Calculus 1 answers two questions:
How fast is something changing right now? (Derivatives)
How much total change has accumulated? (Integrals)
These two ideas are connected by the Fundamental Theorem of Calculus, which says differentiation and integration are inverse operations. That single insight ties the entire course together.
run rise f(x) tangent line slope = derivative
The derivative (slope of the tangent line) tells you the instantaneous rate of change. The integral (area under the curve) tells you the total accumulation. Every topic in this course builds on one of these two ideas.
Limits are the foundation. The derivative is a limit. The integral is a limit. Without limits, calculus doesn’t exist.
Key skills: evaluate limits by substitution, factoring, and rationalizing. Recognize indeterminate forms (0/0). Check continuity using the three conditions. Identify removable, jump, and infinite discontinuities.
The derivative of f at x = a is:
f ′ ( a ) = lim h → 0 f ( a + h ) − f ( a ) h f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} f ′ ( a ) = h → 0 lim h f ( a + h ) − f ( a ) The rules that make this fast:
Power rule: derivative of x to the n is n times x to the (n-1)
Product rule: (uv)’ = u’v + uv’
Quotient rule: (u/v)’ = (u’v - uv’) / v²
Chain rule: derivative of f(g(x)) is f’(g(x)) times g’(x)
Trig: sin becomes cos, cos becomes negative sin, tan becomes sec²
Exponential: derivative of e to the x is e to the x
Logarithmic: derivative of ln(x) is 1/x
Derivatives solve real problems:
Related rates: when two quantities change together, differentiate with respect to time
Implicit differentiation: find dy/dx when y isn’t isolated
Linear approximation: use the tangent line to estimate function values near a known point
Maxima and minima: set f’ = 0, classify with the first or second derivative test
Mean Value Theorem: somewhere between a and b, the instantaneous rate equals the average rate
Curve sketching: combine f’ (increasing/decreasing) and f” (concavity) to draw the graph
Optimization: maximize or minimize a quantity under constraints
Local Maximum f' + then f' − Local Minimum f' − then f' +
Integration is differentiation in reverse.
The indefinite integral gives a family of antiderivatives (don’t forget + C). The definite integral gives a number: the net signed area under the curve.
∫ a b f ( x ) d x = F ( b ) − F ( a ) \int_a^b f(x)\,dx = F(b) - F(a) ∫ a b f ( x ) d x = F ( b ) − F ( a ) Key techniques: power rule for integration, basic trig and exponential integrals, u-substitution (chain rule in reverse).
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Applications: area under curves, area between curves, average value of a function, net change vs total change.
Power rule (derivative):
d d x [ x n ] = n x n − 1 \frac{d}{dx}[x^n] = nx^{n-1} d x d [ x n ] = n x n − 1 Power rule (integral):
∫ x n d x = x n + 1 n + 1 + C ( n ≠ − 1 ) \int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1) ∫ x n d x = n + 1 x n + 1 + C ( n = − 1 ) FTC:
∫ a b f ( x ) d x = F ( b ) − F ( a ) \int_a^b f(x)\,dx = F(b) - F(a) ∫ a b f ( x ) d x = F ( b ) − F ( a ) Average value:
f avg = 1 b − a ∫ a b f ( x ) d x f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx f avg = b − a 1 ∫ a b f ( x ) d x Derivative with product and chain rule:
Find the derivative of x² sin(3x).
Product rule: u = x², v = sin(3x).
d d x [ x 2 sin ( 3 x ) ] = 2 x sin ( 3 x ) + x 2 ⋅ 3 cos ( 3 x ) = 2 x sin ( 3 x ) + 3 x 2 cos ( 3 x ) \frac{d}{dx}[x^2\sin(3x)] = 2x\sin(3x) + x^2 \cdot 3\cos(3x) = 2x\sin(3x) + 3x^2\cos(3x) d x d [ x 2 sin ( 3 x )] = 2 x sin ( 3 x ) + x 2 ⋅ 3 cos ( 3 x ) = 2 x sin ( 3 x ) + 3 x 2 cos ( 3 x ) Definite integral:
∫ 0 π / 2 sin x d x = [ − cos x ] 0 π / 2 = − cos ( π / 2 ) + cos ( 0 ) = 0 + 1 = 1 \int_0^{\pi/2}\sin x\,dx = [-\cos x]_0^{\pi/2} = -\cos(\pi/2) + \cos(0) = 0 + 1 = 1 ∫ 0 π /2 sin x d x = [ − cos x ] 0 π /2 = − cos ( π /2 ) + cos ( 0 ) = 0 + 1 = 1 Optimization:
Maximize area of a rectangle with 400 ft of fencing against a wall. Let x = width, y = 400 - 2x.
A ( x ) = x ( 400 − 2 x ) , A ′ ( x ) = 400 − 4 x = 0 ⇒ x = 100 , y = 200 A(x) = x(400 - 2x), \quad A'(x) = 400 - 4x = 0 \quad \Rightarrow \quad x = 100, \; y = 200 A ( x ) = x ( 400 − 2 x ) , A ′ ( x ) = 400 − 4 x = 0 ⇒ x = 100 , y = 200 Maximum area = 20,000 sq ft.
Average value:
Average of x² on [0, 3]:
1 3 ∫ 0 3 x 2 d x = 1 3 ⋅ 9 = 3 \frac{1}{3}\int_0^3 x^2\,dx = \frac{1}{3}\cdot 9 = 3 3 1 ∫ 0 3 x 2 d x = 3 1 ⋅ 9 = 3
You’ve gone from arithmetic all the way through Calculus 1. That’s years of building, and every layer matters. Derivatives answer “how fast?” Integrals answer “how much?” The Fundamental Theorem connects them. You now have the tools to model change, optimize systems, and calculate totals from rates. When you’re ready, Calculus 2 builds directly on this foundation. Take a breath. You earned this.
The derivative represents: A. Total accumulation B. Instantaneous rate of change C. Average value D. Net signed area
The Chain Rule is used when: A. Two functions are multiplied B. Functions are composed (nested) C. One is divided by another D. The function is linear
The FTC says the definite integral equals: A. F(a) - F(b) B. The derivative at b C. F(b) - F(a) D. F(x) + C
In related rates, you differentiate with respect to: A. x only B. y only C. t (time) D. The radius
The average value formula is: A. f(b) - f(a) B. (1/(b-a)) times the integral C. The maximum value D. The slope at the midpoint
If f'(c) = 0 and f''(c) is positive, then c is a: A. Local maximum B. Inflection point C. Local minimum D. Vertical asymptote
U-substitution is the reverse of: A. The power rule B. The chain rule C. The product rule D. The quotient rule
The definite integral (without absolute values) gives: A. Total unsigned area B. The derivative C. The limit D. Net signed area
The MVT guarantees a point where: A. The derivative is zero B. Instantaneous rate equals average rate C. The function is constant D. f'' is positive
For optimization on a closed interval, check: A. Only the midpoint B. Only critical points C. Critical points and endpoints D. Every point
The derivative of e to the kx is: A. e to the kx B. k C. ln(k) times e to the kx D. k times e to the kx
Implicit differentiation is needed when: A. The equation is linear B. y is already isolated C. y is not solved for explicitly D. Only polynomials are involved
Area between f(x) and g(x) where f is on top: A. The integral of f + g B. The average of f and g C. The integral of (f - g) D. Only the upper curve
The linear approximation formula is: A. f(x) B. f'(x) C. f(a) + f'(a)(x - a) D. f(a)
Calculus 1 primarily covers: A. Only integration B. Derivatives, their applications, and beginning integration C. Only limits D. Only abstract theory
Integrating velocity gives: A. Acceleration B. Net displacement C. Average speed D. Maximum velocity
The Second Derivative Test is inconclusive when: A. f''(c) is positive B. f''(c) is negative C. f'(c) is not zero D. f''(c) = 0
The power rule for integration gives: A. n times x to the (n-1) B. x to the (n+1) divided by (n+1), plus C C. Only for n = 1 D. Only for constants
The most important connection in Calculus 1 is: A. Limits are unnecessary B. Derivatives and integrals are inverse operations C. Only polynomials matter D. Graphing is optional
After Calculus 1, you should be confident with: A. Only memorizing formulas B. Rates of change, optimization, and basic integration C. Only graphing D. Only abstract proofs
Retry Quiz
