Indefinite Integrals

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# Indefinite Integrals ## ๐ŸŽฏ Learning Objectives By the end of this topic, you will be able to: 1. **Define** an anti-derivative and the indefinite integral 2.

Indefinite Integrals

๐ŸŽฏ Learning Objectives

By the end of this topic, you will be able to:
  1. Define an anti-derivative and the indefinite integral
  2. Apply basic integration rules (power, exponential, trigonometric)
  3. Find indefinite integrals using substitution method
  4. Recognize the constant of integration CC

๐Ÿ“‹ Prerequisites

  • Derivatives โ€” differentiation rules (integration is the reverse)

๐Ÿ“– Core Content

15.1 Intuition: The Reverse of Differentiation

If differentiation tells you the rate of change, integration tells you the original quantity given its rate. If velocity v(t)v(t) is the derivative of position s(t)s(t), then position is the integral of velocity. An anti-derivative of f(x)f(x) is a function F(x)F(x) such that Fโ€ฒ(x)=f(x)F'(x) = f(x). The indefinite integral is the set of all anti-derivatives:
โˆซf(x)โ€‰dx=F(x)+C\int f(x)\,dx = F(x) + C
where CC is the constant of integration (since derivative of a constant is 00).
๐Ÿ’ก Why this matters: Integration is used everywhere: computing areas, volumes, work, probability (cumulative distribution functions), and solving differential equations.

15.2 Basic Integration Rules

Derivative RuleIntegral Rule
ddxC=0\frac{d}{dx}C = 0โˆซ0โ€‰dx=C\int 0\,dx = C
ddxxn=nxnโˆ’1\frac{d}{dx}x^n = nx^{n-1}โˆซxnโ€‰dx=xn+1n+1+Cย (nโ‰ โˆ’1)\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\ (n \neq -1)
$\frac{d}{dx}\lnx
ddxex=ex\frac{d}{dx}e^x = e^xโˆซexโ€‰dx=ex+C\int e^x\,dx = e^x + C
ddxax=axlnโกa\frac{d}{dx}a^x = a^x\ln aโˆซaxโ€‰dx=axlnโกa+C\int a^x\,dx = \frac{a^x}{\ln a} + C
ddxsinโกx=cosโกx\frac{d}{dx}\sin x = \cos xโˆซcosโกxโ€‰dx=sinโกx+C\int \cos x\,dx = \sin x + C
ddxcosโกx=โˆ’sinโกx\frac{d}{dx}\cos x = -\sin xโˆซsinโกxโ€‰dx=โˆ’cosโกx+C\int \sin x\,dx = -\cos x + C
ddxtanโกx=secโก2x\frac{d}{dx}\tan x = \sec^2 xโˆซsecโก2xโ€‰dx=tanโกx+C\int \sec^2 x\,dx = \tan x + C
Linearity: โˆซ[f(x)+g(x)]โ€‰dx=โˆซf(x)โ€‰dx+โˆซg(x)โ€‰dx\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx โˆซcf(x)โ€‰dx=cโˆซf(x)โ€‰dx\int cf(x)\,dx = c\int f(x)\,dx

15.3 Worked Examples

Example 1.1: Find โˆซ(3x2โˆ’2x+1)โ€‰dx\int (3x^2 - 2x + 1)\,dx. \int (3x^2 - 2x + 1)\,dx &= 3\int x^2\,dx - 2\int x\,dx + \int 1\,dx \\ &= 3\cdot\frac{x^3}{3} - 2\cdot\frac{x^2}{2} + x + C \\ &= x^3 - x^2 + x + C \end{aligned}
\\boxed{x^3 - x^2 + x + C} **Example 1.2:** Find $\\int \\left(e^x + \\frac{1}{x}\\right)\\,dx$. \\int e^x,dx + \\int\\frac{1}{x},dx = e^x + \\ln|x| + C \\boxed{e^x + \\ln|x| + C} **Example 1.3:** Find $\\int \\sqrt\[3\]{x^2}\\,dx$. \\int x^{2/3},dx = \\frac{x^{5/3}}{5/3} + C = \\frac{3}{5}x^{5/3} + C \\boxed{\\frac{3}{5}x^{5/3} + C} \--- ### 15.4 Substitution (Reverse Chain Rule) When the integrand has the form $f'(g(x))\\cdot g'(x)$, substitute $u = g(x)$: \\int f'(g(x)) \\cdot g'(x),dx = \\int f'(u),du = f(u) + C = f(g(x)) + C **Recipe:** 1. Choose $u = g(x)$ (the "inner function") 2. Compute $du = g'(x)\\,dx$ 3. Rewrite integral in terms of $u$ 4. Integrate 5. Substitute back **Example 2.1:** Find $\\int 2x\\,e^{x^2}\\,dx$. Let $u = x^2$, $du = 2x\\,dx$. $\\int e^u\\,du = e^u + C = e^{x^2} + C$ \\boxed{e^{x^2} + C} **Example 2.2:** Find $\\int \\frac{2x}{x^2+1}\\,dx$. Let $u = x^2+1$, $du = 2x\\,dx$. $\\int \\frac{1}{u}\\,du = \\ln|u| + C = \\ln|x^2+1| + C$ \\boxed{\\ln(x^2+1) + C} **Example 2.3:** Find $\\int \\cos(3x)\\,dx$. Let $u = 3x$, $du = 3\\,dx \\implies dx = du/3$. $\\int \\cos u \\cdot \\frac{du}{3} = \\frac{1}{3}\\sin u + C = \\frac{1}{3}\\sin(3x) + C$ \\boxed{\\frac{1}{3}\\sin(3x) + C} \--- ## ๐Ÿ“ Key Formulas โ€” Summary Table | Integral | Result | Condition | |:---|:---|:---| | $\\int x^n\\,dx$ | $\\frac{x^{n+1}}{n+1} + C$ | $n \\neq -1$ | | $\\int \\frac{1}{x}\\,dx$ | $\\ln|x| + C$ | | | $\\int e^x\\,dx$ | $e^x + C$ | | | $\\int a^x\\,dx$ | $\\frac{a^x}{\\ln a} + C$ | $a > 0, a \\neq 1$ | | $\\int \\sin x\\,dx$ | $-\\cos x + C$ | | | $\\int \\cos x\\,dx$ | $\\sin x + C$ | | | $\\int \\sec^2 x\\,dx$ | $\\tan x + C$ | | | Substitution | $\\int f(g(x))g'(x)\\,dx = \\int f(u)\\,du$ | $u = g(x)$ | --- ## โš ๏ธ Common Pitfalls ### Pitfall 1: Forgetting the Constant $C$ Every indefinite integral must include $+C$. Differentiating any constant gives $0$, so the integral is only determined up to a constant. ### Pitfall 2: Misapplying Power Rule for $n = -1$ $\\int x^{-1}\\,dx = \\ln|x| + C$, NOT $\\frac{x^0}{0}$ (which is undefined). ### Pitfall 3: Forgetting to Divide by the Coefficient in Substitution When substituting $u = ax$, $dx = du/a$. **The $1/a$ factor is essential.** --- ## ๐Ÿ“ Practice Questions > **Q1: Find $\\int x^4\\,dx$.** > > $\\frac{x^5}{5} + C$ > >
\boxed{\frac{x^5}{5} + C}
> **Q2: Find $\\int (4x^3 - 2x + 7)\\,dx$.** > > $x^4 - x^2 + 7x + C$ > >
\boxed{x^4 - x^2 + 7x + C}
> **Q3: Find $\\int \\frac{1}{x^2}\\,dx$.** > > $\\int x^{-2}\\,dx = \\frac{x^{-1}}{-1} + C = -\\frac{1}{x} + C$ > >
\boxed{-\frac{1}{x} + C}
> **Q4: Find $\\int e^{2x}\\,dx$.** > > Let $u = 2x$, $du = 2\\,dx$, $dx = du/2$. > $\\frac{1}{2}\\int e^u\\,du = \\frac{1}{2}e^{2x} + C$ > >
\boxed{\frac{1}{2}e^{2x} + C}
> **Q5: Find $\\int 3\\sin x\\,dx$.** > > $-3\\cos x + C$ > >
\boxed{-3\cos x + C}
> **Q6: Find $\\int \\frac{2x}{x^2+1}\\,dx$.** > > Let $u = x^2+1$, $du = 2x\\,dx$. $\\int \\frac{du}{u} = \\ln|u| + C = \\ln(x^2+1) + C$ > >
\boxed{\ln(x^2+1) + C}
> **Q7: Find $\\int \\sqrt{x}\\,dx$.** > > $\\int x^{1/2}\\,dx = \\frac{x^{3/2}}{3/2} + C = \\frac{2}{3}x^{3/2} + C$ > >
\boxed{\frac{2}{3}x^{3/2} + C}
> **Q8: Find $\\int (x+1)^4\\,dx$.** > > Let $u = x+1$, $du = dx$. $\\int u^4\\,du = \\frac{u^5}{5} + C = \\frac{(x+1)^5}{5} + C$ > >
\boxed{\frac{(x+1)^5}{5} + C}
> **Q9: Find $\\int \\frac{1}{3x+2}\\,dx$.** > > Let $u = 3x+2$, $du = 3\\,dx$, $dx = du/3$. > $\\frac{1}{3}\\int \\frac{du}{u} = \\frac{1}{3}\\ln|3x+2| + C$ > >
\boxed{\frac{1}{3}\ln|3x+2| + C}
> **Q10: Find $\\int \\sin(2x)\\cos x\\,dx$.** > > This requires trig identity: $\\sin(2x)\\cos x = 2\\sin x\\cos^2 x$, then substitution $u = \\cos x$. > Let $u = \\cos x$, $du = -\\sin x\\,dx$. > $-2\\int u^2\\,du = -2\\cdot\\frac{u^3}{3} + C = -\\frac{2}{3}\\cos^3 x + C$ > >
\boxed{-\frac{2}{3}\cos^3 x + C}$$ --- ## ๐Ÿ”— Cross-References - Next topic: [Definite Integrals & FTC](/courses/bsma1001/notes/integration-definite) - Related: [Integration Techniques](/courses/bsma1001/notes/integration-techniques) - Across courses: BSMA1003 Maths 2 (multivariable integration); BSMA1002 Stats 1 (probability densities) Join Discord Previous8.2 Applications of DerivativesNext9.2 Definite Integrals & FTC
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