Calc stuff

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Last updated 8:08 PM on 6/25/26
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91 Terms

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xn+1n+1+C\frac{x^{n+1}}{n+1}+C

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ln|x| + C

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ekxk+C\frac{e^{kx}}{k}+C

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<p>Chain Rule</p>

Chain Rule

f’(g) g’(x)

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∫cos(kx) dx =

1k\frac{1}{k} sin(kx) + c

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∫sin(kx) dx =

1k\frac{1}{k} cos(kx) + c

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∫sec2 x dx =

tan(x) + c

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∫csc(x) cot(x) dx =

-csc(x) + c

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∫sec(x) tan(x) dx =

sec(x) + C

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∫csc2x dx =

-cot(x) + C

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A measure of the slope of the tangent line, or the rate of change, of f(x)f(x) at the given point (a,f(a))(a, f(a)) calculated as f(x)f(a)xa\frac{f(x) - f(a)}{x - a} .

Slope of the Secant Line

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The limiting values of the average velocities over shorter and shorter time periods.

Instantaneous Velocity

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To say that limxaf(x)=L\lim_{x \to a} f(x) = L means that as xx approaches aa, but xax \neq a, then f(x)f(x) must approach LL.

Limit of a Function

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A limit where the value is different when approaching from either the positive side (xa+x \to a^+) or the negative side (xax \to a^-).

One-Sided Limits

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If nn is a positive even integer, then limxa+1(xa)n=\lim_{x \to a^+} \frac{1}{(x-a)^n} = \infty, limxa1(xa)n=\lim_{x \to a^-} \frac{1}{(x-a)^n} = \infty, and limxa1(xa)n=\lim_{x \to a} \frac{1}{(x-a)^n} = \infty.

Infinite Limit Theorem (Positive Even Integer)

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If nn is a positive odd integer, then limxa+1(xa)n=\lim_{x \to a^+} \frac{1}{(x-a)^n} = \infty and limxa1(xa)n=\lim_{x \to a^-} \frac{1}{(x-a)^n} = -\infty, hence the two-sided limit is DNE (Does Not Exist).

Infinite Limit Theorem (Positive Odd Integer)

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The limit law stating limxa[f(x)+g(x)]=limxaf(x)+limxag(x)\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x).

Sum Law

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The limit law stating limxa[f(x)g(x)]=(limxaf(x))(limxag(x))\lim_{x \to a} [f(x) g(x)] = (\lim_{x \to a} f(x)) (\lim_{x \to a} g(x)).

Product Law

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If ff is a function such that aa is in the domain of ff, then limxaf(x)=f(a)\lim_{x \to a} f(x) = f(a).

Direct Substitution Property

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If f(x)g(x)h(x)f(x) \leq g(x) \leq h(x) when xx is near aa (but not necessarily equal to aa) and limxaf(x)=L=limxah(x)\lim_{x \to a} f(x) = L = \lim_{x \to a} h(x), then limxag(x)=L\lim_{x \to a} g(x) = L.

Squeeze Theorem

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A function ff is continuous at aa if: 1. f(a)f(a) is defined; 2. limxaf(x)\lim_{x \to a} f(x) exists; 3. limxaf(x)=f(a)\lim_{x \to a} f(x) = f(a).

Continuity at a Point

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The derivative of a function at a number aa, denoted by f(a)f'(a), is defined as f(a)=limh0f(a+h)f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}.

Limit Definition of Derivative

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A function ff is differentiable at a number aa if the limit f(a)=limh0f(a+h)f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} exists.

Differentiable at a Number

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If f(x)f(x) is differentiable at aa, then ff is continuous at aa.

Differentiability Implies Continuity

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The differentiation rule stating that for any real number nn, ddx[xn]=nxn1\frac{d}{dx} [x^n] = n x^{n-1}.

The Power Rule

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The differentiation rule stating that ddx[f(x)g(x)]=f(x)g(x)+f(x)g(x)\frac{d}{dx} [f(x) g(x)] = f'(x) g(x) + f(x) g'(x).

The Product Rule

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The differentiation rule stating that ddx[f(x)g(x)]=g(x)f(x)f(x)g(x)[g(x)]2\frac{d}{dx} [\frac{f(x)}{g(x)}] = \frac{g(x) f'(x) - f(x) g'(x)}{[g(x)]^2}.

The Quotient Rule

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The state of a moving particle when its velocity at time tt is equal to zero (v(t)=0v(t) = 0).

Particle at Rest

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A line on a function f(x)f(x) where the slope is zero, found by setting the derivative f(x)=0f'(x) = 0.

Horizontal Tangent Line

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limx0sin(ax)ax=1\lim_{x\to0}\frac{\sin(ax)}{ax}=1 ; 2. limx01cos(ax)ax=1\lim_{x\to0}\frac{1-\cos\left(ax\right)}{ax}=1 ; 3. limx0tan(x)x=1\lim_{x \to 0} \frac{\tan(x)}{x} = 1 .

Big Three Limits

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g(x)g(x)\frac{g^{\prime}\left(x\right)}{g\left(x\right)}

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\frac{1}{x\ln\left(a\right)},x>0 .

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g(x)g(x)lna\frac{g^{\prime}\left(x\right)}{g\left(x\right)\ln a}

34
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exe^{x}

35
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axln(a)a^{x}\ln\left(a\right)

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eg(x)g(x)e^{g\left(x\right)}g^{\prime}\left(x\right)

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ln(a)ag(x)g(x)\ln\left(a\right)a^{g\left(x\right)}g^{\prime}\left(x\right)

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\frac{d}{\differentialD x}\cos^{-1}\left(x\right) =

11x2\frac{-1}{\sqrt{1-x^2}}

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\frac{d}{\differentialD x}\sin^{-1}\left(x\right) =

11x2\frac{1}{\sqrt{1-x^2}}

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\frac{d}{\differentialD x}\cot^{-1}\left(x\right) =

11+x2\frac{-1}{1+x^2}

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\frac{d}{\differentialD x}\csc^{-1}\left(x\right) =

1x1x2\frac{-1}{|x|\sqrt{1-x^2}}

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\frac{d}{\differentialD x}\tan^{-1}\left(x\right) =

11+x2\frac{1}{1+x^2}

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\frac{d}{\differentialD x}\sec^{-1}\left(x\right)

1x1x2\frac{1}{\left|x\right|\sqrt{1-x^2}}

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sec2(x) =

1 + tan2(x)

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\frac{d}{\differentialD x}\left(\frac{1}{x}\right)=

1x2-\frac{1}{x^2}

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\frac{d}{\differentialD x}\left(\sqrt{x}\right)=

12x\frac{1}{2\sqrt{x}}

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v(t) and a(t) have different signs

Particle is slowing down:

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v(t) < 0 (negative)

Particle is moving left/down:

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v(t) > 0 (positive)

Particle is moving right/up:

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v(t) and a(t) have same sign

Particle is speeding up: (|velocity| is getting bigger)

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If f is continuous on [a,b] and k is any number between f(a) and f(b), then there is at least one number c between a and b such that f(c) = k

Intermediate Value Theorem

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  1. Find critical numbers

  2. Identify endpoints

  3. Find f(critical numbers) and f(endpoints)

  4. determine absolute max/min values by comparing the y-values; stated in a sentence.

Steps for Absolute Mins/Max:

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If f is continuous on [a,] and differentiable on (a,b), then there exists a number “c” on (a,b) such that

f(c)=f(b)f(a)baf^{\prime}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}

Mean Value Theorem

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x = c if a critical number because

f’(x) = 0 or f’(x) is undefined

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f(x) is ____ on [a,b] because f’(x) > 0

increasing

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f(x) is ____ on [a,b] because f’(x) < 0

decreasing

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f(x) is ____ on (a,b) because f’’(x) > 0

concave up

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f(x) is ____ on (a,b) because f’’(x) < 0

concave down

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f’’(x) = 0 or = DNE and f'“(x) changes sign around point c. (Or if f’(x) slope changes sign and f’’(x) = 0 or DNE)

Inflection point

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tanxdx=∫\tan xdx=

lncosx+C-\ln\left|\cos x\right|+C

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∫\frac{1}{x^2+a^2}\differentialD x=

1atan1(xa)+C\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C

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F.T.O.C :ab ⁣f(x)dx=\int_{a}^{b}\!f^{\prime}\left(x\right)\,dx=

f(b)f(a)f\left(b\right)-f\left(a\right)

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ab ⁣f(x)+g(x)dx=\int_{a}^{b}\!f\left(x\right)+g\left(x\right)\,dx=

\int_{a}^{b}\!f\left(x\right)dx+\int_{a}^{b}\!g\left(x\right)\differentialD x

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ab ⁣f(x)g(x)dx=\int_{a}^{b}\!f\left(x\right)-g\left(x\right)\,dx=

\int_{a}^{b}\!f\left(x\right)\,dx-\int_{a}^{b}\!g\left(x\right)\differentialD x

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ab ⁣cf(x)dx=\int_{a}^{b}\!cf\left(x\right)\,dx=

cab ⁣f(x)dxc\int_{a}^{b}\!f\left(x\right)\,dx

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ab ⁣f(x)dx=\int_{a}^{b}\!f\left(x\right)\,dx=

ba ⁣f(x)dx-\int_{b}^{a}\!f\left(x\right)\,dx

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aa ⁣f(x)dx=\int_{a}^{a}\!f\left(x\right)\,dx=

0

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1baab ⁣f(x)dx\frac{1}{b-a}\int_{a}^{b}\!f\left(x\right)\,dx

Average Value of a Function

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Second F.T.O.C. P1: \frac{d}{\differentialD x}\int_{a}^{x}\!f\left(t\right)\,dt=

f(x)

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Second F.T.O.C. P2: \frac{d}{\differentialD x}\int_{a}^{g\left(x\right)}\!f\left(t\right)\,dt=

 ⁣f(g(x))g(x)\!f\left(g\left(x\right)\right)\cdot\,g^{\prime}\left(x\right)

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ab ⁣f(x)dx\int_{a}^{b}\!f\left(x\right)\,dx represents the ___ ______ in the function f from time a to b.

net change

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  1. separate variables

  2. integrate each side

  3. make sure to place C on side with independent variable (usually x)

  4. plug in initial condition for C (if given)

  5. solve for dependent variable (usually y)

Steps to solve differential equations.

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y=Cekty=Ce^{kt} or y=y0ekty=y_0e^{kt}

Exponential growth

y= end amount

C or y0 = initial amount

k = growth constant/growth rate

t = time elapsed

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y=y0ekty=y_0e^{-kt}

Exponential decay

negative sign = decays over time, rather than growing.

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v(t) = s’(t)

Velocity

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a(t) = v’(t) = s’’(t)

acceleration

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|v'(t)|

speed

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s(b)s(a)ba\frac{s\left(b\right)-s\left(a\right)}{b-a} (given v(t)); or 1baab ⁣v(t)dt\frac{1}{b-a}\int_{a}^{b}\!v\left(t\right)\,dt (given a(t))

average velocity

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v(b)v(a)ba\frac{v\left(b\right)-v\left(a\right)}{b-a} (given v(t)); or 1baab ⁣a(t)dt\frac{1}{b-a}\int_{a}^{b}\!a\left(t\right)\,dt (given a(t))

average acceleration

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ab ⁣v(t)dt\int_{a}^{b}\!v\left(t\right)\,dt

Displacement

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ab ⁣v(t)dt\int_{a}^{b}\left|\!v\left(t\right)\right|\,dt

total distance

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s(b)=s(a)+ab ⁣v(t)dts\left(b\right)=s\left(a\right)+\int_{a}^{b}\!v\left(t\right)\,dt

position at b

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ab ⁣[f(x)g(x)]dx=\int_{a}^{b}\!\left\lbrack f\left(x\right)-g\left(x\right)\right\rbrack\,dx=

Area perpendicular to x-axis; f(x) is top curve, g(x) is bottom, a and b are x-coordinates of point of intersection.

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\int_{a}^{b}\!\left\lbrack f\left(y\right)-g\left(y\right)]\right.\differentialD y=

Area perpendicular to y-axis; f(y) is right curve, g(y) is left curve, a and b are y-coordinates of point of intersection.

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  1. decide whether its dx or dy

  2. find formula in terms of x or y

  3. find the limits (make sure they match x or y)

  4. integrate and evaluate

Volume = ∫Area

Volume steps

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V = ab ⁣πr2dx\int_{a}^{b}\!\pi r^2dx

Volume of a disc around a Horizontal Axis of rotation

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\int_{a}^{b}\!\left\lbrack\pi R^2-\pi r^2\right\rbrack\differentialD x

Volume of a washer around a horizontal axis of rotation.

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Volume of a slab (cross-section) around a horizontal axis of rotation. A(x) is the area for the cross section.

ab ⁣A(x)dx\int_{a}^{b}\!A\left(x\right)\,dx

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V = ab ⁣πr2dy\int_{a}^{b}\!\pi r^2dy

Volume of a disc around a vertical axis of rotation

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\int_{a}^{b}\!\left\lbrack\pi R^2-\pi r^2\right\rbrack\differentialD y

Volume of a washer around a vertical axis of rotation

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Volume of a slab (cross-section) around a vertical axis of rotation. A(y) is the area for the cross section.

ab ⁣A(y)dy\int_{a}^{b}\!A\left(y\right)\,dy