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← Calculus I: Change & Accumulation levels

Calculus I: Change & Accumulation / LEVEL 1 · DIFFICULTY 4/5

Calculus I: Change & Accumulation

Read the prerequisite, study a worked example and justify each step in these selected university topics.

4 stages · 32 practice problems · two 8-question assessment forms

STAGE UC1 1.1 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Limits & Continuity

Useful preparation: Functions, Domains & Transformations

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

A limit describes nearby behavior, not necessarily the point value. As an input window shrinks, all allowed inputs near its center must force outputs near one common number. Numerical tables suggest answers; algebra or theorems justify them.

Two-sided limit
The number approached from both sides of an interior input. Left and right approaches must agree; the function need not be defined at the input itself.
Continuity at a point
The function is defined there, its relevant limit exists, and the limit equals the actual value. At a domain endpoint, use the approach available within the domain.
Removable discontinuity
A missing or incorrect point value with an existing finite limit; assigning that limit repairs continuity at the point.

Conditions before calculation

  • The quotient limit law requires the denominator's limit to be nonzero. A zero denominator requires further analysis, not division by zero.
  • Algebraic cancellation may be used near the target when the canceled factor is nonzero there; preserve the original function's domain.
  • The intermediate value theorem requires continuity on the entire closed interval; opposite endpoint signs alone are insufficient.

Why the method works

1. Make closeness precise

Epsilon controls output error. Delta must work for every permitted nearby input, excluding the center. This requirement is stronger than checking selected numerical values.

lim⁡x→af(x)=L  ⟺  ∀ε>0 ∃δ>0:0<∣x−a∣<δ ⟹ ∣f(x)−L∣<ε\lim_{x\to a}f(x)=L\iff \forall\varepsilon>0\ \exists\delta>0:\quad 0<|x-a|<\delta\ \Longrightarrow\ |f(x)-L|<\varepsilon

2. Simplify a punctured neighborhood

For a zero-over-zero indeterminate form, try factoring or rationalizing. Equality away from the center suffices to transfer the limit.

x−ax−a=1x+a(x≠a, a>0)\frac{\sqrt{x}-\sqrt a}{x-a}=\frac1{\sqrt x+\sqrt a}\qquad(x\ne a,\ a>0)

3. Separate existence from point assignment

Compute the one-sided limits first. Only when they agree can assigning that common value make the function continuous.

lim⁡x→a−f(x)=lim⁡x→a+f(x)=L,f(a)=L\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=L,\qquad f(a)=L

Worked examples

EXAMPLE 1

Evaluate the radical limit without decimal estimates.

  1. Direct substitution gives zero over zero, which does not determine the answer. The square root is defined on a neighborhood of 4.

    lim⁡x→4x−2x−4\lim_{x\to4}\frac{\sqrt x-2}{x-4}

  2. Multiply numerator and denominator by the conjugate. For nearby inputs other than 4, the common factor can be canceled.

    x−2x−4⋅x+2x+2=1x+2(x≠4)\frac{\sqrt x-2}{x-4}\cdot\frac{\sqrt x+2}{\sqrt x+2}=\frac1{\sqrt x+2}\quad(x\ne4)

  3. The new denominator tends to 4, so the quotient limit law now applies.

    lim⁡x→41x+2=14\lim_{x\to4}\frac1{\sqrt x+2}=\frac14

Conclusion: The limit is one quarter even though the original quotient is undefined at 4.

EXAMPLE 2

Choose the point value that makes the displayed piecewise function continuous.

  1. The formula away from 1 determines nearby behavior independently of k.

    f(x)={x2−1x−1,x≠1,k,x=1.f(x)=\begin{cases}\dfrac{x^2-1}{x-1},&x\ne1,\\k,&x=1.\end{cases}

  2. Factor on the punctured neighborhood. The same linear expression describes both sides.

    (x−1)(x+1)x−1=x+1(x≠1),lim⁡x→1f(x)=2\frac{(x-1)(x+1)}{x-1}=x+1\quad(x\ne1),\qquad \lim_{x\to1}f(x)=2

  3. Continuity requires the assigned center value to match this limit. Changing k cannot alter the limit because it changes only one input.

    f(1)=k=2f(1)=k=2

Conclusion: Exactly one assignment, k equal to 2, repairs the discontinuity.

Pitfalls & counterexamples

Tempting claim: If substitution gives zero over zero, the limit equals zero.

Correction: Zero over zero is an indeterminate form: different nearby formulas can produce different limits or no limit.

Test the claim: The radical quotient above tends to one quarter, while the quotient of the absolute value of x by x has unequal left and right limits at zero.

Check your understanding

If a function's left limit is 3 and right limit is 5, can changing its value at the point make it continuous?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

No.

The two-sided limit fails to exist, and a point assignment cannot change either one-sided approach.

Explore an interactive illustration
A secant becomes a tangent

Parent curve y=x2\displaystyle y = x^{2}. Fix A at x=a\displaystyle x = a. Bring B closer to A to explore an instantaneous rate.

Watch the secant approach the tangent
0%
AB

Red secant: (a+h)2−a2h=2 a+h=3\displaystyle \frac{\left(a + h\right)^{2} - a^{2}}{h} = 2\, a + h = 3. Dashed green tangent: limit as h→0=2 a=2\displaystyle h \to 0 = 2\, a = 2. The limit avoids dividing by zero: simplify for h≠0\displaystyle h \ne 0, then let h approach zero.

Slope gap: 1 · The animation approaches h=0\displaystyle h = 0; it never divides by zero.

Connect this idea

Limits define derivatives and improper integrals. Continuity provides the hypotheses behind existence theorems used in root finding and optimization.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    Find the limit of x2−4x−2\displaystyle \frac{x^{2} - 4}{x - 2} as x approaches 2.

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  2. Problem 2

    For continuity, define f (x)=x2−9x−3\displaystyle f\, \left(x\right) = \frac{x^{2} - 9}{x - 3} for x≠3\displaystyle x \ne 3, and f (3)=b\displaystyle f\, \left(3\right) = b. Find b.

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  3. Problem 3

    The left and right limits at x=1\displaystyle x = 1 equal 2 and 5. Does a two-sided finite limit exist? Enter 1 for yes, 0 for no.

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  4. Problem 4

    Find the limit of 2 x2+13 x2+x\displaystyle \frac{2\, x^{2} + 1}{3\, x^{2} + x} as x tends to positive infinity.

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  5. Problem 5New context

    A model has a removable hole: find the limit of x2−9x−3\displaystyle \frac{x^{2} - 9}{x - 3} as x approaches 3.

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  6. Problem 6New context

    To remove a sensor-model hole, define f (x)=x2−16x−4\displaystyle f\, \left(x\right) = \frac{x^{2} - 16}{x - 4} for x≠4\displaystyle x \ne 4, and f (4)=b\displaystyle f\, \left(4\right) = b. Find b.

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  7. Problem 7New context

    A piecewise response has left and right limits at x=2\displaystyle x = 2 equal 3 and 6. Does a two-sided finite limit exist? Enter 1 for yes, 0 for no.

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  8. Problem 8New context

    For a long-run ratio, find the limit of 3 x2+14 x2+x\displaystyle \frac{3\, x^{2} + 1}{4\, x^{2} + x} as x tends to positive infinity.

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Open stage UC1 1.1 in the student workspace →

STAGE UC1 1.2 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Derivatives, Chain Rule & Implicit Change

Useful preparation: Limits & Continuity

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

A derivative describes a function's local linear behavior. Shrinking an input interval turns average change into instantaneous change. Nested functions transmit change through layers, while implicit equations couple the changes of their variables.

Derivative
The finite limit of the difference quotient at an input. It gives the tangent slope and the instantaneous output change per unit input.
Linear approximation
The tangent-line estimate near a chosen input. Differentiability means the approximation error becomes negligible relative to the input increment.
Implicit differentiation
Differentiating an equation while treating one variable as a local function of another, so every occurrence of that dependent variable carries its own derivative.

Conditions before calculation

  • For an ordinary two-sided derivative, the function must be defined on a neighborhood of the input and the difference-quotient limit must be finite.
  • The chain rule requires the inner function to be differentiable at the input and the outer function to be differentiable at the resulting inner value.
  • For a continuously differentiable implicit equation, a nonzero partial derivative with respect to the dependent variable guarantees a local differentiable branch; zero requires separate analysis.

Why the method works

1. Start from the local rate

The difference quotient compares output change with input change. Its limit gives the tangent slope and local linear coefficient; h denotes the increment.

f′(a)=lim⁡h→0f(a+h)−f(a)h,f(a+h)=f(a)+f′(a)h+o(h)f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}h,\qquad f(a+h)=f(a)+f'(a)h+o(h)

2. Multiply the rates through a composition

An input change first changes g, then that intermediate change changes f. Their local conversion factors multiply.

ddxf(g(x))=f′(g(x))g′(x)\frac{d}{dx}f(g(x))=f'(g(x))g'(x)

3. Differentiate each dependency

Along a solution branch, F remains zero, so its total rate vanishes. Solve for the unknown slope when its coefficient is nonzero.

F(x,y(x))=0 ⟹ Fx+Fyy′=0,y′=−FxFy(Fy≠0)F(x,y(x))=0\ \Longrightarrow\ F_x+F_y y'=0,\qquad y'=-\frac{F_x}{F_y}\quad(F_y\ne0)

Worked examples

EXAMPLE 1

Differentiate the nested polynomial and find its tangent slope at input 1.

  1. Name the inner expression and differentiate the outer cube with respect to that expression.

    y=(1+2x2)3,u=1+2x2,dydu=3u2y=(1+2x^2)^3,\qquad u=1+2x^2,\qquad \frac{dy}{du}=3u^2

  2. Differentiate the inner expression with respect to x and multiply the two rates.

    dudx=4x,y′=3(1+2x2)2(4x)=12x(1+2x2)2\frac{du}{dx}=4x,\qquad y'=3(1+2x^2)^2(4x)=12x(1+2x^2)^2

  3. Evaluate only after keeping both factors. The point value also supplies the tangent-line anchor.

    y(1)=27,y′(1)=108,y−27=108(x−1)y(1)=27,\qquad y'(1)=108,\qquad y-27=108(x-1)

Conclusion: The tangent slope is 108; omitting the inner derivative would give the wrong local rate.

EXAMPLE 2

Find the tangent to the implicit curve at the point with coordinates 1 and 2.

  1. Verify the point lies on the curve. The partial derivative with respect to y is nonzero there, so a local y branch is justified.

    x2+xy+y2=7,1+2+4=7,Fy(1,2)=1+4=5x^2+xy+y^2=7,\qquad 1+2+4=7,\qquad F_y(1,2)=1+4=5

  2. Use the product rule on the mixed term and the chain rule on the squared dependent variable.

    2x+y+xy′+2yy′=02x+y+xy'+2yy'=0

  3. Collect the dependent-variable derivatives before dividing, then evaluate at the stated point.

    y′=−2x+yx+2y,y′(1)=−45,y−2=−45(x−1)y'=-\frac{2x+y}{x+2y},\qquad y'(1)=-\frac45,\qquad y-2=-\frac45(x-1)

Conclusion: The negative slope describes the local branch without solving the entire curve explicitly.

Pitfalls & counterexamples

Tempting claim: Every continuous function has a derivative.

Correction: Continuity controls function values; differentiability additionally requires a consistent finite local slope.

Test the claim: The absolute-value function is continuous at zero, but its left slope is negative one and its right slope is positive one.

Check your understanding

A derivative has units of meters per second. What units can its input and output have?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

Time in seconds and position in meters.

The derivative's units are output units divided by input units, matching the difference quotient.

Explore an interactive illustration
A secant becomes a tangent

Parent curve y=x2\displaystyle y = x^{2}. Fix A at x=a\displaystyle x = a. Bring B closer to A to explore an instantaneous rate.

Watch the secant approach the tangent
0%
AB

Red secant: (a+h)2−a2h=2 a+h=3\displaystyle \frac{\left(a + h\right)^{2} - a^{2}}{h} = 2\, a + h = 3. Dashed green tangent: limit as h→0=2 a=2\displaystyle h \to 0 = 2\, a = 2. The limit avoids dividing by zero: simplify for h≠0\displaystyle h \ne 0, then let h approach zero.

Slope gap: 1 · The animation approaches h=0\displaystyle h = 0; it never divides by zero.

Connect this idea

Derivatives lead to optimization and differential equations. The same dependency bookkeeping becomes the multivariable chain rule and gradient.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    For f (x)=2 x3\displaystyle f\, \left(x\right) = 2\, x^{3}, find f′ (2)\displaystyle f'\, \left(2\right).

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  2. Problem 2

    For f (x)=(2 x+1)2\displaystyle f\, \left(x\right) = \left(2\, x + 1\right)^{2}, find f′ (1)\displaystyle f'\, \left(1\right).

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  3. Problem 3

    On the circle x2+y2=25\displaystyle x^{2} + y^{2} = 25, find dydx\displaystyle \frac{dy}{dx} at (3, 4).

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  4. Problem 4

    A square has side length 3 cm increasing at 2 cm/s. At that instant, find dAdt\displaystyle \frac{dA}{dt} in cm²/s.

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  5. Problem 5New context

    Position in meters is f (x)=3 x3\displaystyle f\, \left(x\right) = 3\, x^{3}, find f′ (2)\displaystyle f'\, \left(2\right).

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  6. Problem 6New context

    A squared response follows f (x)=(3 x+1)2\displaystyle f\, \left(x\right) = \left(3\, x + 1\right)^{2}, find f′ (1)\displaystyle f'\, \left(1\right).

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  7. Problem 7New context

    A path follows the circle x2+y2=100\displaystyle x^{2} + y^{2} = 100, find dydx\displaystyle \frac{dy}{dx} at (6, 8).

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  8. Problem 8New context

    A growing square tile has side length 4 cm increasing at 2 cm/s. At that instant, find dAdt\displaystyle \frac{dA}{dt} in cm²/s.

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Open stage UC1 1.2 in the student workspace →

STAGE UC1 1.3 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Derivative Tests & Optimization

Useful preparation: Derivatives, Chain Rule & Implicit Change

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Optimization combines a model, a feasible domain, and comparison. Derivatives identify candidates. Explain the objective, find all relevant candidates, and verify which attains the best allowed value.

Local and absolute extrema
A local extremum compares nearby feasible inputs; an absolute extremum compares the entire specified domain.
Critical number
An interior domain input where the derivative is zero or does not exist. Such an input is a candidate, not a guaranteed extremum.
Feasible domain
All inputs satisfying the problem's mathematical and contextual restrictions, including any boundary points that are genuinely allowed.

Conditions before calculation

  • A continuous real function on a closed bounded interval attains an absolute maximum and minimum. Open or unbounded domains need separate existence analysis.
  • An interior differentiable local extremum has zero derivative, but the converse fails. Nondifferentiable points and endpoints need attention too.
  • The second-derivative test applies at a stationary point when the second derivative is continuous nearby; a zero second derivative leaves this test inconclusive.

Why the method works

1. Reduce the model without losing feasibility

Use constraints to reduce the objective to one variable, retaining its feasible interval. Physically impossible stationary points cannot solve the problem.

x+y=s,x,y≥0 ⟹ A(x)=x(s−x),0≤x≤sx+y=s,\quad x,y\ge0\ \Longrightarrow\ A(x)=x(s-x),\quad 0\le x\le s

2. Classify local behavior

A positive-to-negative derivative change means rise then fall; the reverse means fall then rise. At a stationary point, nonzero curvature supplies the second-derivative test.

f′(c)=0:f′′(c)>0⇒local minimum,f′′(c)<0⇒local maximumf'(c)=0:\qquad f''(c)>0\Rightarrow\text{local minimum},\quad f''(c)<0\Rightarrow\text{local maximum}

3. Make the global comparison

Evaluate critical numbers and endpoints, comparing function values rather than derivative values. On an open domain, a limiting value may never be attained.

{f(a),f(b)}∪{f(c):c∈(a,b), f′(c)=0 or f′(c) undefined}\{f(a),f(b)\}\cup\{f(c):c\in(a,b),\ f'(c)=0\text{ or }f'(c)\text{ undefined}\}

Worked examples

EXAMPLE 1

A rectangle uses 36 meters of boundary material. Find the largest area.

  1. Let x and y be side lengths. The perimeter constraint leaves one free variable; zero lengths describe degenerate boundary cases used for comparison.

    2x+2y=36,y=18−x,A(x)=18x−x2,0≤x≤182x+2y=36,\qquad y=18-x,\qquad A(x)=18x-x^2,\quad 0\le x\le18

  2. Set the derivative to zero. Its sign is positive before 9 and negative after 9, so area increases then decreases.

    A′(x)=18−2x,A′(x)=0  ⟺  x=9A'(x)=18-2x,\qquad A'(x)=0\iff x=9

  3. Compare the candidate with both endpoints, and recover the other dimension.

    A(0)=A(18)=0,A(9)=81,y=9A(0)=A(18)=0,\qquad A(9)=81,\qquad y=9

Conclusion: A square with side length 9 meters attains the maximum area of 81 square meters.

EXAMPLE 2

Find all locations of the absolute extrema of the cubic on the displayed closed interval.

  1. Continuity guarantees extrema, so gather interior stationary points and endpoints.

    f(x)=x3−3x,−2≤x≤3;f′(x)=3(x2−1)=0  ⟺  x=±1f(x)=x^3-3x,\quad -2\le x\le3;\qquad f'(x)=3(x^2-1)=0\iff x=\pm1

  2. Evaluate every candidate. This step captures both boundary effects and ties.

    f(−2)=−2,f(−1)=2,f(1)=−2,f(3)=18f(-2)=-2,\qquad f(-1)=2,\qquad f(1)=-2,\qquad f(3)=18

  3. Compare the four values. The interior local maximum is smaller than the right endpoint value; the minimum occurs at two distinct inputs.

    max⁡f=18 at x=3,min⁡f=−2 at x=−2,1\max f=18\text{ at }x=3,\qquad \min f=-2\text{ at }x=-2,1

Conclusion: A local maximum need not be the absolute maximum, and an absolute extremum need not have a unique location.

Pitfalls & counterexamples

Tempting claim: Solving for zero derivative completes an optimization problem.

Correction: Stationarity only produces candidates; domain restrictions, nondifferentiable points, endpoints, and attainment must still be checked.

Test the claim: The cubing function has zero derivative at zero but increases through that point, so there is no local extremum there.

Check your understanding

Does the function equal to its input have an absolute maximum on the open interval from zero to one?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

No.

Its values approach one but never reach it; any allowed input can be increased while remaining inside the interval.

Connect this idea

The same candidate-and-comparison logic extends to constrained multivariable optimization. Existence theorems explain why checking finitely many candidates can certify a global answer.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    For f (x)=(x−2)2+5\displaystyle f\, \left(x\right) = \left(x - 2\right)^{2} + 5 on the real line. At what x is it minimized?

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  2. Problem 2

    A rectangle has perimeter 12 meters. What width maximizes its area?

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  3. Problem 3

    For f (x)=x3−3 x\displaystyle f\, \left(x\right) = x^{3} - 3\, x, find its positive critical input.

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  4. Problem 4

    Find the maximum value of f (x)=x2\displaystyle f\, \left(x\right) = x^{2} on the closed interval [−4, 2].

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  5. Problem 5New context

    A cost function is f (x)=(x−3)2+5\displaystyle f\, \left(x\right) = \left(x - 3\right)^{2} + 5 on the real line. At what x is it minimized?

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  6. Problem 6New context

    A rectangular garden has perimeter 16 meters. What width maximizes its area?

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  7. Problem 7New context

    A performance measure is f (x)=x3−12 x\displaystyle f\, \left(x\right) = x^{3} - 12\, x, find its positive critical input.

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  8. Problem 8New context

    A bounded process needs the maximum value of f (x)=x2\displaystyle f\, \left(x\right) = x^{2} on the closed interval [−5, 3].

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Open stage UC1 1.3 in the student workspace →

STAGE UC1 1.4 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Integrals & the Fundamental Theorem

Useful preparation: Derivative Tests & Optimization

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Integration adds small signed contributions. Rate times interval approximates change; refining partitions produces an integral. The Fundamental Theorem connects accumulation to differentiation and explains antiderivative evaluation.

Definite integral
The limit of weighted sums over an interval when that limit exists independently of sample choices. Its units are integrand units multiplied by input units.
Antiderivative
A function whose derivative equals the integrand on an interval. Any two antiderivatives on the same interval differ by a constant.
Accumulation function
An integral whose upper or lower endpoint varies. It records how the accumulated signed quantity depends on that endpoint.

Conditions before calculation

  • Continuity on a closed interval is a sufficient condition for integrability and for the standard Fundamental Theorem statements used here.
  • The derivative of an accumulation function equals the integrand at interior points where the integrand is continuous; moving endpoints also require the chain rule.
  • A signed integral measures net change. Total geometric area or total distance requires integrating an appropriate nonnegative magnitude.

Why the method works

1. Build accumulation from small pieces

Each rectangle approximates a contribution using a subinterval sample. The largest width tends to zero; negative heights contribute negative quantities.

∫abf(t) dt=lim⁡max⁡Δti→0∑if(ti∗)Δti\int_a^b f(t)\,dt=\lim_{\max\Delta t_i\to0}\sum_i f(t_i^*)\Delta t_i

2. Recover the local rate

Moving the accumulation endpoint adds a thin strip. Continuity makes its average height approach the endpoint height.

A(x)=∫axf(t) dt,A′(x)=f(x)A(x)=\int_a^x f(t)\,dt,\qquad A'(x)=f(x)

3. Evaluate through an antiderivative

An antiderivative and accumulation have the same derivative, so they differ by a constant. Endpoint subtraction eliminates it; a definite value needs no arbitrary constant.

F′=f⟹∫abf(t) dt=F(b)−F(a)F'=f\quad\Longrightarrow\quad\int_a^b f(t)\,dt=F(b)-F(a)

Worked examples

EXAMPLE 1

A particle has the displayed velocity in meters per second. Find its displacement and total distance over the first four seconds.

  1. Locate the change of direction. Velocity is negative before the zero and positive afterward.

    v(t)=3t−6,0≤t≤4;v(t)=0  ⟺  t=2v(t)=3t-6,\quad 0\le t\le4;\qquad v(t)=0\iff t=2

  2. Integrate velocity for signed displacement. The two directional contributions cancel.

    ∫04(3t−6) dt=[32t2−6t]04=0\int_0^4(3t-6)\,dt=\left[\frac32t^2-6t\right]_0^4=0

  3. Distance adds magnitudes, so reverse the sign on the interval where velocity is negative and split at the zero.

    ∫04∣v(t)∣ dt=−∫02(3t−6) dt+∫24(3t−6) dt=6+6=12\int_0^4|v(t)|\,dt=-\int_0^2(3t-6)\,dt+\int_2^4(3t-6)\,dt=6+6=12

Conclusion: Displacement is zero meters, while total distance is 12 meters; returning to the start does not mean no motion occurred.

EXAMPLE 2

Differentiate the integral with two moving endpoints without finding an antiderivative.

  1. The integrand is continuous for every real input, so accumulation is defined even when the endpoints reverse order.

    B(x)=∫xx2e−t2 dtB(x)=\int_x^{x^2}e^{-t^2}\,dt

  2. Write the interval as a difference of two accumulations from the same fixed base. This exposes both endpoint dependencies.

    B(x)=∫0x2e−t2 dt−∫0xe−t2 dtB(x)=\int_0^{x^2}e^{-t^2}\,dt-\int_0^x e^{-t^2}\,dt

  3. Apply the Fundamental Theorem and multiply by each endpoint's derivative. The lower endpoint contributes with a minus sign.

    B′(x)=2xe−x4−e−x2,B(1)=0,B′(1)=e−1B'(x)=2x e^{-x^4}-e^{-x^2},\qquad B(1)=0,\quad B'(1)=e^{-1}

Conclusion: An integral can be differentiated exactly even when an elementary antiderivative is unavailable.

Pitfalls & counterexamples

Tempting claim: A zero definite integral means the integrand is zero everywhere.

Correction: Positive and negative contributions may cancel. The integral only reports their signed total.

Test the claim: The particle's velocity integral is zero even though its velocity is nonzero except at the turning time.

Check your understanding

If a continuous inflow rate is positive throughout an interval, what happens to its accumulation function?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

It is strictly increasing.

Every extension of the endpoint adds a positive contribution, and the accumulation derivative equals the positive rate.

Connect this idea

The theorem powers substitution and integration by parts. The distinction between net change and total magnitude reappears in line integrals, flux, and probability densities.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    Evaluate ∫\displaystyle \int from 0 to 2 of 2x dx.

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  2. Problem 2

    Let F (x)=∫\displaystyle F\, \left(x\right) = \int from 0 to x of (3 t2+2) dt\displaystyle \left(3\, t^{2} + 2\right)\, dt. Find F′ (2)\displaystyle F'\, \left(2\right).

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  3. Problem 3

    Find ∫\displaystyle \int from 2 to 0 of 3 dx.

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  4. Problem 4

    For G (x)=∫\displaystyle G\, \left(x\right) = \int from 0 to x2\displaystyle x^{2} of 2t dt. Find G′ (2)\displaystyle G'\, \left(2\right).

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  5. Problem 5New context

    A velocity accumulation requires ∫\displaystyle \int from 0 to 3 of 2x dx.

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  6. Problem 6New context

    A cumulative flow is F (x)=∫\displaystyle F\, \left(x\right) = \int from 0 to x of (3 t2+3) dt\displaystyle \left(3\, t^{2} + 3\right)\, dt. Find F′ (2)\displaystyle F'\, \left(2\right).

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  7. Problem 7New context

    A net-flow calculation asks for ∫\displaystyle \int from 3 to 0 of 3 dx.

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  8. Problem 8New context

    An accumulated signal is G (x)=∫\displaystyle G\, \left(x\right) = \int from 0 to x2\displaystyle x^{2} of 3t dt. Find G′ (2)\displaystyle G'\, \left(2\right).

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