Mathematics · EUEE explainer

Calculus: limits, derivatives, and integrals — EUEE explained

Explainer · sample exam · 12 worked examples

The Ethiopian Grade 9–12 path into calculus: functions from Grades 10–11, then Grade 12 Unit 2 (Introduction to Calculus) — rates of change, limits, derivatives, applications, and integration. Worked examples plus one 2015 E.C. EUEE limit, solved step by step.

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Explainer for Calculus: limits, derivatives, and integrals. Hosted on this page — not uploaded to YouTube.

In plain language

A limit is the value a function approaches as the input approaches a number. The derivative is that instantaneous rate of change — the slope of the tangent — f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h} when the limit exists. The definite integral ∫abf(x) dx\int_a^b f(x)\,dx is the signed area under y=f(x)y=f(x), built as a limit of Riemann sums.

Grade 12 Mathematics Unit 2 is the calculus unit in the new Ethiopian curriculum: Introduction to Derivatives, Applications of Derivative, and Introduction to Integration. EUEE Mathematics keeps asking limits, tangent slopes, and rates of change. The algebra you need starts in Grades 9–11.

Grade 9 — the algebra calculus sits on

Grade 9 · Unit 2: The Number System

The new Grade 9 Mathematics book does not name calculus. It builds the language you will differentiate later: Unit 2 The Number System, Unit 3 Solving Equations, Unit 4 Solving Inequalities, and Unit 5 Introduction to Trigonometry.

Before a limit or a derivative, you need to simplify (x2−4)/(x−2)(x^2-4)/(x-2), solve f′(x)=0f'(x)=0, and read a sine or cosine later in Grade 10. That is why this explainer starts here instead of jumping to f′(x)f'(x).

Secant becomes the tangent

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}

On y=x2y=x^2, the dashed secant through x=ax=a and x=a+hx=a+h rotates onto the gold tangent as h→0h\to 0. Drag hh (or watch it animate). Reduced-motion users see the tight hh.

Instantaneous rate of change

Average rate =f(b)−f(a)b−a=\frac{f(b)-f(a)}{b-a}; instantaneous rate =f′(a)=f'(a).

The moving point carries its tangent. The slope at that instant is the derivative — not the average rise over a long interval.

Area as a Riemann sum

∑i=1nf(xi) Δx  →  ∫abf(x) dx\sum_{i=1}^{n} f(x_i)\,\Delta x \;\to\; \int_a^b f(x)\,dx

Left rectangles under y=0.35x2+0.25y=0.35x^2+0.25 on [0,2][0,2]. More strips, closer to the true area. The Grade 12 book uses left endpoints for simplicity.

Grade 10 Unit 1 — relations and functions

Grade 10 · Unit 1: Relations and Functions · p. 10–30

Grade 10 Unit 1 opens the function idea the derivative needs: define a relation, define a function, identify types of functions, and sketch their graphs.

A few pages later the same unit defines domain as the set of possible inputs. Without domain you cannot say “ff is differentiable at aa” — aa has to be an input where the difference quotient has a limit.

Later Grade 10 units stay in the calculus chain even though they are not named calculus: Unit 2 Polynomial Functions, Unit 3 Exponential and Logarithmic Functions, and Unit 4 Trigonometric Functions. Those are the families you differentiate in Grade 12.

“Define relation. Define function. Identify types of functions. Sketch graphs of various types of relations and functions.”

Grade 10 Maths, Unit 1

Secant becomes the tangent

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}

On y=x2y=x^2, the dashed secant through x=ax=a and x=a+hx=a+h rotates onto the gold tangent as h→0h\to 0. Drag hh (or watch it animate). Reduced-motion users see the tight hh.

Instantaneous rate of change

Average rate =f(b)−f(a)b−a=\frac{f(b)-f(a)}{b-a}; instantaneous rate =f′(a)=f'(a).

The moving point carries its tangent. The slope at that instant is the derivative — not the average rise over a long interval.

Area as a Riemann sum

∑i=1nf(xi) Δx  →  ∫abf(x) dx\sum_{i=1}^{n} f(x_i)\,\Delta x \;\to\; \int_a^b f(x)\,dx

Left rectangles under y=0.35x2+0.25y=0.35x^2+0.25 on [0,2][0,2]. More strips, closer to the true area. The Grade 12 book uses left endpoints for simplicity.

Domain and range — the inputs a derivative is allowed to use

Grade 10 · Unit 1: Relations and Functions · p. 30

If f(x)=xf(x)=\sqrt{x}, the difference quotient at a=0a=0 from the left is not even defined in the reals. Limits from both sides need a neighbourhood in the domain.

For a rational function p(x)q(x)\frac{p(x)}{q(x)}, first find where q(x)=0q(x)=0. Those xx-values are not in the domain. A limit there may still exist (a hole) even though f(a)f(a) does not.

“Domain – The set of all possible values which qualify as inputs to a function is known as the domain of the function.”

Grade 10 Maths, Unit 1

Secant becomes the tangent

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}

On y=x2y=x^2, the dashed secant through x=ax=a and x=a+hx=a+h rotates onto the gold tangent as h→0h\to 0. Drag hh (or watch it animate). Reduced-motion users see the tight hh.

Instantaneous rate of change

Average rate =f(b)−f(a)b−a=\frac{f(b)-f(a)}{b-a}; instantaneous rate =f′(a)=f'(a).

The moving point carries its tangent. The slope at that instant is the derivative — not the average rise over a long interval.

Area as a Riemann sum

∑i=1nf(xi) Δx  →  ∫abf(x) dx\sum_{i=1}^{n} f(x_i)\,\Delta x \;\to\; \int_a^b f(x)\,dx

Left rectangles under y=0.35x2+0.25y=0.35x^2+0.25 on [0,2][0,2]. More strips, closer to the true area. The Grade 12 book uses left endpoints for simplicity.

Grade 11 Units 1–2 — functions, then rational functions

Grade 11 · Unit 1: Relations and functions · p. 5

The Grade 11 Mathematics table of contents lists Unit 1: relations, inverse of relations, types of functions, composition, inverse functions, and applications — then Unit 2 Rational Expressions and Rational Functions.

Composition is the chain rule waiting to happen: if h=f∘gh=f\circ g, then h′(x)=f′(g(x)) g′(x)h'(x)=f'(g(x))\,g'(x). Rational functions are where EUEE limit items like x2−1x2+1\frac{x^2-1}{x^2+1} come from.

A worked relation in that unit lists ordered pairs, then takes domain as the first coordinates and range as the second.

“Unit 1 RELATIONS AND FUNCTIONS”

Grade 11 Maths, Unit 1

Secant becomes the tangent

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}

On y=x2y=x^2, the dashed secant through x=ax=a and x=a+hx=a+h rotates onto the gold tangent as h→0h\to 0. Drag hh (or watch it animate). Reduced-motion users see the tight hh.

Instantaneous rate of change

Average rate =f(b)−f(a)b−a=\frac{f(b)-f(a)}{b-a}; instantaneous rate =f′(a)=f'(a).

The moving point carries its tangent. The slope at that instant is the derivative — not the average rise over a long interval.

Area as a Riemann sum

∑i=1nf(xi) Δx  →  ∫abf(x) dx\sum_{i=1}^{n} f(x_i)\,\Delta x \;\to\; \int_a^b f(x)\,dx

Left rectangles under y=0.35x2+0.25y=0.35x^2+0.25 on [0,2][0,2]. More strips, closer to the true area. The Grade 12 book uses left endpoints for simplicity.

Grade 12 Unit 2 — the actual calculus unit

Grade 12 · Unit 2: Introduction to Calculus · p. 72–165

Unit 2 of the new Grade 12 Mathematics book is titled Introduction to Calculus. The contents page splits it as 2.1 Introduction to Derivatives, 2.2 Applications of Derivative, and 2.3 Introduction to Integration.

The same page says calculus deals with quantities that approach other quantities, and that geometrically the derivative is the slope of the tangent. That is the whole course in one paragraph — the rest of this page unpacks it with examples.

The unit introduces differentiation and integration: rate of change, gradients of functions, and areas under curves.

“The derivative is the exact rate at which one quantity changes with respect to another.”

Grade 12 Maths, Unit 2

Secant becomes the tangent

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}

On y=x2y=x^2, the dashed secant through x=ax=a and x=a+hx=a+h rotates onto the gold tangent as h→0h\to 0. Drag hh (or watch it animate). Reduced-motion users see the tight hh.

Instantaneous rate of change

Average rate =f(b)−f(a)b−a=\frac{f(b)-f(a)}{b-a}; instantaneous rate =f′(a)=f'(a).

The moving point carries its tangent. The slope at that instant is the derivative — not the average rise over a long interval.

Area as a Riemann sum

∑i=1nf(xi) Δx  →  ∫abf(x) dx\sum_{i=1}^{n} f(x_i)\,\Delta x \;\to\; \int_a^b f(x)\,dx

Left rectangles under y=0.35x2+0.25y=0.35x^2+0.25 on [0,2][0,2]. More strips, closer to the true area. The Grade 12 book uses left endpoints for simplicity.

Rate of change (Definition 2.1)

Grade 12 · Unit 2: Introduction to Calculus · p. 74

If yy is measured against xx, the average rate on [a,b][a,b] is f(b)−f(a)b−a\frac{f(b)-f(a)}{b-a}. Units are output units per input unit — kilometres per litre, birr per month.

The book notes that a rate can be positive (graph slopes up), negative (slopes down), or zero when the quantity does not change. That sign is the first thing EUEE concept items test before any algebra.

“A rate of change is a rate that describes how one quantity changes in relation to another quantity.”

Grade 12 Maths, Unit 2

Instantaneous rate of change

Average rate =f(b)−f(a)b−a=\frac{f(b)-f(a)}{b-a}; instantaneous rate =f′(a)=f'(a).

The moving point carries its tangent. The slope at that instant is the derivative — not the average rise over a long interval.

Instantaneous rate and the difference quotient

Grade 12 · Unit 2: Introduction to Calculus · p. 80

On the same page the book contrasts the average rate (over an entire period) with the instantaneous rate (at a certain moment). The algebra is the difference quotient f(x0+h)−f(x0)h\frac{f(x_0+h)-f(x_0)}{h} as hh gets closer to zero.

Geometrically that quotient is the slope of the secant. The limit is the slope of the tangent. Watch the animation: as hh shrinks, the dashed secant lies down on the gold tangent.

The book’s own numerical illustration on that page uses f(x)=2x2+9f(x)=2x^2+9 at x=−3x=-3 and obtains −12-12. The worked example below repeats that computation.

“The instantaneous rate of change is the change at that particular moment.”

Grade 12 Maths, Unit 2

Secant becomes the tangent

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}

On y=x2y=x^2, the dashed secant through x=ax=a and x=a+hx=a+h rotates onto the gold tangent as h→0h\to 0. Drag hh (or watch it animate). Reduced-motion users see the tight hh.

Limits and continuity

Grade 12 · Unit 2: Introduction to Calculus

lim⁡x→af(x)=L\lim_{x\to a}f(x)=L means f(x)f(x) can be made as close to LL as you like by taking xx close to aa (but not necessarily equal to aa). Continuity at aa adds three things: f(a)f(a) exists, the limit exists, and they are equal.

A rational function is continuous wherever the denominator is not zero. That is why lim⁡x→1x2−1x2+1\lim_{x\to 1}\frac{x^2-1}{x^2+1} is just substitution — see the sample exam. lim⁡x→2x2−4x−2\lim_{x\to 2}\frac{x^2-4}{x-2} is not: x=2x=2 is a hole, so you cancel first.

One-sided limits matter at endpoints and corners. If the left- and right-hand limits disagree, the two-sided limit does not exist and there is no derivative there either.

Secant becomes the tangent

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}

On y=x2y=x^2, the dashed secant through x=ax=a and x=a+hx=a+h rotates onto the gold tangent as h→0h\to 0. Drag hh (or watch it animate). Reduced-motion users see the tight hh.

Instantaneous rate of change

Average rate =f(b)−f(a)b−a=\frac{f(b)-f(a)}{b-a}; instantaneous rate =f′(a)=f'(a).

The moving point carries its tangent. The slope at that instant is the derivative — not the average rise over a long interval.

Area as a Riemann sum

∑i=1nf(xi) Δx  →  ∫abf(x) dx\sum_{i=1}^{n} f(x_i)\,\Delta x \;\to\; \int_a^b f(x)\,dx

Left rectangles under y=0.35x2+0.25y=0.35x^2+0.25 on [0,2][0,2]. More strips, closer to the true area. The Grade 12 book uses left endpoints for simplicity.

Definition of the derivative (Definition 2.6)

Grade 12 · Unit 2: Introduction to Calculus · p. 90

Section 2.1.4 introduces the derivative so you can talk about slope at every point of a curve. Definition 2.6 names f′(a)f'(a) as the gradient of the tangent to y=f(x)y=f(x) at (a,f(a))(a,f(a)), with hh approaching 00 from both sides.

If that limit exists we say ff is differentiable at aa. Differentiable implies continuous; the converse is false (a corner such as ∣x∣|x| at 00).

Notations you will see: f′(x)f'(x), dydx\frac{dy}{dx}, dfdx\frac{df}{dx}. They all mean the same object — a new function whose value at aa is that tangent slope.

“The slope concept usually pertains to straight lines. When a function is non-linear, its slope may vary from one point to the next.”

Grade 12 Maths, Unit 2

Secant becomes the tangent

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}

On y=x2y=x^2, the dashed secant through x=ax=a and x=a+hx=a+h rotates onto the gold tangent as h→0h\to 0. Drag hh (or watch it animate). Reduced-motion users see the tight hh.

Instantaneous rate of change

Average rate =f(b)−f(a)b−a=\frac{f(b)-f(a)}{b-a}; instantaneous rate =f′(a)=f'(a).

The moving point carries its tangent. The slope at that instant is the derivative — not the average rise over a long interval.

Area as a Riemann sum

∑i=1nf(xi) Δx  →  ∫abf(x) dx\sum_{i=1}^{n} f(x_i)\,\Delta x \;\to\; \int_a^b f(x)\,dx

Left rectangles under y=0.35x2+0.25y=0.35x^2+0.25 on [0,2][0,2]. More strips, closer to the true area. The Grade 12 book uses left endpoints for simplicity.

Differentiation rules Grade 12 actually uses

Grade 12 · Unit 2: Introduction to Calculus · p. 100

Power: (xn)′=nxn−1(x^n)'=n x^{n-1}. Sum/difference: (u±v)′=u′±v′(u\pm v)'=u'\pm v'. Constant multiple: (kf)′=kf′(kf)'=kf'. The book asks you to prove the difference rule from the sum rule and then states the constant-multiple rule (Theorem 2.3).

Product: (uv)′=u′v+uv′(uv)'=u'v+uv'. Quotient: (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}. Chain: (f(g(x)))′=f′(g(x)) g′(x)(f(g(x)))'=f'(g(x))\,g'(x).

Each rule below is a worked calculation you can reuse on the exam.

“Based on the proof of the sum rule of derivative, show the proof of the difference rule.”

Grade 12 Maths, Unit 2

Secant becomes the tangent

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}

On y=x2y=x^2, the dashed secant through x=ax=a and x=a+hx=a+h rotates onto the gold tangent as h→0h\to 0. Drag hh (or watch it animate). Reduced-motion users see the tight hh.

Instantaneous rate of change

Average rate =f(b)−f(a)b−a=\frac{f(b)-f(a)}{b-a}; instantaneous rate =f′(a)=f'(a).

The moving point carries its tangent. The slope at that instant is the derivative — not the average rise over a long interval.

Area as a Riemann sum

∑i=1nf(xi) Δx  →  ∫abf(x) dx\sum_{i=1}^{n} f(x_i)\,\Delta x \;\to\; \int_a^b f(x)\,dx

Left rectangles under y=0.35x2+0.25y=0.35x^2+0.25 on [0,2][0,2]. More strips, closer to the true area. The Grade 12 book uses left endpoints for simplicity.

Applications: increasing, decreasing, critical numbers

Grade 12 · Unit 2: Introduction to Calculus · p. 110

Section 2.2 uses the first derivative to find where ff increases (f′>0f'>0), decreases (f′<0f'<0), and where the slope is zero or undefined — the critical numbers.

A critical number of ff is a domain point cc with f′(c)=0f'(c)=0 or f′(c)f'(c) undefined. Candidates for local max/min live there. The first-derivative test checks the sign of f′f' on each side.

EUEE also asks the mechanical meaning of the derivative (instantaneous rate of change) and equal tangent slopes of two graphs — those sit on the published Calculus bank page.

“Find the critical numbers of the given functions.”

Grade 12 Maths, Unit 2

Secant becomes the tangent

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}

On y=x2y=x^2, the dashed secant through x=ax=a and x=a+hx=a+h rotates onto the gold tangent as h→0h\to 0. Drag hh (or watch it animate). Reduced-motion users see the tight hh.

Instantaneous rate of change

Average rate =f(b)−f(a)b−a=\frac{f(b)-f(a)}{b-a}; instantaneous rate =f′(a)=f'(a).

The moving point carries its tangent. The slope at that instant is the derivative — not the average rise over a long interval.

Area as a Riemann sum

∑i=1nf(xi) Δx  →  ∫abf(x) dx\sum_{i=1}^{n} f(x_i)\,\Delta x \;\to\; \int_a^b f(x)\,dx

Left rectangles under y=0.35x2+0.25y=0.35x^2+0.25 on [0,2][0,2]. More strips, closer to the true area. The Grade 12 book uses left endpoints for simplicity.

Integration and Riemann sums (section 2.3)

Grade 12 · Unit 2: Introduction to Calculus · p. 139–141

Unit 2.3 turns the question around: given a rate, recover the quantity; given y=f(x)y=f(x), find the area under the graph on [a,b][a,b]. The book approximates that area with rectangles.

The book names the sum a Riemann sum and always uses the left endpoint of each subinterval for simplicity. As the number of strips grows, the sum approaches the definite integral.

An antiderivative FF satisfies F′=fF'=f. The fundamental theorem then says ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx=F(b)-F(a) when ff is continuous. That is the computational payoff.

“Example 1 illustrates the first steps of a Riemann summation, a method that allows us to determine the area under a curved graph.”

Grade 12 Maths, Unit 2

Area as a Riemann sum

∑i=1nf(xi) Δx  →  ∫abf(x) dx\sum_{i=1}^{n} f(x_i)\,\Delta x \;\to\; \int_a^b f(x)\,dx

Left rectangles under y=0.35x2+0.25y=0.35x^2+0.25 on [0,2][0,2]. More strips, closer to the true area. The Grade 12 book uses left endpoints for simplicity.

Concept chain

How this idea sits across grades. Each link stays on Abrari (reader or topic page).

  1. prerequisite

    G9 · The Number System / equations

  2. prerequisite

    G10 · Relations and Functions

  3. prerequisite

    G11 · Relations and functions

  4. this

    G12 · Introduction to Calculus

  5. next

    G12 · Differential equations basics

Worked examples

A removable-hole limit

Evaluate lim⁡x→2x2−4x−2\lim_{x \to 2}\frac{x^2-4}{x-2}.

  1. Factor the numerator: x2−4=(x−2)(x+2)x^2-4=(x-2)(x+2).
  2. For x≠2x\neq 2 the fraction simplifies to x+2x+2.
  3. The limit is the value the simplified function approaches: 2+2=42+2=4.

Result: 44

Direct substitution when the function is defined

Evaluate lim⁡x→1x2−1x2+1\lim_{x \to 1}\frac{x^2-1}{x^2+1}.

  1. The denominator at x=1x=1 is 2≠02\neq 0, so the rational function is continuous there.
  2. Substitute: 1−11+1=02=0\frac{1-1}{1+1}=\frac{0}{2}=0.

Result: 00

A limit at infinity

Evaluate lim⁡x→∞3x2+1x2−4\lim_{x \to \infty}\frac{3x^2+1}{x^2-4}.

  1. Divide numerator and denominator by x2x^2: 3+1/x21−4/x2\frac{3+1/x^2}{1-4/x^2}.
  2. As x→∞x\to\infty, the extra terms vanish, so the limit is 33.

Result: 33

Definition of the derivative at a point

Using f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}, find f′(1)f'(1) for f(x)=x2f(x)=x^2.

  1. (1+h)2−1h=1+2h+h2−1h=2+h\frac{(1+h)^2-1}{h}=\frac{1+2h+h^2-1}{h}=2+h.
  2. Let h→0h\to 0 to get 22, which matches 2x2x at x=1x=1.

Result: 22

Power rule

Differentiate f(x)=x3−5x+2f(x)=x^3-5x+2.

  1. ddx[xn]=nxn−1\frac{d}{dx}[x^n]=n x^{n-1} and the derivative of a constant is 00.
  2. f′(x)=3x2−5f'(x)=3x^2-5.

Result: 3x2−53x^2-5

Product rule

Differentiate f(x)=(x2+1)(x−3)f(x)=(x^2+1)(x-3).

  1. (uv)′=u′v+uv′(uv)'=u'v+uv' with u=x2+1u=x^2+1, v=x−3v=x-3.
  2. f′(x)=(2x)(x−3)+(x2+1)(1)=3x2−6x+1f'(x)=(2x)(x-3)+(x^2+1)(1)=3x^2-6x+1.

Result: 3x2−6x+13x^2-6x+1

Quotient rule

Differentiate f(x)=x2−1x2+1f(x)=\frac{x^2-1}{x^2+1}.

  1. (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}.
  2. f′(x)=(2x)(x2+1)−(x2−1)(2x)(x2+1)2=4x(x2+1)2f'(x)=\frac{(2x)(x^2+1)-(x^2-1)(2x)}{(x^2+1)^2}=\frac{4x}{(x^2+1)^2}.

Result: 4x(x2+1)2\dfrac{4x}{(x^2+1)^2}

Chain rule

Differentiate f(x)=(2x+1)3f(x)=(2x+1)^3.

  1. Outer power, inner linear: f′(x)=3(2x+1)2⋅2=6(2x+1)2f'(x)=3(2x+1)^2\cdot 2=6(2x+1)^2.

Result: 6(2x+1)26(2x+1)^2

Instantaneous rate matching the Grade 12 setup

Find the instantaneous rate of change of f(x)=2x2+9f(x)=2x^2+9 at x=−3x=-3.

  1. Difference quotient: 2(−3+h)2+9−(2⋅9+9)h=2(9−6h+h2)−18h=2h−12\frac{2(-3+h)^2+9-(2\cdot 9+9)}{h}=\frac{2(9-6h+h^2)-18}{h}=2h-12.
  2. Let h→0h\to 0 to get −12-12. The power rule f′(x)=4xf'(x)=4x gives the same value at x=−3x=-3.

Result: −12-12

Critical numbers

Find the critical numbers of f(x)=x3−3xf(x)=x^3-3x.

  1. f′(x)=3x2−3=3(x2−1)f'(x)=3x^2-3=3(x^2-1).
  2. f′(x)=0f'(x)=0 at x=±1x=\pm 1. Both are in the domain, so both are critical numbers.

Result: x=±1x=\pm 1

A definite integral as a limit of sums

Evaluate ∫02x2 dx\int_0^2 x^2\,dx.

  1. An antiderivative of x2x^2 is x3/3x^3/3.
  2. [x33]02=83−0=83\left[\frac{x^3}{3}\right]_0^2=\frac{8}{3}-0=\frac{8}{3}.

Result: 83\dfrac{8}{3}

Fundamental theorem, derivative of an integral

If F(x)=∫0xt2 dtF(x)=\int_0^x t^2\,dt, find F′(x)F'(x).

  1. The fundamental theorem says ddx∫axf(t) dt=f(x)\frac{d}{dx}\int_a^x f(t)\,dt=f(x) when ff is continuous.
  2. So F′(x)=x2F'(x)=x^2.

Result: x2x^2

Common mistakes

  • Substituting into x2−4x−2\frac{x^2-4}{x-2} and writing “undefined” instead of taking the limit.
  • Mixing the average rate f(b)−f(a)b−a\frac{f(b)-f(a)}{b-a} with the derivative at a single point.
  • Saying the derivative is “the area under the curve” (that is the integral).
  • Forgetting the chain-rule factor g′(x)g'(x) after differentiating the outer function.
  • Calling every critical number a maximum — check the sign of f′f' on each side.

Sample exam

One past EUEE Mathematics item, worked in full.

2015 E.C. · natural · Q36 · Limits

What is the value of lim⁡x→1x2−1x2+1\lim_{x \to 1} \frac{x^{2} - 1}{x^{2} + 1}?
  1. A. 3
  2. B. 1
  3. C. 2
  4. D. 0
Worked solution

Answer: D

Answer D: 0. The rational function is continuous at x = 1 because the denominator equals 2, so the limit is (1-1)/(1+1) = 0.

Why not A/B/C?

  • A. 3 would require a different numerator or a cancelled factor that is not here.
  • B. 1 is (x^2-1)/(x^2-1) or a mistaken 2/2 — the denominator is x^2+1, not x^2-1.
  • C. 2 is the denominator at x = 1, not the value of the fraction.

Practice this in the app

  1. Read the stem: lim⁡x→1x2−1x2+1\lim_{x\to 1}\frac{x^2-1}{x^2+1}.
  2. Check the denominator at x=1x=1: 1+1=2≠01+1=2\neq 0. The function is defined and continuous at 11.
  3. Substitute: 12−112+1=02=0\frac{1^2-1}{1^2+1}=\frac{0}{2}=0.
  4. Letter D is 00. Letters A–C are 33, 11, and 22 — none equal the substituted value.
  5. Answer: D.

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