∫ Calculus Assessment

Section 1: Limits

A limit describes the value a function approaches as its input approaches some number. We write it as:

read "the limit of f(x) as x approaches a equals L." The function does not have to actually reach that value — or even be defined at x = a — for the limit to exist. Limits are the foundation of both derivatives and integrals.

Method 1 — Direct substitution

For polynomials and most smooth functions, just substitute the value:

Method 2 — Factor and cancel

If substitution gives 0/0, the expression is undefined at that point but the limit may still exist. Factor the numerator, cancel the common factor, then substitute:

Continuity

A function is continuous at x = a when the limit exists and equals the function's actual value there — no jumps, holes, or breaks. Polynomials are continuous everywhere.

Section 2: The Derivative

The derivative of a function measures how fast its output changes as its input changes — the instantaneous rate of change. Graphically, it is the slope of the tangent line to the curve at a point.

From average to instantaneous

The average rate of change between two points is the familiar slope formula, Δy ÷ Δx. Shrink the distance between the points toward zero and the average becomes instantaneous. That is the limit definition of the derivative:

Notation

You will see the derivative written several ways — they all mean the same thing: f′(x), dy/dx, and d/dx [f(x)]. When the input is time, the derivative of a quantity is its rate per second: the derivative of position is velocity, and the derivative of charge is current.

Section 3: Derivative Rules

Working from the limit definition every time would be slow. A short set of rules handles almost every function you will meet:

Rule	Pattern	Example
Constant	d/dx [c] = 0	d/dx [12] = 0
Power	d/dx [xⁿ] = n·xⁿ⁻¹	d/dx [x⁴] = 4x³
Constant multiple	d/dx [c·f] = c·f′	d/dx [5x²] = 10x
Sum / difference	(f ± g)′ = f′ ± g′	d/dx [x² + x] = 2x + 1
Product	(f·g)′ = f′·g + f·g′	see below
Chain	d/dx f(g(x)) = f′(g(x))·g′(x)	see below

The chain rule handles a function inside another function: take the derivative of the outside, leave the inside alone, then multiply by the derivative of the inside.

Section 4: Applications of Derivatives

Rates of change over time

When a quantity depends on time, its derivative tells you how fast it is changing right now.

Capacitors and inductors

Two of the most important equations in electronics are derivatives:

• A capacitor's current is proportional to how fast its voltage changes. If the voltage is steady, no current flows — which is why a capacitor blocks DC.
• An inductor's voltage is proportional to how fast its current changes. Interrupt the current suddenly and di/dt is huge, producing the voltage spike that suppression diodes on relay coils exist to absorb.

Finding a maximum or minimum

At the peak or valley of a smooth curve, the tangent line is flat — the derivative is zero. To find the best (or worst) value, set the derivative equal to 0 and solve.

Section 5: Integrals

Integration is the reverse of differentiation. An antiderivative of f(x) is a function F(x) whose derivative is f(x). Because the derivative of any constant is zero, every antiderivative includes an unknown constant of integration, + C.

Indefinite integrals

Rule	Pattern	Example
Power (n ≠ −1)	∫ xⁿ dx = xⁿ⁺¹ ÷ (n + 1) + C	∫ x³ dx = x⁴/4 + C
Constant	∫ k dx = kx + C	∫ 5 dx = 5x + C
Constant multiple	∫ c·f dx = c ∫ f dx	∫ 4x dx = 2x² + C
Sum	∫ (f + g) dx = ∫ f dx + ∫ g dx	see below

Definite integrals and area

A definite integral has limits a and b and produces a number: the net area between the curve and the x-axis from a to b. The Fundamental Theorem of Calculus says you evaluate it with any antiderivative F:

No + C is needed for a definite integral; it cancels when you subtract.

Section 6: Applications of Integrals

Where a derivative gives a rate, an integral adds up that rate over time to give a total. On a graph, the total is the area under the rate curve.

Charge from current

When the current is constant, the integral is simply current × time:

Battery capacity

Battery capacity in amp-hours is the same integral, measured in hours. A load drawing 0.5 A for 24 hours uses 0.5 × 24 = 12 Ah. When the draw varies — a panel idling at 0.2 A that rises to 1 A in alarm — the amp-hours used are the area under the current-versus-time graph, which you add up section by section.

Energy from power

Capacitor voltage

Integrating the capacitor equation from Section 4 gives the voltage built up by a charging current:

Final Assessment

Answer all 10 questions. You need 80% to pass. Enter your name and email below the questions, then submit.