Calculus 1 Cheat Sheet
1. Limits and Continuity
1.1 Definition of a Limit
The limit of a function \(f(x)\) as \(x\) approaches \(a\) is the value that \(f(x)\) gets closer to as \(x\) gets closer to \(a\). It is denoted as:
\[ \lim_{x \to a} f(x) = L \]
Example:
Find \( \lim_{x \to 2} (3x + 1) \).
- Substitute \( x = 2 \) directly into the function: \( 3(2) + 1 = 6 + 1 = 7 \).
- Result: \( \lim_{x \to 2} (3x + 1) = 7 \).
1.2 Continuity
A function \(f(x)\) is continuous at a point \(x = a\) if the following three conditions are met:
- \(f(a)\) is defined.
- \(\lim_{x \to a} f(x)\) exists.
- \(\lim_{x \to a} f(x) = f(a)\).
Example:
Determine if the function \(f(x) = \frac{x^2 - 1}{x - 1}\) is continuous at \(x = 1\).
- Factor the numerator: \(f(x) = \frac{(x - 1)(x + 1)}{x - 1}\).
- Simplify: \(f(x) = x + 1\) for \(x \neq 1\).
- The limit as \(x\) approaches 1 is \(2\), but \(f(1)\) is undefined.
- Result: The function is not continuous at \(x = 1\).
2. Derivatives and Differentiation
2.1 Definition of the Derivative
The derivative of a function \(f(x)\) at a point \(x = a\) is the slope of the tangent line to the function at that point. It is defined as:
\[ f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \]
Example:
Find the derivative of \(f(x) = x^2\) using the definition of the derivative.
- Apply the definition: \(f'(x) = \lim_{h \to 0} \frac{(x + h)^2 - x^2}{h}\).
- Expand the numerator: \(f'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h}\).
- Simplify: \(f'(x) = \lim_{h \to 0} (2x + h) = 2x\).
- Result: \(f'(x) = 2x\).
2.2 Rules of Differentiation
Key differentiation rules include the power rule, product rule, quotient rule, and chain rule:
- Power Rule: \( \frac{d}{dx} [x^n] = nx^{n-1} \)
- Product Rule: \( \frac{d}{dx} [uv] = u'v + uv' \)
- Quotient Rule: \( \frac{d}{dx} \left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2} \)
- Chain Rule: \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \)
Example:
Differentiate \(f(x) = x^3 \cdot \sin(x)\) using the product rule.
- Let \(u = x^3\) and \(v = \sin(x)\).
- Find \(u' = 3x^2\) and \(v' = \cos(x)\).
- Apply the product rule: \(f'(x) = u'v + uv' = 3x^2 \cdot \sin(x) + x^3 \cdot \cos(x)\).
- Result: \(f'(x) = 3x^2 \sin(x) + x^3 \cos(x)\).
3. Applications of Derivatives
3.1 Finding Tangent Lines
The equation of the tangent line to the function \(f(x)\) at the point \(x = a\) is given by:
\[ y - f(a) = f'(a)(x - a) \]
Example:
Find the equation of the tangent line to the curve \(y = x^2\) at the point \(x = 1\).
- Find the derivative: \(f'(x) = 2x\).
- Evaluate at \(x = 1\): \(f'(1) = 2(1) = 2\).
- Find the point on the curve: \(f(1) = 1^2 = 1\).
- Use the point-slope form: \(y - 1 = 2(x - 1)\).
- Result: The equation of the tangent line is \(y = 2x - 1\).
3.2 Related Rates
Related rates problems involve finding the rate at which one quantity changes with respect to another. Typically, you differentiate an equation with respect to time \(t\) to relate the rates.
Example:
A ladder 10 feet long is leaning against a wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/s, how fast is the top of the ladder sliding down the wall when the bottom is 6 feet from the wall?
- Use the Pythagorean theorem: \(x^2 + y^2 = 10^2\), where \(x\) is the distance from the wall and \(y\) is the height.
- Differentiating with respect to time \(t\): \(2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0\).
- Substitute \(x = 6\), \(y = 8\) (since \(6^2 + 8^2 = 10^2\)), and \(\frac{dx}{dt} = 1\): \(2(6)(1) + 2(8)\frac{dy}{dt} = 0\).
- Solve for \(\frac{dy}{dt}\): \(12 + 16 \frac{dy}{dt} = 0\), \(\frac{dy}{dt} = -\frac{12}{16} = -\frac{3}{4} \text{ ft/s}\).
- Result: The top of the ladder is sliding down at \(\frac{3}{4}\) ft/s.
3.3 Optimization Problems
Optimization involves finding the maximum or minimum value of a function subject to constraints. Typically, you set the derivative equal to zero and solve for critical points, then use the second derivative test or endpoints to determine maxima or minima.
Example:
Find the dimensions of a rectangle with a fixed perimeter of 24 meters that maximize the area.
- Let the length be \(x\) and the width be \(y\). The perimeter is \(2x + 2y = 24\), so \(y = 12 - x\).
- Maximize the area: \(A = xy = x(12 - x) = 12x - x^2\).
- Find the derivative: \(A' = 12 - 2x\). Set \(A' = 0\) to find critical points: \(12 - 2x = 0\), \(x = 6\).
- Since \(A'' = -2 < 0\), \(x = 6\) is a maximum.
- Result: The maximum area is achieved with dimensions \(6 \times 6\), which is a square.
4. The Mean Value Theorem
4.1 Statement of the Theorem
The Mean Value Theorem states that if a function \(f(x)\) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \(c\) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
Example:
Verify the Mean Value Theorem for the function \(f(x) = x^2\) on the interval \([1, 3]\).
- The function \(f(x) = x^2\) is continuous and differentiable everywhere.
- Calculate \(f(3) - f(1)\) and \(3 - 1\): \(f(3) = 9\), \(f(1) = 1\), \(\frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = 4\).
- Find \(f'(x) = 2x\). Set \(f'(c) = 4\): \(2c = 4\), \(c = 2\).
- Result: The Mean Value Theorem is satisfied at \(c = 2\).
5. Integrals and Antiderivatives
5.1 Definition of the Definite Integral
The definite integral of a function \(f(x)\) from \(a\) to \(b\) is the signed area under the curve \(f(x)\) between \(x = a\) and \(x = b\). It is defined as:
\[ \int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x \]
where \( \Delta x = \frac{b - a}{n} \) and \( x_i^* \) is a sample point in the ith subinterval.
Example:
Compute \( \int_{0}^{2} (3x^2 + 2) \, dx \).
- Find the antiderivative of \(3x^2 + 2\): \(F(x) = x^3 + 2x\).
- Evaluate the definite integral: \(F(2) - F(0) = (2^3 + 2 \times 2) - (0 + 0) = 8 + 4 = 12\).
- Result: \( \int_{0}^{2} (3x^2 + 2) \, dx = 12 \).
5.2 Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus connects differentiation and integration. It states:
- Part 1: If \(f(x)\) is continuous on \([a, b]\), then the function \(F(x) = \int_{a}^{x} f(t) \, dt\) is continuous on \([a, b]\), differentiable on \((a, b)\), and \(F'(x) = f(x)\).
- Part 2: If \(F(x)\) is an antiderivative of \(f(x)\), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).
Example:
Find \( \frac{d}{dx} \left( \int_{2}^{x} \sin(t) \, dt \right) \).
- Apply the Fundamental Theorem of Calculus: \( \frac{d}{dx} \left( \int_{2}^{x} \sin(t) \, dt \right) = \sin(x) \).
- Result: \( \frac{d}{dx} \left( \int_{2}^{x} \sin(t) \, dt \right) = \sin(x) \).
5.3 Indefinite Integrals and Basic Integration Rules
An indefinite integral represents a family of functions whose derivative is the integrand. It is expressed as:
\[ \int f(x) \, dx = F(x) + C \]
where \(F(x)\) is an antiderivative of \(f(x)\) and \(C\) is the constant of integration.
Basic Integration Rules:
- Power Rule: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), for \(n \neq -1\)
- Exponential Rule: \( \int e^x \, dx = e^x + C \)
- Trigonometric Rule: \( \int \sin(x) \, dx = -\cos(x) + C \), \( \int \cos(x) \, dx = \sin(x) + C \)
Example:
Compute \( \int (2x^3 - 3x^2 + 4) \, dx \).
- Integrate each term: \( \int 2x^3 \, dx = \frac{2x^4}{4} = \frac{x^4}{2} \), \( \int -3x^2 \, dx = -\frac{3x^3}{3} = -x^3 \), \( \int 4 \, dx = 4x \).
- Combine the results: \( \frac{x^4}{2} - x^3 + 4x + C \).
- Result: \( \int (2x^3 - 3x^2 + 4) \, dx = \frac{x^4}{2} - x^3 + 4x + C \).
6. Techniques of Integration
6.1 Integration by Substitution
Integration by substitution is used to simplify integrals by making a substitution that transforms the integral into a simpler form. It is often used when the integrand is a composite function.
General Formula:
\[ \int f(g(x))g'(x) \, dx = \int f(u) \, du \]
where \(u = g(x)\) and \(du = g'(x) \, dx\).
Example:
Compute \( \int (3x^2 \cdot \sin(x^3)) \, dx \) using substitution.
- Let \(u = x^3\), then \(du = 3x^2 \, dx\).
- Substitute into the integral: \( \int \sin(u) \, du \).
- Integrate: \( -\cos(u) + C \).
- Substitute back: \( -\cos(x^3) + C \).
- Result: \( \int (3x^2 \cdot \sin(x^3)) \, dx = -\cos(x^3) + C \).
6.2 Integration by Parts
Integration by parts is based on the product rule for differentiation. It is used when the integrand is a product of two functions. The formula is:
\[ \int u \, dv = uv - \int v \, du \]
Example:
Compute \( \int x \cdot e^x \, dx \) using integration by parts.
- Let \(u = x\) and \(dv = e^x \, dx\).
- Then \(du = dx\) and \(v = e^x\).
- Apply the formula: \( \int x \cdot e^x \, dx = x \cdot e^x - \int e^x \, dx \).
- Integrate: \( x \cdot e^x - e^x + C \).
- Result: \( \int x \cdot e^x \, dx = e^x(x - 1) + C \).
7. Applications of Integration
7.1 Area Under a Curve
The area under a curve \(y = f(x)\) from \(x = a\) to \(x = b\) is given by the definite integral:
\[ \text{Area} = \int_{a}^{b} f(x) \, dx \]
Example:
Find the area under the curve \(y = x^2\) from \(x = 0\) to \(x = 2\).
- Set up the integral: \( \int_{0}^{2} x^2 \, dx \).
- Find the antiderivative: \( \frac{x^3}{3} \).
- Evaluate at the bounds: \( \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \).
- Result: The area is \( \frac{8}{3} \) square units.
7.2 Volume of a Solid of Revolution
The volume of a solid of revolution generated by rotating a curve \(y = f(x)\) around the x-axis from \(x = a\) to \(x = b\) is given by:
\[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]
Example:
Find the volume of the solid generated by rotating the curve \(y = x^2\) around the x-axis from \(x = 0\) to \(x = 1\).
- Set up the integral: \( V = \pi \int_{0}^{1} (x^2)^2 \, dx = \pi \int_{0}^{1} x^4 \, dx \).
- Find the antiderivative: \( \frac{x^5}{5} \).
- Evaluate at the bounds: \( \pi \left(\frac{1^5}{5} - \frac{0^5}{5}\right) = \frac{\pi}{5} \).
- Result: The volume is \( \frac{\pi}{5} \) cubic units.
7.3 Average Value of a Function
The average value of a continuous function \(f(x)\) on the interval \([a, b]\) is given by:
\[ f_{\text{avg}} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx \]
Example:
Find the average value of the function \(f(x) = 4 - x^2\) on the interval \([-2, 2]\).
- Set up the integral: \( \int_{-2}^{2} (4 - x^2) \, dx \).
- Find the antiderivative: \( 4x - \frac{x^3}{3} \).
- Evaluate at the bounds: \( \left[4(2) - \frac{(2)^3}{3}\right] - \left[4(-2) - \frac{(-2)^3}{3}\right] = 8 - \frac{8}{3} + 8 - \frac{8}{3} = 16 - \frac{16}{3} \).
- Divide by the interval length: \( f_{\text{avg}} = \frac{1}{4} \left(16 - \frac{16}{3}\right) = \frac{1}{4} \cdot \frac{32}{3} = \frac{8}{3} \).
- Result: The average value is \( \frac{8}{3} \).