Foundations & Precalculus Review
Calculus is the study of how functions change and accumulate, so every later idea leans on fluency with functions themselves. This chapter gathers the algebra, function families, and notation the rest of the course quietly assumes.
A function \( f \) assigns to each input \( x \) in its domain exactly one output \( f(x) \). The set of resulting outputs is the range. We write \( f:\mathbb{R}\to\mathbb{R} \) for a real-valued function of a real variable.
The function toolkit
A handful of parent functions, combined and transformed, generate almost everything in single-variable calculus.
From a parent \( f \), the graph of \( y=a\,f(x-h)+k \) is shifted right by \( h \), up by \( k \), and scaled vertically by \( a \) (reflected across the \( x \)-axis when \( a\lt 0 \)).
| x | x² | g(x) |
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The faint grey curve is the parent \(y=x^2\); the bold curve is \(y=a(x-h)^2\). Watch the vertex slide and flip.
Exponentials and logarithms
These are inverse operations, and their algebra appears constantly in growth models and integration. The key laws: The natural base \( e\approx 2.71828 \) makes \( \dfrac{d}{dx}e^{x}=e^{x} \), which is why it dominates calculus.
Find the domain of \( \displaystyle f(x)=\frac{\sqrt{x-2}}{x-5} \). The square root needs \( x-2\ge 0 \), so \( x\ge 2 \); the denominator forbids \( x=5 \). Combining,
Describe \( y=-2(x-3)^{2}+1 \) relative to \( y=x^{2} \). It is shifted right \( 3 \) and up \( 1 \), stretched vertically by \( 2 \), and reflected (since \( a=-2\lt 0 \)). The vertex is \( (3,1) \) and the parabola opens downward.
Limits & Continuity
Calculus rests on a single idea: what value does a function approach as its input creeps toward a target? It need not ever reach that value — what matters is the trend.
We write \( \displaystyle \lim_{x \to c} f(x) = L \) if the outputs \( f(x) \) can be made as close to \( L \) as we like by taking \( x \) sufficiently close to \( c \) (but not equal to \( c \)).
The function may even be undefined exactly at \( c \). Consider \( \displaystyle f(x)=\frac{x^2-1}{x-1} \). At \( x=1 \) the formula gives \( \tfrac{0}{0} \), yet factoring reveals \( f(x)=x+1 \) everywhere else, so the curve heads straight for \( 2 \).
| x | f(x) |
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Slide from either side: \(f(x)\) marches to \(2\), even though the point at \(x=1\) is a hole.
One-sided limits
Sometimes the approach differs by direction. The left-hand limit \( \displaystyle \lim_{x\to c^-} f(x) \) and right-hand limit \( \displaystyle \lim_{x\to c^+} f(x) \) must agree for the two-sided limit to exist:
A function \( f \) is continuous at \( c \) when all three hold: (i) \( f(c) \) is defined, (ii) \( \displaystyle\lim_{x\to c} f(x) \) exists, and (iii) the two are equal, \( \displaystyle\lim_{x\to c} f(x)=f(c) \). Informally: you can draw it through \( c \) without lifting the pen.
Two limits worth memorising
Evaluate \( \displaystyle \lim_{x\to 3}\frac{x^2-9}{x-3} \). Direct substitution gives \( \tfrac00 \), so factor the numerator: \( x^2-9=(x-3)(x+3) \). Cancel the common factor (valid since \( x\neq 3 \) in the limit):
Derivatives
The derivative is the first question of calculus made precise: the instantaneous rate of change of \( f \), equivalently the slope of the tangent line to its graph.
The derivative of \( f \) at \( x \) is the limit of slopes of secant lines: when this limit exists. It is also written \( \dfrac{dy}{dx} \) or \( \dfrac{d}{dx}f(x) \).
As \( h\to 0 \), the secant through \( \big(x,f(x)\big) \) and \( \big(x+h,f(x+h)\big) \) pivots into the tangent. Watch it happen — drag the point of tangency and the slope readout below updates live.
| x | f(x) | tangent y |
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The rules of differentiation
We rarely use the limit definition directly. These rules build every elementary derivative:
\( \dfrac{d}{dx}\sin x=\cos x \), \(\quad \dfrac{d}{dx}\cos x=-\sin x \), \(\quad \dfrac{d}{dx}\tan x=\sec^2 x \), \( \dfrac{d}{dx}e^{x}=e^{x} \), \(\quad \dfrac{d}{dx}\ln x=\dfrac1x \quad (x>0). \)
Differentiate \( y=\sin\!\big(3x^2\big) \). Outer function \( \sin u \) with \( u=3x^2 \). Then \( \dfrac{dy}{du}=\cos u \) and \( \dfrac{du}{dx}=6x \). By the chain rule,
Differentiate \( y=x^2 e^{x} \). With \( f=x^2 \) (so \( f'=2x \)) and \( g=e^x \) (so \( g'=e^x \)):
Integrals
The integral is the second question of calculus: how much accumulates. Geometrically, the definite integral measures the signed area between a curve and the horizontal axis.
Partition \( [a,b] \) into \( n \) strips of width \( \Delta x=\frac{b-a}{n} \). Sampling a height in each strip and summing the rectangle areas gives a Riemann sum; the integral is its limit:
The explorer below builds that limit by hand. Increase the number of rectangles and watch the approximation tighten onto the true area — the error column shrinks toward zero.
| i | [x, x+Δx] | sample x* | f(x*) | area |
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Antiderivatives & the indefinite integral
Reversing differentiation, \( F \) is an antiderivative of \( f \) if \( F'=f \). Since constants vanish under differentiation, antiderivatives carry a \( +C \).
Find the area under \( f(x)=x^2 \) from \( 0 \) to \( 2 \). An antiderivative is \( F(x)=\tfrac{x^3}{3} \). Evaluating (using the Fundamental Theorem, next chapter):
The Fundamental Theorem
Here the two halves meet. The slope problem (indigo) and the area problem (amber) turn out to be inverses of one another — arguably the central result of all calculus.
Part I. If \( \displaystyle A(x)=\int_a^x f(t)\,dt \), then \( A \) is differentiable and \( A'(x)=f(x) \). Differentiating an accumulated area recovers the original function. Part II. If \( F \) is any antiderivative of \( f \), then \( \displaystyle\int_a^b f(x)\,dx=F(b)-F(a). \) This is what makes integration computable.
The twin explorer makes Part I visible. The amber area on the upper graph is exactly the indigo height of the accumulation graph below. Drag \( x \): the rate at which the lower curve rises equals the upper curve's height at that instant — that is, \( A'(x)=f(x) \).
| t | f(t) | A(t) |
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Evaluate \( \displaystyle\int_0^{\pi} \sin x\,dx \). An antiderivative of \( \sin x \) is \( -\cos x \). By Part II,
Part I: If \( F(x)=\int_a^x f(t)\,dt \), then \( F'(x)=f(x) \).
Part II: \( \int_a^b f(x)\,dx = F(b)-F(a) \) where \( F'=f \).
Evaluate \( \int_1^4 2x\,dx \).
Find \( \dfrac{d}{dx}\int_0^{x^2}\cos t\,dt \).
Applications of Derivatives
The sign of the derivative tells us where a function rises or falls; the sign of the second derivative tells us how it bends. Together they locate maxima, minima, and inflection points — the heart of optimization.
A critical point of \( f \) is a value \( c \) where \( f'(c)=0 \) or \( f'(c) \) does not exist. Local maxima and minima can occur only at critical points.
On an interval: if \( f'>0 \) then \( f \) is increasing; if \( f'<0 \) then \( f \) is decreasing. If \( f''>0 \) the graph is concave up; if \( f''<0 \) it is concave down. An inflection point is where concavity changes.
Second derivative test. At a critical point \( c \): if \( f''(c)>0 \) then \( c \) is a local minimum; if \( f''(c)<0 \) then \( c \) is a local maximum.
Optimization puts this to work: model the quantity, differentiate, set the derivative to zero, and confirm the extremum. The explorer animates the classic box problem — the volume rises to a single peak exactly where \( V'(x)=0 \).
| x | V(x) |
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Cut squares of side \(x\) from a \(10\times 8\) sheet and fold up: \(V(x)=x(10-2x)(8-2x)\). The amber marker is the maximum.
Find and classify the extrema of \( f(x)=x^3-3x \). Critical points: \( f'(x)=3x^2-3=3(x-1)(x+1)=0 \Rightarrow x=\pm 1 \). Use \( f''(x)=6x \): So a local minimum at \( (1,-2) \) and a local maximum at \( (-1,2) \).
A rectangular pen is built against a wall using \( 100 \) m of fence on the other three sides. Maximise the area. Let the two perpendicular sides be \( x \); the parallel side is \( 100-2x \). Then \( A(x)=x(100-2x)=100x-2x^2 \), so \( A'(x)=100-4x=0 \Rightarrow x=25 \).
Related rates
When two quantities are tied together by an equation and both vary with time, differentiate the relation with respect to \( t \) (chain rule) to connect their rates.
A spherical balloon is inflated so its radius grows at \( \dfrac{dr}{dt}=2 \) cm/s. How fast is the volume increasing when \( r=5 \) cm? From \( V=\tfrac{4}{3}\pi r^3 \), differentiate with respect to \( t \): \( \dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt} \). At \( r=5 \):
L'Hôpital's Rule
If \( \displaystyle\lim_{x\to a}\dfrac{f(x)}{g(x)} \) has indeterminate form \( \tfrac00 \) or \( \tfrac{\infty}{\infty} \), then provided the right-hand limit exists.
Evaluate \( \displaystyle\lim_{x\to\infty}\frac{\ln x}{x} \) (form \( \tfrac{\infty}{\infty} \)). Differentiate top and bottom: \( \dfrac{d}{dx}\ln x=\tfrac1x \), \( \dfrac{d}{dx}x=1 \):
Derivatives measure rates of change. The first derivative test locates local extrema; the second derivative test distinguishes minima from maxima and reveals concavity.
A farmer has 200 m of fencing to enclose a rectangle against a wall. Find dimensions maximizing area. Let width \( = x \), length \( = 200-2x \). Area \( A=x(200-2x)=200x-2x^2 \).
A ladder 10 m long slides down a wall. When the base is 6 m from the wall and moving out at 2 m/s, how fast is the top falling? At \( x=6 \), \( y=8 \): \( 12(2)+16\dfrac{dy}{dt}=0 \Rightarrow \dfrac{dy}{dt}=-1.5 \) m/s.
Techniques of Integration
Most integrals are not in a table. Two techniques reverse the differentiation rules you already know and unlock the majority of exam problems.
If \( u=g(x) \) then \( du=g'(x)\,dx \), and Choose \( u \) so that its derivative also appears (up to a constant) in the integrand.
From the product rule, \( \displaystyle\int u\,dv = uv-\int v\,du. \) Pick \( u \) by LIATE (logarithms, inverse-trig, algebraic, trig, exponential — earlier letters make better \( u \)).
Evaluate \( \displaystyle\int 2x\cos\!\big(x^2\big)\,dx \). Let \( u=x^2 \), so \( du=2x\,dx \). The integral becomes \( \int \cos u\,du \):
Evaluate \( \displaystyle\int x\,e^{x}\,dx \). Let \( u=x,\ dv=e^{x}dx \), so \( du=dx,\ v=e^{x} \). Then \( uv-\int v\,du \):
Beyond basic antiderivatives: substitution reverses the chain rule; integration by parts reverses the product rule; partial fractions decompose rational integrands.
Evaluate \( \int 2x e^{x^2}\,dx \). Let \( u=x^2 \), \( du=2x\,dx \): \( \int e^u\,du = e^{x^2}+C \).
Evaluate \( \int x\cos x\,dx \). Let \( u=x \), \( dv=\cos x\,dx \). Then \( du=dx \), \( v=\sin x \):
Applications of Integration
Definite integrals measure accumulated quantities: the area trapped between two curves, the volume swept out by rotating a region, or the average height of a function.
Area between curves: \( \displaystyle A=\int_a^b \big[\,f(x)-g(x)\,\big]\,dx \) where \( f \) is the upper curve. Volume of revolution (disks): \( \displaystyle V=\pi\int_a^b \big[R(x)\big]^2\,dx. \) Average value: \( \displaystyle \bar f=\frac{1}{b-a}\int_a^b f(x)\,dx. \)
Drag the line in the explorer: the amber region is the area enclosed between the parabola and the line, recomputed live between their intersection points.
| x | f(x) = 4 − x² | g(x) |
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Find the area between \( y=x \) and \( y=x^2 \) on \( [0,1] \). Here \( x\geq x^2 \) on \( [0,1] \), so integrate the difference:
The region under \( y=\sqrt{x} \) on \( [0,4] \) is revolved about the \( x \)-axis. Find the volume. Disk method with \( R(x)=\sqrt{x} \), so \( R^2=x \):
Definite integrals compute areas between curves, volumes of solids of revolution, arc lengths, and work done by variable forces.
Find the area between \( y=x^2 \) and \( y=x \) from \( x=0 \) to \( x=1 \).
Find the volume when \( y=\sqrt{x} \) on \( [0,4] \) is revolved about the \( x \)-axis.
Differential Equations
A differential equation relates a function to its own rate of change. Solving one means finding every function whose derivative behaves as prescribed — the language in which growth, cooling, and motion are written.
An equation involving an unknown function and its derivatives, such as \( \dfrac{dy}{dx}=f(x,y) \). A solution is a function \( y(x) \) that satisfies it; an initial condition picks out one particular solution.
Separable equations. If \( \dfrac{dy}{dx}=g(x)\,h(y) \), gather each variable on its own side and integrate:
The equation \( \dfrac{dy}{dt}=ky \) has solution \( y=y_0\,e^{kt} \): exponential growth when \( k>0 \) and decay when \( k<0 \).
A slope field draws the prescribed slope at a grid of points; solution curves simply flow along it. Pick an equation and slide the initial value — the amber curve threads through the field from \( (0,\,y_0) \).
| x | y |
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Solve \( \dfrac{dy}{dx}=2xy \). Separate and integrate: \( \displaystyle\int\frac{dy}{y}=\int 2x\,dx \), so \( \ln|y|=x^2+C \). Exponentiating,
Solve \( \dfrac{dy}{dx}=3y \) with \( y(0)=4 \). By the exponential model \( y=y_0 e^{kt} \) with \( k=3 \), and \( y(0)=4 \):
A differential equation relates a function to its derivatives. Separable ODEs are solved by separating variables and integrating both sides.
Solve \( \dfrac{dy}{dx}=2xy \), \( y(0)=3 \).
Parametric & Polar
Not every curve is a function \( y=f(x) \). Parametric equations trace a path through time, and polar coordinates describe position by distance and angle — both unlock curves that loop, spiral, and cross themselves.
For a curve \( \big(x(t),\,y(t)\big) \), the slope and arc length are
With \( x=r\cos\theta,\ y=r\sin\theta \), the area swept by a polar curve is
Drag the trace slider to watch each curve drawn by its parameter — the indigo portion is traced so far, the amber dot is the current point.
| t | x | y |
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For \( x=t^3,\ y=t^2 \), find \( \dfrac{dy}{dx} \). Differentiate each: \( \dfrac{dx}{dt}=3t^2 \), \( \dfrac{dy}{dt}=2t \). Then
Find the area enclosed by the cardioid \( r=1+\cos\theta \). Apply \( A=\tfrac12\int_0^{2\pi} r^2\,d\theta \) with \( r^2=1+2\cos\theta+\cos^2\theta \):
Parametric curves express \( x \) and \( y \) in terms of a parameter \( t \). Polar coordinates use \( (r,\theta) \). Both require adapted formulas for slopes, arc lengths, and areas.
Find \( \dfrac{dy}{dx} \) for \( x=t^2 \), \( y=t^3 \).
Area enclosed by \( r=2\cos\theta \), a circle of radius 1:
Sequences & Series
A series adds infinitely many terms. Remarkably, many familiar functions can be written as “infinite polynomials” — and a short stretch of that polynomial already approximates the function beautifully near a point.
An infinite series \( \sum_{n=1}^{\infty} a_n \) converges if its partial sums \( S_N=\sum_{n=1}^{N} a_n \) approach a finite limit as \( N\to\infty \); otherwise it diverges.
For \( |r|<1 \), the geometric series converges: If \( |r|\geq 1 \) it diverges. The \( p \)-series \( \sum 1/n^{p} \) converges exactly when \( p>1 \).
Near \( x=0 \), a smooth function equals its Maclaurin series \( \displaystyle f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n} \). Three to know:
The explorer builds the Maclaurin polynomial for \( \sin x \) one term at a time. With a single term it is just the line \( y=x \); add more and the amber curve wraps itself around \( \sin x \) across an ever-wider interval.
| x | sin(x) | Taylor(x) |
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\(N\) nonzero terms match \(\sin x\) through degree \(2N-1\). Far from \(0\) the polynomial eventually peels away.
Evaluate \( \displaystyle\sum_{n=0}^{\infty}\Big(\tfrac{1}{2}\Big)^{n} \). Here \( a=1,\ r=\tfrac12 \), and \( |r|<1 \), so the series converges:
Approximate \( e^{0.1} \) with a third-degree Taylor polynomial. Using \( e^{x}\approx 1+x+\tfrac{x^2}{2}+\tfrac{x^3}{6} \) at \( x=0.1 \):
Convergence Tests & Power Series
A series \( \sum a_n \) converges when its partial sums \( S_N=\sum_{n=1}^{N}a_n \) approach a finite limit. Since we rarely sum infinitely many terms by hand, a battery of tests decides convergence from the shape of \( a_n \) alone.
If \( \displaystyle\lim_{n\to\infty}a_n\neq 0 \), then \( \sum a_n \) diverges. Caution: \( a_n\to 0 \) does not guarantee convergence — the harmonic series \( \sum \tfrac1n \) diverges even though its terms vanish.
The p-series \( \displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{p}} \) converges if and only if \( p\gt 1 \). For positive terms, if \( 0\le a_n\le b_n \) and \( \sum b_n \) converges, so does \( \sum a_n \) (and the contrapositive for divergence).
| n | S(n) |
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The geometric series \(\sum (\tfrac12)^n\) sums to \(1\). Each new term halves the remaining gap — visible convergence.
The ratio and root tests
For series with factorials or exponentials, compare a term to the one before it. Let \( \displaystyle L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| \). Then the series converges absolutely if \( L\lt 1 \), diverges if \( L\gt 1 \), and the test is inconclusive if \( L=1 \).
A power series \( \displaystyle\sum_{n=0}^{\infty}c_n(x-a)^{n} \) converges for all \( x \) with \( |x-a|\lt R \) and diverges for \( |x-a|\gt R \), where \( R \) is the radius of convergence. The endpoints \( x=a\pm R \) must be tested separately.
Test \( \displaystyle\sum_{n=1}^{\infty}\frac{n}{2^{n}} \) for convergence. Form the ratio of successive terms and take the limit: so the series converges absolutely.
Find the radius of convergence of \( \displaystyle\sum_{n=0}^{\infty}\frac{x^{n}}{n!} \). The ratio of terms is \( \left|\dfrac{x^{n+1}/(n+1)!}{x^{n}/n!}\right|=\dfrac{|x|}{n+1}\to 0 \) for every \( x \). Since \( L=0\lt 1 \) always, \( R=\infty \) — this series (which is \( e^{x} \)) converges everywhere.
Vectors & Vector-Valued Functions
Multivariable calculus begins with vectors — quantities carrying both magnitude and direction. Letting a vector depend on a parameter traces a curve through space, and differentiating it recovers velocity and acceleration.
A vector \( \mathbf{v}=\langle v_1,v_2,v_3\rangle \) is an ordered triple with length \( \displaystyle|\mathbf{v}|=\sqrt{v_1^{2}+v_2^{2}+v_3^{2}} \). Vectors add componentwise, and a unit vector is \( \hat{\mathbf{v}}=\mathbf{v}/|\mathbf{v}| \).
The dot product \( \mathbf{u}\cdot\mathbf{v}=|\mathbf{u}|\,|\mathbf{v}|\cos\theta \) measures alignment (zero when perpendicular). The cross product \( \mathbf{u}\times\mathbf{v} \) has length \( |\mathbf{u}|\,|\mathbf{v}|\sin\theta \) and points perpendicular to both.
| vector | ⟨x, y⟩ | magnitude |
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Blue is \(\mathbf u=\langle3,1\rangle\), amber is \(\mathbf v\), dark is the resultant \(\mathbf u+\mathbf v\) — the parallelogram diagonal.
Vector-valued functions
A path in space is \( \mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle \). Differentiating component-by-component gives the velocity \( \mathbf{r}'(t) \), whose length \( |\mathbf{r}'(t)| \) is the speed; differentiating again gives acceleration.
Find the angle between \( \mathbf{u}=\langle 1,2,2\rangle \) and \( \mathbf{v}=\langle 2,0,1\rangle \). Here \( \mathbf{u}\cdot\mathbf{v}=2+0+2=4 \), \( |\mathbf{u}|=3 \), \( |\mathbf{v}|=\sqrt5 \), so
For the helix \( \mathbf{r}(t)=\langle\cos t,\sin t,t\rangle \), find the velocity and speed. Differentiating each component, \( \mathbf{r}'(t)=\langle-\sin t,\cos t,1\rangle \). Its length is constant:
Partial Derivatives & the Gradient
For a surface \( z=f(x,y) \), there is no single slope — it depends on the direction you walk. Partial derivatives measure the slope along each axis, and the gradient bundles them into the direction of steepest ascent.
The partial derivative \( \displaystyle f_x=\frac{\partial f}{\partial x} \) differentiates \( f \) with respect to \( x \) while holding \( y \) constant; \( f_y \) does the reverse. Geometrically, \( f_x \) is the slope of the curve where the surface is sliced by a plane of fixed \( y \).
The gradient \( \nabla f=\langle f_x,f_y\rangle \) points in the direction of greatest increase, with rate \( |\nabla f| \). The slope in a unit direction \( \mathbf{u} \) is the directional derivative \( D_{\mathbf{u}}f=\nabla f\cdot\mathbf{u} \).
| x | g(x) | tangent y |
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Fixing \(y=c\) gives the curve \(g(x)=f(x,c)\). Its slope at \(x=1\) is the partial \(f_x(1,c)\) — and it shifts as \(c\) moves.
Tangent planes and local extrema
Near a point, a smooth surface looks flat. The tangent plane at \( (a,b) \) is Local maxima and minima occur where \( \nabla f=\mathbf{0} \); the second-derivative test uses \( D=f_{xx}f_{yy}-f_{xy}^{2} \) to classify them (\( D\gt 0 \) and \( f_{xx}\gt 0 \) gives a minimum).
For \( f(x,y)=x^{2}y+\sin(xy) \), find \( f_x \) and \( f_y \). Treating \( y \) as constant for \( f_x \), then \( x \) as constant for \( f_y \):
For \( f(x,y)=x^{2}+y^{2} \), find the gradient at \( (1,2) \) and the direction of fastest increase. Since \( \nabla f=\langle 2x,2y\rangle \), at \( (1,2) \) we get \( \nabla f=\langle 2,4\rangle \). Fastest increase is along this vector, i.e. the unit direction
Multiple Integrals
A single integral accumulates over an interval; a double integral accumulates a function over a two-dimensional region — most concretely, the volume trapped under a surface \( z=f(x,y) \).
Partition a region \( R \) into small cells of area \( \Delta A \), sample \( f \) in each, and sum. The double integral is the limit as the cells shrink:
Over a rectangle \( R=[a,b]\times[c,d] \), a double integral equals an iterated integral, and the order may be swapped:
| quantity | value |
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Each cell is shaded by \(f=x+y\) at its midpoint over \([0,2]^2\). Refine the grid and the midpoint sum converges to \(8\).
Double integrals in polar coordinates
For circular regions, polar coordinates simplify everything — but the area element gains a factor of \( r \): That extra \( r \) is the Jacobian of the polar change of variables, and forgetting it is the most common error in the topic.
Evaluate \( \displaystyle\int_0^1\!\!\int_0^2 (x+2y)\,dx\,dy \). Integrate in \( x \) first, treating \( y \) as constant:
Confirm the area of the unit disk \( x^{2}+y^{2}\le 1 \) using a polar double integral.
Vector Calculus
The capstone of Calc III integrates vector fields — assignments of a vector to every point — along curves and across surfaces. Three great theorems (Green, Stokes, Divergence) tie these integrals back to the derivatives \( \nabla\cdot\mathbf{F} \) and \( \nabla\times\mathbf{F} \).
The line integral \( \displaystyle\int_C \mathbf{F}\cdot d\mathbf{r} \) measures the work a field \( \mathbf{F} \) does along a curve \( C \). A field is conservative when \( \mathbf{F}=\nabla f \) for some \( f \); then the integral is path-independent and equals \( f(\text{end})-f(\text{start}) \).
For a positively oriented simple closed curve \( C \) bounding a region \( R \) in the plane,
| (x, y) | F(x, y) |
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At \(t=1\) the field is \(\langle -y,x\rangle\) (pure rotation, curl \(2\)); at \(t=0\) it is \(\langle x,y\rangle\) (pure source, divergence \(2\)).
Stokes' and the Divergence theorems
Green's theorem generalizes in two directions. Stokes' theorem relates circulation around a boundary curve to the curl over a surface; the Divergence theorem relates outward flux through a closed surface to the divergence inside:
Evaluate \( \displaystyle\oint_C (-y\,dx+x\,dy) \) around the unit circle, counterclockwise. Here \( P=-y,\ Q=x \), so \( Q_x-P_y=1-(-1)=2 \). By Green's theorem,
Find the outward flux of \( \mathbf{F}=\langle x,y,z\rangle \) through the unit sphere. Since \( \nabla\cdot\mathbf{F}=1+1+1=3 \), the Divergence theorem turns the surface flux into a volume integral over the unit ball \( E \):
Practice set
Fifty mixed problems spanning all ten chapters, ordered roughly from limits to series and tagged by difficulty. Click any problem to reveal a complete worked solution.
Evaluate \( \displaystyle\lim_{x\to 2}\frac{x^2-4}{x-2} \).
BasicFactor: \( x^2-4=(x-2)(x+2) \). Cancel \( (x-2) \):
\[ \lim_{x\to 2}(x+2)=4. \]Differentiate \( f(x)=5x^4-3x^2+7 \).
BasicApply the power and sum rules term by term:
\[ f'(x)=20x^3-6x. \]Differentiate \( y=\dfrac{x}{x^2+1} \).
IntermediateQuotient rule with \( f=x,\ g=x^2+1 \), so \( f'=1,\ g'=2x \):
\[ y'=\frac{(1)(x^2+1)-(x)(2x)}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}. \]Differentiate \( y=e^{2x}\cos x \).
IntermediateProduct rule; the first factor needs the chain rule, \( \tfrac{d}{dx}e^{2x}=2e^{2x} \):
\[ y'=2e^{2x}\cos x - e^{2x}\sin x = e^{2x}\big(2\cos x-\sin x\big). \]Evaluate \( \displaystyle\int_1^{3}\big(2x+1\big)\,dx \).
BasicAntiderivative \( x^2+x \). By the Fundamental Theorem:
\[ \big[x^2+x\big]_1^{3}=(9+3)-(1+1)=12-2=10. \]Find the area under \( f(x)=\sin x \) over \( \big[0,\tfrac{\pi}{2}\big] \).
IntermediateUse \( \int \sin x\,dx=-\cos x \):
\[ \int_0^{\pi/2}\sin x\,dx=\big[-\cos x\big]_0^{\pi/2}=0-(-1)=1. \]Differentiate \( y=\big(3x^2+1\big)^{5} \).
AdvancedChain rule: outer power \( u^5 \) gives \( 5u^4 \); inner derivative \( \tfrac{d}{dx}(3x^2+1)=6x \):
\[ y'=5\big(3x^2+1\big)^{4}\cdot 6x=30x\big(3x^2+1\big)^{4}. \]A particle's position is \( s(t)=t^3-6t^2+9t \). Find the times when it is momentarily at rest.
AdvancedVelocity is the derivative; rest means \( v(t)=0 \):
\[ v(t)=s'(t)=3t^2-12t+9=3(t-1)(t-3). \]So the particle is at rest at \( t=1 \) and \( t=3 \).
Find and classify the local extrema of \( f(x)=2x^3-9x^2+12x \).
Advanced\( f'(x)=6x^2-18x+12=6(x-1)(x-2) \), so critical points at \( x=1,2 \). With \( f''(x)=12x-18 \):
\[ f''(1)=-6<0\ (\text{max}),\quad f''(2)=6>0\ (\text{min}). \]Local maximum \( (1,5) \), local minimum \( (2,4) \).
Evaluate \( \displaystyle\int x\sin\!\big(x^2\big)\,dx \).
IntermediateLet \( u=x^2 \), so \( du=2x\,dx \) and \( x\,dx=\tfrac12\,du \):
\[ \int x\sin\!\big(x^2\big)\,dx=\tfrac12\int\sin u\,du=-\tfrac12\cos\!\big(x^2\big)+C. \]Find the area enclosed between \( y=4-x^2 \) and \( y=x+2 \).
AdvancedIntersections: \( 4-x^2=x+2\Rightarrow x^2+x-2=0\Rightarrow x=-2,1 \). The parabola is on top, so
\[ \int_{-2}^{1}\big[(4-x^2)-(x+2)\big]\,dx=\int_{-2}^{1}\big(2-x-x^2\big)\,dx=\frac{9}{2}. \]Evaluate \( \displaystyle\sum_{n=1}^{\infty} 3\Big(\tfrac{1}{4}\Big)^{n} \).
IntermediateThe first term (at \( n=1 \)) is \( a=\tfrac34 \) and the ratio is \( r=\tfrac14 \):
\[ \sum_{n=1}^{\infty} 3\Big(\tfrac14\Big)^{n}=\frac{a}{1-r}=\frac{3/4}{3/4}=1. \]Solve the separable equation \( \dfrac{dy}{dx}=\dfrac{x}{y} \).
IntermediateSeparate: \( y\,dy=x\,dx \). Integrate both sides:
\[ \tfrac{y^2}{2}=\tfrac{x^2}{2}+C\ \Longrightarrow\ y^2-x^2=C. \]For \( x=t^2,\ y=t^3 \), find \( \dfrac{dy}{dx} \).
IntermediateEvaluate \( \displaystyle\lim_{x\to 0}\frac{1-\cos x}{x^2} \).
AdvancedForm \( \tfrac00 \); apply L'Hôpital twice:
\[ \lim_{x\to 0}\frac{1-\cos x}{x^2}=\lim_{x\to 0}\frac{\sin x}{2x}=\lim_{x\to 0}\frac{\cos x}{2}=\frac12. \]Evaluate \( \displaystyle\lim_{x\to 2}\big(3x^2-x+1\big) \).
BasicThe polynomial is continuous, so substitute directly:
\[ 3(2)^2-2+1=11. \]Evaluate \( \displaystyle\lim_{x\to 0}\frac{\sin 3x}{x} \).
BasicUse \( \lim_{u\to 0}\tfrac{\sin u}{u}=1 \):
\[ \lim_{x\to 0}\frac{\sin 3x}{x}=3\lim_{x\to 0}\frac{\sin 3x}{3x}=3. \]Evaluate \( \displaystyle\lim_{x\to\infty}\frac{2x^2-3x+1}{5x^2+x} \).
IntermediateDivide by \( x^2 \); lower-order terms vanish, leaving the ratio of leading coefficients:
\[ \frac{2}{5}. \]Evaluate \( \displaystyle\lim_{x\to 0}\frac{\sqrt{x+4}-2}{x} \).
IntermediateMultiply by the conjugate \( \sqrt{x+4}+2 \):
\[ \lim_{x\to 0}\frac{(x+4)-4}{x\big(\sqrt{x+4}+2\big)}=\lim_{x\to 0}\frac{1}{\sqrt{x+4}+2}=\frac14. \]Find \( k \) so that \( f(x)=\begin{cases} x^2,&x\le 1\\ kx,&x>1\end{cases} \) is continuous at \( x=1 \).
IntermediateThe two pieces must agree at \( x=1 \): \( 1^2=k(1) \).
\[ k=1. \]Differentiate \( f(x)=7x^3-4x+9 \).
BasicDifferentiate \( y=\cos 5x \).
BasicChain rule with inner \( 5x \):
\[ y'=-5\sin 5x. \]Differentiate \( y=e^{2x} \).
BasicDifferentiate \( y=\ln(3x) \).
BasicChain rule: \( \tfrac{1}{3x}\cdot 3 \).
\[ y'=\frac1x. \]Differentiate \( y=x^3\ln x \).
IntermediateProduct rule: \( 3x^2\ln x + x^3\cdot\tfrac1x \).
\[ y'=x^2\big(3\ln x+1\big). \]Differentiate \( y=\dfrac{x}{x+1} \).
IntermediateQuotient rule:
\[ y'=\frac{(1)(x+1)-x(1)}{(x+1)^2}=\frac{1}{(x+1)^2}. \]Differentiate \( y=\sqrt{x^2+1} \).
IntermediateChain rule on \( (x^2+1)^{1/2} \):
\[ y'=\frac{x}{\sqrt{x^2+1}}. \]Differentiate \( y=\tan 3x \).
IntermediateFind \( \dfrac{dy}{dx} \) if \( x^2+y^2=25 \).
AdvancedDifferentiate implicitly: \( 2x+2y\,y'=0 \).
\[ \frac{dy}{dx}=-\frac{x}{y}. \]Find \( \dfrac{dy}{dx} \) if \( xy+y^2=6 \).
AdvancedDifferentiate implicitly (product rule on \( xy \)): \( y+x\,y'+2y\,y'=0 \).
\[ \frac{dy}{dx}=-\frac{y}{x+2y}. \]Find the tangent line to \( y=x^2 \) at \( x=3 \).
IntermediatePoint \( (3,9) \); slope \( y'=2x=6 \). Then \( y-9=6(x-3) \):
\[ y=6x-9. \]On what intervals is \( f(x)=x^3-3x \) increasing?
Intermediate\( f'(x)=3x^2-3>0 \Rightarrow x^2>1 \):
\[ x<-1 \ \text{or}\ x>1. \]A rectangular field is enclosed with 60 m of fencing on all four sides. What dimensions maximise the area?
AdvancedWith \( l+w=30 \), area \( A=l(30-l) \), so \( A'(l)=30-2l=0\Rightarrow l=15 \).
\[ 15\times 15,\qquad A_{\max}=225\ \text{m}^2. \]A 10 m ladder leans on a wall; its base slides away at 1 m/s. How fast is the top sliding down when the base is 6 m from the wall?
AdvancedFrom \( x^2+y^2=100 \), at \( x=6 \) we get \( y=8 \). Differentiate: \( 2x\,x'+2y\,y'=0 \):
\[ y'=-\frac{x\,x'}{y}=-\frac{6(1)}{8}=-0.75\ \text{m/s}. \]The top slides down at \( 0.75 \) m/s.
Evaluate \( \displaystyle\lim_{x\to 0}\frac{e^{x}-1}{x} \).
IntermediateForm \( \tfrac00 \); by L'Hôpital, \( \lim e^{x}/1 \):
\[ =1. \]Evaluate \( \displaystyle\int \big(6x^2-4x+3\big)\,dx \).
BasicEvaluate \( \displaystyle\int_0^{1} e^{x}\,dx \).
BasicEvaluate \( \displaystyle\int_1^{e}\frac{1}{x}\,dx \).
BasicEvaluate \( \displaystyle\int_0^{\pi/2}\cos x\,dx \).
IntermediateEvaluate \( \displaystyle\int_0^{4}\sqrt{x}\,dx \).
IntermediateWrite \( \sqrt{x}=x^{1/2} \):
\[ \Big[\tfrac{2}{3}x^{3/2}\Big]_0^{4}=\tfrac{2}{3}(8)=\frac{16}{3}. \]Evaluate \( \displaystyle\int 2x\,e^{x^2}\,dx \).
IntermediateLet \( u=x^2 \), \( du=2x\,dx \):
\[ \int e^{u}\,du=e^{x^2}+C. \]Evaluate \( \displaystyle\int x\cos x\,dx \).
IntermediateBy parts with \( u=x,\ dv=\cos x\,dx \):
\[ x\sin x-\int\sin x\,dx=x\sin x+\cos x+C. \]Evaluate \( \displaystyle\int \ln x\,dx \).
AdvancedBy parts with \( u=\ln x,\ dv=dx \):
\[ x\ln x-\int x\cdot\tfrac1x\,dx=x\ln x-x+C. \]Evaluate \( \displaystyle\int \frac{2x+1}{x^2+x}\,dx \).
IntermediateThe numerator is the derivative of the denominator, so let \( u=x^2+x \):
\[ \int\frac{du}{u}=\ln\big|x^2+x\big|+C. \]Find the area under \( y=4-x^2 \) from \( x=-2 \) to \( x=2 \).
IntermediateThe region under \( y=x \) from \( 0 \) to \( 2 \) is revolved about the \( x \)-axis. Find the volume.
AdvancedDisk method with \( R(x)=x \):
\[ \pi\int_0^{2}x^2\,dx=\pi\Big[\tfrac{x^3}{3}\Big]_0^{2}=\frac{8\pi}{3}. \]Find the average value of \( f(x)=x^2 \) on \( [0,3] \).
IntermediateSolve \( \dfrac{dy}{dx}=2y \) with \( y(0)=5 \).
IntermediateExponential model \( y=y_0e^{kt} \) with \( k=2 \), \( y_0=5 \):
\[ y=5e^{2x}. \]For \( x=t^2,\ y=2t \), find \( \dfrac{dy}{dx} \).
IntermediateEvaluate \( \displaystyle\sum_{n=0}^{\infty}\Big(\tfrac{2}{3}\Big)^{n} \).
IntermediateGeometric with \( a=1,\ r=\tfrac23 \) and \( |r|<1 \):
\[ \frac{1}{1-\tfrac23}=3. \]Test generator
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