Interactive AP Calculus Formula Sheet

§1

Limits & Continuity

The behavior of functions as inputs approach a value.

Explanation

Imagine zooming in on the graph as x slides toward a from the left and from the right. If the y-value heads to the same number L from both sides, that number is the limit. The function doesn't even have to be defined at a — limits care about what's happening near a, not at it.

When to Use It

Use whenever a free-response problem asks for a limit value. Try direct substitution first — only escalate to algebra or L'Hôpital if you hit an indeterminate form.

Worked Example

Evaluate the limit of 3x + 1 as x approaches 2.

x \to 2 lim (3 x + 1)

1
The function 3x + 1 is a polynomial, so it's continuous everywhere — direct substitution works.
2
Substitute x = 2 into the expression.
$3 (2) + 1$
3
Simplify.
$6 + 1 = 7$

Answer

lim_{x \to 2} (3 x + 1) = 7

Explanation

If you can sandwich a hard-to-evaluate function between two friendlier ones that converge to the same value, the middle function is forced to converge there too. It's how you tame wild oscillating functions like x²sin(1/x) near zero.

When to Use It

Reach for this when a function involves bounded oscillation (typically sin or cos of something wild) multiplied by a factor heading to 0.

Worked Example

Find the limit of x² sin(1/x) as x approaches 0.

x \to 0 lim x^{2} sin (1/ x)

1
Sine of anything is bounded between -1 and 1.
$- 1 \leq sin (1/ x) \leq 1$
2
Multiply through by x², which is non-negative.
$- x^{2} \leq x^{2} sin (1/ x) \leq x^{2}$
3
Both outer bounds go to 0 as x → 0.
$lim_{x \to 0} (- x^{2}) = 0, lim_{x \to 0} x^{2} = 0$
4
By the Squeeze Theorem, the middle expression is forced to the same limit.

Answer

lim_{x \to 0} x^{2} sin (1/ x) = 0

Explanation

When plugging in gives you 0/0 or ∞/∞, the ratio of the original functions behaves like the ratio of their slopes near that point. Take the derivative of the top and the derivative of the bottom (separately — this is not the quotient rule) and try the limit again.

When to Use It

Use the instant a limit produces 0/0 or ∞/∞. You must show the indeterminate form first — graders deduct points if you skip that check.

Worked Example

Evaluate the limit of sin(x)/x as x approaches 0.

x \to 0 lim \frac{sin x}{x}

1
Plugging in x = 0 gives 0/0, an indeterminate form — L'Hôpital applies.
2
Differentiate the top: d/dx[sin x] = cos x.
3
Differentiate the bottom: d/dx[x] = 1.
4
Rewrite the limit using those derivatives.
$lim_{x \to 0} \frac{c o s x}{1}$
5
Substitute x = 0.
$\frac{c o s 0}{1} = \frac{1}{1} = 1$

Answer

lim_{x \to 0} \frac{s i n x}{x} = 1

Explanation

A function is continuous at a point when three things agree: f(a) exists, the limit exists, and they're equal. Visually, you can draw the graph through that point without lifting your pencil — no holes, no jumps, no vertical asymptotes.

When to Use It

Use on multiple-choice continuity questions and on piecewise problems asking you to solve for a constant that makes the function continuous.

Worked Example

Is f(x) = (x² − 4)/(x − 2) continuous at x = 2? If not, can it be made continuous?

1
Check f(2): plug in to get 0/0 — f(2) is undefined, so f is not continuous at 2.
2
Factor and simplify the expression.
$\frac{( x - 2 ) ( x + 2 )}{x - 2} = x + 2 for x \neq = 2$
3
Take the limit as x → 2.
$lim_{x \to 2} (x + 2) = 4$
4
The limit exists, so we can patch the hole by defining f(2) = 4.

Answer

f (2) = 4 makes it continuous

Explanation

If a continuous function starts at one height and ends at another, it must pass through every height in between somewhere along the way. This is what lets you prove an equation has a solution without actually solving it — just show the function changes sign.

When to Use It

Use to justify that an equation has a solution on an interval, especially when a free-response prompt says "explain why there must exist a value c such that…".

Worked Example

Show that f(x) = x³ − x − 1 has a root in the interval [1, 2].

1
f is a polynomial, so it's continuous everywhere — IVT applies on [1, 2].
2
Compute f at the endpoints.
$f (1) = 1 - 1 - 1 = - 1$
3
And the other endpoint.
$f (2) = 8 - 2 - 1 = 5$
4
0 lies between −1 and 5, so by IVT some c in (1, 2) gives f(c) = 0.

Answer

\exists c \in (1, 2) : f (c) = 0

§2

Derivatives

Instantaneous rates of change and slopes of tangent lines.

Explanation

Pick two points on the graph that are h apart. The fraction is the slope of the line connecting them — the average rate of change. As h shrinks to zero, that secant line tilts into the tangent line at x, and its slope becomes the instantaneous rate of change.

When to Use It

Use when a problem explicitly says "using the definition of the derivative" or when a limit matches the difference-quotient form — that's a hidden derivative in disguise.

Worked Example

Use the definition to find f'(x) for f(x) = x².

1
Write f(x + h).
$f (x + h) = (x + h)^{2} = x^{2} + 2 x h + h^{2}$
2
Form the difference quotient.
$\frac{( x ^{2} + 2 x h + h ^{2} ) - x ^{2}}{h} = \frac{2 x h + h ^{2}}{h}$
3
Cancel an h from the numerator.
$2 x + h$
4
Take the limit as h → 0.
$lim_{h \to 0} (2 x + h) = 2 x$

Answer

f^{'} (x) = 2 x

Explanation

The fastest derivative rule there is. Multiply by the current exponent, then knock the exponent down by one. Works for any real exponent — integers, fractions, negatives — as long as it's a constant power of x.

When to Use It

Use on essentially every polynomial, root, or reciprocal — rewrite radicals as fractional exponents first so the rule applies.

Worked Example

Differentiate f(x) = x⁵.

1
Identify the exponent: n = 5.
2
Multiply by the exponent.
$5 \cdot x^{5}$
3
Reduce the exponent by one.
$5 x^{5 - 1} = 5 x^{4}$

Answer

f^{'} (x) = 5 x^{4}

Explanation

When two functions are multiplied, you can't just multiply their derivatives. Instead, take the derivative of the first times the second, plus the first times the derivative of the second. Each piece gets its turn being differentiated.

When to Use It

Use when two functions of x are multiplied and neither is a constant. Don't apply it to constants or to a single function raised to a power.

Worked Example

Differentiate h(x) = x² · sin(x).

1
Let f = x² and g = sin(x).
2
Compute the derivatives.
$f^{'} = 2 x, g^{'} = cos x$
3
Apply the product rule.
$h^{'} = f^{'} g + f g^{'} = (2 x) (sin x) + (x^{2}) (cos x)$

Answer

h^{'} (x) = 2 x sin x + x^{2} cos x

Explanation

For a fraction of two functions: bottom times derivative of top, minus top times derivative of bottom, all divided by the bottom squared. The order matters because of the minus sign — flipping it gives the wrong answer.

When to Use It

Use for explicit fractions where both top and bottom contain x. If the denominator is just a constant, factor it out instead — it's faster.

Worked Example

Differentiate h(x) = x² / (x + 1).

1
Let f = x², g = x + 1, so f' = 2x and g' = 1.
2
Apply the quotient rule.
$h^{'} = \frac{( 2 x ) ( x + 1 ) - ( x ^{2} ) ( 1 )}{( x + 1 ) ^{2}}$
3
Expand the numerator.
$h^{'} = \frac{2 x ^{2} + 2 x - x ^{2}}{( x + 1 ) ^{2}}$
4
Simplify.
$h^{'} = \frac{x ^{2} + 2 x}{( x + 1 ) ^{2}}$

Answer

h^{'} (x) = \frac{x ^{2} + 2 x}{( x + 1 ) ^{2}}

Explanation

For nested functions, peel them like an onion. Differentiate the outer function while leaving the inner function untouched inside, then multiply by the derivative of the inner function. Miss that final multiplication and you've made the most common calculus mistake.

When to Use It

Use whenever you see a composition (something inside something else). This is the single most-tested derivative rule on the AP exam.

Worked Example

Differentiate h(x) = sin(3x² + 1).

1
Outer function: sin(u). Inner function: u = 3x² + 1.
2
Derivative of the outer (leave inner alone).
$cos (3 x^{2} + 1)$
3
Derivative of the inner.
$\frac{d}{d x} (3 x^{2} + 1) = 6 x$
4
Multiply them.
$h^{'} (x) = cos (3 x^{2} + 1) \cdot 6 x$

Answer

h^{'} (x) = 6 x cos (3 x^{2} + 1)

Explanation

Sine and cosine cycle into each other (with a minus sign appearing when you differentiate cosine). Tangent's derivative is secant squared. These three show up constantly — memorize them cold and the rest of the trig derivatives follow from the quotient rule.

When to Use It

Use any time you differentiate a trig function; combine with the chain rule when the argument isn't just x.

Worked Example

Find the slope of f(x) = sin(x) at x = π/3.

1
Differentiate.
$f^{'} (x) = cos x$
2
Evaluate at x = π/3.
$f^{'} (π /3) = cos (π /3)$
3
Recall the unit-circle value.
$cos (π /3) = \frac{1}{2}$

Answer

f^{'} (π /3) = \frac{1}{2}

Explanation

e^x is the function that's its own derivative — that's literally what makes e special. The natural log's derivative is 1/x. For other bases, you pick up a factor of ln(a), which is why mathematicians prefer base e: it's the only base with no extra constant.

When to Use It

Use for e^x, a^x, ln(x), and log_a(x). Pair with the chain rule when there's anything more complex than x inside.

Worked Example

Differentiate f(x) = e^(2x) + ln(x).

1
For e^(2x), apply the chain rule.
$\frac{d}{d x} e^{2 x} = e^{2 x} \cdot 2 = 2 e^{2 x}$
2
Differentiate ln(x).
$\frac{d}{d x} ln x = \frac{1}{x}$
3
Add the two results.

Answer

f^{'} (x) = 2 e^{2 x} + \frac{1}{x}

Explanation

Inverse trig derivatives are surprisingly algebraic — no trig functions appear in the answer. Recognizing the patterns 1/√(1-x²) and 1/(1+x²) when integrating is just as important as knowing them as derivatives.

When to Use It

Use when you differentiate arcsin or arctan, and recognize these forms on integration problems too — they're frequent disguises in u-sub.

Worked Example

Find the derivative of f(x) = arctan(x) at x = 1.

1
Use the inverse-trig rule.
$f^{'} (x) = \frac{1}{1 + x ^{2}}$
2
Substitute x = 1.
$f^{'} (1) = \frac{1}{1 + 1 ^{2}} = \frac{1}{2}$

Answer

f^{'} (1) = \frac{1}{2}

Explanation

If you drove 60 miles in one hour, at some instant your speedometer must have read exactly 60. More formally: on any smooth, continuous interval, there's at least one point where the tangent slope matches the slope of the line connecting the endpoints.

When to Use It

Use on free-response questions asking you to guarantee a specific instantaneous rate exists, or to find a c value where it does. Always state that f is continuous and differentiable first.

Worked Example

Find c guaranteed by the MVT for f(x) = x² on [1, 3].

1
Compute the average rate of change.
$\frac{f ( 3 ) - f ( 1 )}{3 - 1} = \frac{9 - 1}{2} = 4$
2
Differentiate.
$f^{'} (x) = 2 x$
3
Set f'(c) equal to the average rate.
$2 c = 4 \Rightarrow c = 2$
4
Verify c = 2 lies in the open interval (1, 3). ✓

Answer

c = 2

§3

Integrals

Accumulation, area, and antiderivatives.

Explanation

If you build a new function by accumulating area under f from a fixed point a out to a moving endpoint x, then the rate that area is piling up — the derivative — is just the height of f at x. Integration and differentiation are inverse operations.

When to Use It

Use when a problem defines a function as an integral with a variable upper limit and asks for its derivative — combine with the chain rule if the upper limit isn't just x.

Worked Example

If F(x) = ∫₀ˣ cos(t) dt, find F'(x).

1
F is an accumulation function with a fixed lower bound (0) and variable upper bound (x).
2
Part 1 of the FTC says the derivative is just the integrand evaluated at x.
3
Replace t with x in cos(t).
$F^{'} (x) = cos x$

Answer

F^{'} (x) = cos x

Explanation

To find the total accumulated area from a to b, don't slice it into a million rectangles — just find any antiderivative F of f, plug in the endpoints, and subtract. This single shortcut is what makes calculus practical instead of tedious.

When to Use It

Use to evaluate every definite integral where you can find an antiderivative. This is the workhorse of integral free-response.

Worked Example

Evaluate the integral of x² from 1 to 3.

\int_{1}^{3} x^{2} d x

1
Find an antiderivative using the power rule.
$F (x) = \frac{x ^{3}}{3}$
2
Evaluate at the upper limit.
$F (3) = \frac{27}{3} = 9$
3
Evaluate at the lower limit.
$F (1) = \frac{1}{3}$
4
Subtract.
$9 - \frac{1}{3} = \frac{26}{3}$

Answer

\int_{1}^{3} x^{2} d x = \frac{26}{3}

Explanation

The reverse of the power rule for derivatives: raise the exponent by one and divide by the new exponent. The exception is n = -1, because dividing by zero is forbidden — that case becomes the natural log instead. And never forget the +C; antiderivatives come in infinite families.

When to Use It

Use on every polynomial antiderivative. Rewrite radicals and reciprocals as powers of x first.

Worked Example

Integrate ∫ x³ dx.

1
Identify n = 3.
2
Add one to the exponent.
$x^{3 + 1} = x^{4}$
3
Divide by the new exponent.
$\frac{x ^{4}}{4}$
4
Add the constant of integration.

Answer

\int x^{3} d x = \frac{x ^{4}}{4} + C

Explanation

When you spot a function and its derivative tangled together inside an integral, rename the inner function u. The messy integral collapses into a clean one in u. It's the chain rule running backward, and it's the workhorse of integration.

When to Use It

Use when the integrand contains a function and (a constant multiple of) its derivative. If straightforward antidifferentiation fails, this is your first move.

Worked Example

Evaluate ∫ 2x · cos(x²) dx.

1
Let u = x² (the inner function).
2
Differentiate to find du.
$d u = 2 x d x$
3
The 2x dx in the original integral becomes du.
$\int cos (u) d u$
4
Integrate.
$sin u + C$
5
Replace u with x².

Answer

\int 2 x cos (x^{2}) d x = sin (x^{2}) + C

Explanation

Just the trig derivatives read in reverse. Watch the sign on sine — integrating sine gives negative cosine, because differentiating cosine gives negative sine. The minus signs flip when you go the other direction.

When to Use It

Use on direct trig antiderivatives and as the inner step after a u-substitution that lands on sin, cos, or sec².

Worked Example

Evaluate ∫₀^(π/2) sin(x) dx.

1
An antiderivative of sin(x) is −cos(x).
2
Evaluate at the upper limit.
$- cos (π /2) = 0$
3
Evaluate at the lower limit.
$- cos (0) = - 1$
4
Subtract.
$0 - (- 1) = 1$

Answer

\int_{0}^{π /2} sin x d x = 1

Explanation

e^x integrates to itself — same as its derivative. And 1/x, the one case the power rule can't touch, integrates to the natural log. The absolute value matters: 1/x is defined for negative x too, and ln|x| handles both sides.

When to Use It

Use for ∫e^x and ∫1/x. Remember the absolute value on ln|x| — it matters when the bounds include negative numbers.

Worked Example

Evaluate ∫₁ᵉ (1/x) dx.

1
The antiderivative of 1/x is ln|x|.
2
Evaluate at the upper limit.
$ln e = 1$
3
Evaluate at the lower limit.
$ln 1 = 0$
4
Subtract.
$1 - 0 = 1$

Answer

\int_{1}^{e} \frac{1}{x} d x = 1

Explanation

Add up all the y-values across the interval (that's the integral) and divide by the width of the interval. It's the height of a rectangle with the same width and the same total area as the region under the curve.

When to Use It

Use when a problem asks for the average value, average rate, or average temperature/velocity/etc. of a continuous function over an interval.

Worked Example

Find the average value of f(x) = x² on [0, 3].

1
Width of the interval: b − a = 3.
2
Integrate f.
$\int_{0}^{3} x^{2} d x = \frac{x ^{3}}{3}_{0}^{3} = 9$
3
Divide by the width.
$\overset{ˉ}{f} = \frac{1}{3} (9) = 3$

Answer

\overset{ˉ}{f} = 3

§4

Applications

Motion, areas, volumes, and accumulated change.

Explanation

Position tells you where, velocity tells you how fast where is changing, and acceleration tells you how fast velocity is changing. Each one is the derivative of the one before it — and integrating runs the chain backward.

When to Use It

Use on particle-motion problems whenever you need to convert between position, velocity, and acceleration via differentiation or integration.

Worked Example

A particle has position s(t) = t³ − 6t². Find velocity and acceleration at t = 2.

1
Differentiate s to get velocity.
$v (t) = 3 t^{2} - 12 t$
2
Differentiate again for acceleration.
$a (t) = 6 t - 12$
3
Plug in t = 2.
$v (2) = 12 - 24 = - 12$
4
And for acceleration.
$a (2) = 12 - 12 = 0$

Answer

v (2) = - 12, a (2) = 0

Explanation

Displacement is net movement — backwards trips cancel forwards trips. Distance is total movement, which is why it uses the absolute value of velocity: every step counts, regardless of direction. Drive 10 miles east then 10 miles west: displacement 0, distance 20.

When to Use It

Use when a problem distinguishes between "displacement" or "net change in position" (no absolute value) and "total distance traveled" (absolute value).

Worked Example

A particle has velocity v(t) = t − 2 on [0, 3]. Find displacement and distance.

1
Displacement: integrate v directly.
$\int_{0}^{3} (t - 2) d t = \frac{t ^{2}}{2} - 2 t_{0}^{3} = \frac{9}{2} - 6 = - \frac{3}{2}$
2
v(t) = 0 at t = 2; v is negative on [0, 2] and positive on [2, 3].
3
Distance from 0 to 2.
$\int_{0}^{2} - (t - 2) d t = 2$
4
Distance from 2 to 3.
$\int_{2}^{3} (t - 2) d t = \frac{1}{2}$
5
Add them.
$2 + \frac{1}{2} = \frac{5}{2}$

Answer

Δ s = - \frac{3}{2}, d = \frac{5}{2}

Explanation

The height of the region at each x is the upper curve minus the lower curve. Integrate that height across the interval to add up all the thin vertical strips. If the curves cross, split the integral and switch which one is on top.

When to Use It

Use when a problem asks for the area of a region bounded by two curves. Find the intersection points first to set your limits.

Worked Example

Find the area between y = x² and y = x from x = 0 to x = 1.

1
On [0, 1], x ≥ x², so the line y = x is on top.
2
Set up the integral.
$A = \int_{0}^{1} (x - x^{2}) d x$
3
Integrate.
$\frac{x ^{2}}{2} - \frac{x ^{3}}{3}_{0}^{1}$
4
Evaluate.
$\frac{1}{2} - \frac{1}{3} = \frac{1}{6}$

Answer

A = \frac{1}{6}

Explanation

Spin a region under f(x) around the x-axis and it sweeps out a solid made of thin circular disks. Each disk has radius f(x), so area πf(x)², and integrating those areas across the interval stacks them into the full volume.

When to Use It

Use when a region touching the axis is revolved around that axis. Each cross section is a solid disk.

Worked Example

Find the volume when y = √x on [0, 4] is rotated about the x-axis.

1
Radius of each disk is f(x) = √x.
2
Square the radius.
$[f (x)]^{2} = x$
3
Set up the integral.
$V = π \int_{0}^{4} x d x$
4
Integrate.
$π \cdot \frac{x ^{2}}{2}_{0}^{4} = π \cdot 8$

Answer

V = 8 π

Explanation

When the region you're spinning doesn't touch the axis, each slice is a washer — a disk with a smaller disk punched out. Take the area of the outer circle, subtract the area of the inner circle, and integrate. The outer radius is the curve farther from the axis.

When to Use It

Use when the region being revolved does not touch the axis of rotation, so each cross section has a hole.

Worked Example

Rotate the region between y = x and y = x² on [0, 1] about the x-axis. Find the volume.

1
Outer radius R(x) = x (the line is farther from the axis here).
2
Inner radius r(x) = x².
3
Set up the washer integral.
$V = π \int_{0}^{1} (x^{2} - x^{4}) d x$
4
Integrate.
$π [\frac{x ^{3}}{3} - \frac{x ^{5}}{5}]_{0}^{1} = π (\frac{1}{3} - \frac{1}{5})$
5
Simplify.
$π \cdot \frac{2}{15}$

Answer

V = \frac{2 π}{15}

Explanation

For solids built by stacking shapes (squares, triangles, semicircles) along an axis, write A(x) for the area of a single slice at position x, then integrate to add them all up. The shape of the slice changes the formula for A(x), but the structure is always the same.

When to Use It

Use when a solid is described by stacking shapes (squares, triangles, semicircles) along an axis — not by revolution.

Worked Example

A solid sits on the region under y = √x on [0, 4]. Each cross section perpendicular to the x-axis is a square. Find the volume.

1
Side length of each square is the height of the region: s = √x.
2
Area of one square slice.
$A (x) = s^{2} = x$
3
Set up the integral.
$V = \int_{0}^{4} x d x$
4
Integrate.
$\frac{x ^{2}}{2}_{0}^{4} = 8$

Answer

V = 8

§5

Differential Equations

Equations relating a function to its derivatives — and how to solve them.

Explanation

If a differential equation can be written so that everything with y (including dy) is on one side and everything with x (including dx) is on the other, you can integrate each side independently. Don't forget a single +C — usually attached to the x-side — and use the initial condition to pin it down.

When to Use It

Use whenever dy/dx factors as a function of x times a function of y. This is the most common AP-style DE problem.

Worked Example

Solve dy/dx = xy with y(0) = 2.

\frac{d y}{d x} = x y

1
Separate variables — y's on the left, x's on the right.
$\frac{d y}{y} = x d x$
2
Integrate both sides.
$ln ∣ y ∣ = \frac{x ^{2}}{2} + C$
3
Exponentiate to solve for y.
$y = A e^{x^{2} /2}$
4
Apply y(0) = 2 to find A.
$2 = A e^{0} \Rightarrow A = 2$

Answer

y = 2 e^{x^{2} /2}

Explanation

A slope field is a visual representation of dy/dx. At each grid point (x, y), you draw a tiny segment with slope equal to dy/dx evaluated there. A solution curve through a point follows those segments — it's tangent to them. Slope fields let you sketch solutions without solving the DE.

When to Use It

Use to sketch a particular solution through a given point, or to identify which DE matches a given slope-field picture on multiple choice.

Worked Example

Describe the slope field for dy/dx = x at the points (0,0), (1,2), and (-2,1).

1
The slope depends only on x, not y — segments in the same vertical column are parallel.
2
At (0,0): slope = 0 (horizontal segment).
3
At (1,2): slope = 1 (segment rises one unit per unit right).
4
At (-2,1): slope = -2 (segment falls steeply).

Answer

Slopes: 0, 1, - 2 respectively.

Explanation

Any quantity whose rate of change is proportional to itself grows or decays exponentially. k > 0 means growth (population, compound interest); k < 0 means decay (radioactive material, cooling). The constant C is the initial value y(0), since y(0) = Ce^0 = C.

When to Use It

Use whenever a problem says 'the rate of change is proportional to the amount present' — population, radioactive decay, Newton's law of cooling, continuously compounded interest.

Worked Example

A bacterial culture grows so that dy/dt = 0.3y. If y(0) = 500, find y(t) and y(5).

\frac{d y}{d t} = 0.3 y

1
This is the exponential growth model with k = 0.3, so the solution form is y = Ce^{0.3t}.
2
Apply y(0) = 500.
$500 = C e^{0} \Rightarrow C = 500$
3
Write the particular solution.
$y (t) = 500 e^{0.3 t}$
4
Evaluate at t = 5.
$y (5) = 500 e^{1.5} \approx 2241$

Answer

y (t) = 500 e^{0.3 t}, y (5) \approx 2241

The interactiveformula sheet.

Limits & Continuity

Definition of a Limit

Squeeze Theorem

L'Hôpital's Rule

Continuity at a Point

Intermediate Value Theorem

Derivatives

Definition of the Derivative

Power Rule

Product Rule

Quotient Rule

Chain Rule

Trig Derivatives

Exponential & Log

Inverse Trig

Mean Value Theorem

Integrals

Fundamental Theorem (Part 1)

Fundamental Theorem (Part 2)

Power Rule (Integration)

u-Substitution

Common Trig Integrals

Exponential & Log Integrals

Average Value

Applications

Position, Velocity, Acceleration

Displacement vs Distance

Area Between Curves

Volume — Disk Method

Volume — Washer Method

Volume — Known Cross Sections

Differential Equations

Separation of Variables

Slope Fields

Exponential Growth & Decay