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Summary Algebra and Integration in Calculus

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Algebra and Integration in Calculus This is your go-to online store for premium study materials, lecture notes, research papers, and exam prep guides from Stanford University. Whether you're a student looking for in-depth course summaries, past exams, or well-structured notes from top Stanford scholars, we've got you covered. Our collection includes: 
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 Subject-specific summaries Designed to help you excel in your studies, our resources provide valuable insights from one of the world’s leading universities. Get the edge you need with high-quality academic materials today.

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Algebra and Integration in Calculus
Algebraic functions and integration are key concepts in calculus. Integration is the process of
finding the antiderivative or the area under a curve. It is the reverse process of
differentiation and has numerous applications in science, engineering, and economics.

Key Concepts in Algebraic Integration
1. Indefinite Integral (Antiderivative):
o The indefinite integral of a function is the general form of its antiderivative. It is
denoted as:∫f(x) dx∫f(x)dx Where f(x)f(x) is the function being integrated, and
indicates the variable of integration.
o Constant of Integration: When integrating, always add a constant CC because
an indefinite integral represents a family of functions.
2. ∫f(x) dx=F(x)+C∫f(x)dx=F(x)+C
Example:
∫xn dx=xn+1n+1+C,(where n≠−1)∫xndx=n+1xn+1+C,(where n=−1)
3. Definite Integral:
o A definite integral represents the area under a curve between two limits, aa and
bb, and is written as:∫abf(x) dx∫abf(x)dx
o It calculates the total accumulated area from x=ax=a to x=bx=b. The result of a
definite integral is a numerical value.
4. Example:
∫13(2x+1) dx∫13(2x+1)dx
Step-by-step:
o Find the antiderivative:∫(2x+1) dx=x2+x+C∫(2x+1)dx=x2+x+C
o Apply limits a=1a=1 and
b=3b=3:[x2+x]13=(32+3)−(12+1)=(9+3)−(1+1)=12−2=10[x2+x]13
=(32+3)−(12+1)=(9+3)−(1+1)=12−2=10
o The area under the curve is 10.
5. Properties of Integrals:
o Linearity: Integration is linear,
meaning:∫[af(x)+bg(x)] dx=a∫f(x) dx+b∫g(x) dx∫[af(x)+bg(x)]dx=a∫f(x)dx+b∫g(x)dx
o Additivity: If the interval of integration is split into smaller subintervals,
then:∫abf(x) dx=∫acf(x) dx+∫cbf(x) dx∫abf(x)dx=∫acf(x)dx+∫cbf(x)dx



Common Algebraic Functions and Their Integrals:
1. Power Rule:
∫xn dx=xn+1n+1+C,n≠−1∫xndx=n+1xn+1+C,n=−1
Example:
∫x3 dx=x44+C∫x3dx=4x4+C
2. Exponential Functions:
∫ex dx=ex+C∫exdx=ex+C
Example:
∫e2x dx=e2x2+C∫e2xdx=2e2x+C

, 3. Logarithmic Functions:
∫1x dx=ln⁡∣x∣+C∫x1dx=ln∣x∣+C
Example:
∫1x+1 dx=ln⁡∣x+1∣+C∫x+11dx=ln∣x+1∣+C
4. Trigonometric Functions:
o ∫sin⁡(x) dx=−cos⁡(x)+C∫sin(x)dx=−cos(x)+C
o ∫cos⁡(x) dx=sin⁡(x)+C∫cos(x)dx=sin(x)+C
o ∫sec⁡2(x) dx=tan⁡(x)+C∫sec2(x)dx=tan(x)+C
5. Inverse Trigonometric Functions:
o ∫11−x2 dx=arcsin⁡(x)+C∫1−x21dx=arcsin(x)+C
o ∫11+x2 dx=arctan⁡(x)+C∫1+x21dx=arctan(x)+C




Integration Techniques:
1. Substitution Rule:
o Use when the integrand is a composite function. If u=g(x)u=g(x),
then:∫f(g(x))⋅g′(x) dx=∫f(u) du∫f(g(x))⋅g′(x)dx=∫f(u)du
2. Example:
∫2x⋅ex2 dx(let u=x2,du=2x dx)∫2x⋅ex2dx(let
u=x2,du=2xdx)∫eu du=eu+C=ex2+C∫eudu=eu+C=ex2+C
3. Integration by Parts:
o This is based on the product rule for differentiation and is useful when integrating
a product of two functions.∫u dv=uv−∫v du∫udv=uv−∫vdu
4. Example:
∫xcos⁡(x) dx∫xcos(x)dx
Let u=xu=x, and dv=cos⁡(x) dxdv=cos(x)dx. Then du=dxdu=dx, and v=sin⁡(x)v=sin(x).
Applying the integration by parts formula:
∫xcos⁡(x) dx=xsin⁡(x)−∫sin⁡(x) dx=xsin⁡(x)+cos⁡(x)+C∫xcos(x)dx=xsin(x)−∫sin(x)dx=xsi
n(x)+cos(x)+C
5. Partial Fraction Decomposition:
o Useful for rational functions where the degree of the numerator is less than the
degree of the denominator. Express the rational function as a sum of simpler
fractions. Example:
6. ∫1x2−1 dx=∫(12(x−1)−12(x+1)) dx∫x2−11dx=∫(2(x−1)1−2(x+1)1)dx
Then integrate each term:
12ln⁡∣x−1∣−12ln⁡∣x+1∣+C21ln∣x−1∣−21ln∣x+1∣+C
7. Trigonometric Substitution:
o Used when the integrand contains square roots of quadratic expressions. Use a
trigonometric identity to simplify the integrand. Example:
8. ∫11−x2 dx∫1−x21dx
Let x=sin⁡(θ)x=sin(θ), so dx=cos⁡(θ) dθdx=cos(θ)dθ. The integral becomes:
∫1cos⁡(θ)⋅cos⁡(θ) dθ=∫dθ=θ+C=arcsin⁡(x)+C∫cos(θ)1⋅cos(θ)dθ=∫dθ=θ+C=arcsin(x)+C



Applications of Integration:
1. Finding Area Under a Curve:

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This is your go-to online store for premium study materials, lecture notes, research papers, and exam prep guides from Stanford University. Whether you're a student looking for in-depth course summaries, past exams, or well-structured notes from top Stanford scholars, we've got you covered. Our collection includes: 
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