing. A power series Σc_k(x − x₀)ᵏ converges for |x − x₀| < R with R = lim |c_k/c_k+1|; an alternating series with terms decreasing to 0 converges.

Ratio of consecutive terms

|a_k+1/a_k| = 2·(k/(k + 1))ᵏ = 2/(1 + 1/k)ᵏ → 2/e

Conclusion

L = 0.7358 < 1

Answer

L = 0.7358

L < 1, so the series converges.

4.Sequences & series

Answer: |R₅| = 2.17 × 10⁻⁵

Taylor polynomials: near x = a a smooth function is approximated by Pₙ(x) = f(a) + f′(a)(x − a) + f″(a)(x − a)²/2! + … + f⁽ⁿ⁾(a)(x − a)ⁿ/n!; for a = 0 this is the Maclaurin polynomial. Standard series: eˣ = Σxᵏ/k!, sin x = x − x³/3! + x⁵/5! − …, cos x = 1 − x²/2! + x⁴/4! − …, ln(1 + x) = x − x²/2 + x³/3 − …, 1/(1 − x) = Σxᵏ. Lagrange error bound: |f(x) − Pₙ(x)| ≤ M·|x − a|ⁿ⁺¹/(n + 1)!, where M is the largest value of |f⁽ⁿ⁺¹⁾| between a and x.

Bound on the derivative

M = 1

Lagrange bound

|R₅| ≤ 1·0.5⁶/6! = 2.17 × 10⁻⁵

Answer

|R₅| = 2.17 × 10⁻⁵

The true error is smaller: |f(−0.5) − P₅(−0.5)| = 2.024 × 10⁻⁵.

5.Multivariable calculus

Answer: dz/dt = 72

Gradient: ∇f = (f_x, f_y) points in the direction of steepest ascent, and |∇f| is the largest rate of change. The directional derivative along a direction v is D_uf = ∇f·u with the unit vector u = v/|v|. The tangent plane at (a, b) is z = f(a, b) + f_x(a, b)(x − a) + f_y(a, b)(y − b); it gives the linear approximation of f near that point. Chain rule: dz/dt = f_x·dx/dt + f_y·dy/dt.

Point and rates

x(0) = −2, y(0) = 1

dx/dt = 2, dy/dt = −2t = 0

Partial derivatives

f_x = −8xy³ − 10x = 36

f_y = −12x²y² = −48

Chain rule

dz/dt = 36·2 + (−48)·0 = 72

Answer

dz/dt = 72

6.Multivariable calculus

Answer: f_max = 58

Critical points and double integrals: a critical point has f_x = f_y = 0. Second-derivative test with D = f_xxf_yy − f_xy²: D > 0 and f_xx > 0 is a local minimum, D > 0 and f_xx < 0 a local maximum, D < 0 a saddle point, and D = 0 decides nothing. For a double integral over a general region, let the inner limits be the curves that bound the region; for discs and sectors use polar coordinates, x = r cos θ, y = r sin θ and dA = r dr dθ (do not forget the factor r).

Critical points

Solve f_x = 0 and f_y = 0 together.

f_x = −(3x² − 27) = 0

f_y = −(2y − 4) = 0

(3, 2), (−3, 2)

Second-derivative test

(3, 2): f_xx = −18, f_yy = −2, f_xy = 0, D = 36 → local maximum

(−3, 2): f_xx = 18, f_yy = −2, f_xy = 0, D = −36 → saddle point

Value at the extremum

f(3, 2) = 58

Answer

f_max = 58

7.Differential equations

Answer: t = 40.41 min

First-order differential equations: dy/dt = k·y has the solution y = y₀·e^kt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e^−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e^{∫p dx}: (μy)′ = μq, so μy = ∫μq dx + C.

Cooling constant

k = ln(335/248)/15 = 0.02005 min⁻¹

Time to reach the target

t = ln(335/149)/0.02005 = 40.41 min

Answer

t = 40.41 min

8.Differential equations

Answer: y(1) = 1.581

Integrating factor

μ = e^3t

(e^3ty)′ = 9e^2t

General solution

y = 9/2·e^−t + C·e^−3t = 4.5·e^−t + C·e^−3t

Initial condition

3 = 4.5 + C → C = −1.5

Evaluate

y(1) = 4.5·0.3679 + (−1.5)·0.04979 = 1.581

Answer

y(1) = 1.581

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