4.3 Indefinite and definite Integration

Detailed Theory: Indefinite and Definite Integration

1. Introduction to Integration

1.1 What is Integration?

Integration is the reverse process of differentiation. If differentiation gives us the rate of change, integration gives us the accumulation of quantities.

Two main types:

  1. Indefinite Integration: Finds antiderivatives (general form)

  2. Definite Integration: Computes accumulated quantity over an interval

1.2 The Integral Symbol

The integral symbol \int comes from the elongated "S" meaning "sum."

  • Indefinite integral: f(x)dx=F(x)+C\int f(x) dx = F(x) + C

  • Definite integral: abf(x)dx=[F(x)]ab=F(b)F(a)\int_{a}^{b} f(x) dx = [F(x)]_{a}^{b} = F(b) - F(a)

1.3 Relationship with Derivatives

If F(x)=f(x)F'(x) = f(x), then:

f(x)dx=F(x)+C\int f(x) dx = F(x) + C

where CC is the constant of integration.

Example: Since ddx(x3)=3x2\frac{d}{dx}(x^3) = 3x^2, then:

3x2dx=x3+C\int 3x^2 dx = x^3 + C

2. Indefinite Integration

2.1 Basic Integration Formulas

Power Rule (for n ≠ -1)

xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C

Examples:

  • x3dx=x44+C\int x^3 dx = \frac{x^4}{4} + C

  • xdx=x1/2dx=x3/23/2+C=23x3/2+C\int \sqrt{x} dx = \int x^{1/2} dx = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C

  • 1x2dx=x2dx=x11+C=1x+C\int \frac{1}{x^2} dx = \int x^{-2} dx = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C

Exponential Functions

  1. exdx=ex+C\int e^x dx = e^x + C

  2. axdx=axlna+C\int a^x dx = \frac{a^x}{\ln a} + C for a>0a > 0, a1a \neq 1

Trigonometric Functions

  1. sinxdx=cosx+C\int \sin x dx = -\cos x + C

  2. cosxdx=sinx+C\int \cos x dx = \sin x + C

  3. sec2xdx=tanx+C\int \sec^2 x dx = \tan x + C

  4. csc2xdx=cotx+C\int \csc^2 x dx = -\cot x + C

  5. secxtanxdx=secx+C\int \sec x \tan x dx = \sec x + C

  6. cscxcotxdx=cscx+C\int \csc x \cot x dx = -\csc x + C

Inverse Trigonometric Forms

  1. 11x2dx=sin1x+C\int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x + C

  2. 11+x2dx=tan1x+C\int \frac{1}{1+x^2} dx = \tan^{-1} x + C

  3. 1xx21dx=sec1x+C\int \frac{1}{|x|\sqrt{x^2-1}} dx = \sec^{-1} x + C

2.2 Properties of Indefinite Integrals

Linearity Properties

  1. [f(x)+g(x)]dx=f(x)dx+g(x)dx\int [f(x) + g(x)] dx = \int f(x) dx + \int g(x) dx

  2. [f(x)g(x)]dx=f(x)dxg(x)dx\int [f(x) - g(x)] dx = \int f(x) dx - \int g(x) dx

  3. kf(x)dx=kf(x)dx\int k f(x) dx = k \int f(x) dx for constant kk

Constant Multiple Rule

For constant cc:

cf(x)dx=cf(x)dx\int c f(x) dx = c \int f(x) dx

Sum/Difference Rule

[f(x)±g(x)]dx=f(x)dx±g(x)dx\int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx

2.3 Example: Using Basic Rules

Find (3x22x+5)dx\int (3x^2 - 2x + 5) dx

Solution:

(3x22x+5)dx=3x2dx2xdx+5dx\int (3x^2 - 2x + 5) dx = 3\int x^2 dx - 2\int x dx + 5\int dx
=3x332x22+5x+C= 3 \cdot \frac{x^3}{3} - 2 \cdot \frac{x^2}{2} + 5x + C
=x3x2+5x+C= x^3 - x^2 + 5x + C

3. Integration by Substitution

3.1 The Method

Substitution is the reverse of chain rule. For integral f(g(x))g(x)dx\int f(g(x))g'(x) dx:

Let u=g(x)u = g(x), then du=g(x)dxdu = g'(x) dx

The integral becomes: f(u)du\int f(u) du

3.2 Steps for Substitution

  1. Choose substitution u=g(x)u = g(x)

  2. Compute du=g(x)dxdu = g'(x) dx

  3. Rewrite integral in terms of uu

  4. Integrate with respect to uu

  5. Substitute back u=g(x)u = g(x)

3.3 Examples

Example 1: Find 2x(x2+1)3dx\int 2x(x^2 + 1)^3 dx

Let u=x2+1u = x^2 + 1, then du=2xdxdu = 2x dx

The integral becomes: u3du=u44+C\int u^3 du = \frac{u^4}{4} + C

Substitute back: (x2+1)44+C\frac{(x^2 + 1)^4}{4} + C

Example 2: Find sin(3x)dx\int \sin(3x) dx

Let u=3xu = 3x, then du=3dxdu = 3 dx, so dx=du3dx = \frac{du}{3}

sin(3x)dx=sinudu3=13sinudu\int \sin(3x) dx = \int \sin u \cdot \frac{du}{3} = \frac{1}{3} \int \sin u du
=13cosu+C=13cos(3x)+C= -\frac{1}{3} \cos u + C = -\frac{1}{3} \cos(3x) + C

Example 3: Find xex2dx\int x e^{x^2} dx

Let u=x2u = x^2, then du=2xdxdu = 2x dx, so xdx=du2x dx = \frac{du}{2}

xex2dx=eudu2=12eudu\int x e^{x^2} dx = \int e^u \cdot \frac{du}{2} = \frac{1}{2} \int e^u du
=12eu+C=12ex2+C= \frac{1}{2} e^u + C = \frac{1}{2} e^{x^2} + C

3.4 Trigonometric Substitution

Used for integrals containing a2x2\sqrt{a^2 - x^2}, a2+x2\sqrt{a^2 + x^2}, or x2a2\sqrt{x^2 - a^2}.

Three main cases:

  1. For a2x2\sqrt{a^2 - x^2}: Use x=asinθx = a\sin\theta, dx=acosθdθdx = a\cos\theta d\theta

    Example: dx4x2\int \frac{dx}{\sqrt{4-x^2}}

    Let x=2sinθx = 2\sin\theta, dx=2cosθdθdx = 2\cos\theta d\theta

    2cosθdθ44sin2θ=2cosθdθ2cosθ=dθ=θ+C\int \frac{2\cos\theta d\theta}{\sqrt{4-4\sin^2\theta}} = \int \frac{2\cos\theta d\theta}{2\cos\theta} = \int d\theta = \theta + C

    Back substitute: θ=sin1(x2)\theta = \sin^{-1}\left(\frac{x}{2}\right)

    So dx4x2=sin1(x2)+C\int \frac{dx}{\sqrt{4-x^2}} = \sin^{-1}\left(\frac{x}{2}\right) + C

  2. For a2+x2\sqrt{a^2 + x^2}: Use x=atanθx = a\tan\theta, dx=asec2θdθdx = a\sec^2\theta d\theta

  3. For x2a2\sqrt{x^2 - a^2}: Use x=asecθx = a\sec\theta, dx=asecθtanθdθdx = a\sec\theta\tan\theta d\theta


4. Integration by Parts

4.1 Formula

Based on product rule: ddx(uv)=uv+uv\frac{d}{dx}(uv) = u'v + uv'

Integration gives: udv=uvvdu\int u dv = uv - \int v du

4.2 Choosing u and dv

Use LIATE rule for choosing uu (in order of preference):

  • L: Logarithmic functions (lnx\ln x)

  • I: Inverse trigonometric functions

  • A: Algebraic functions (polynomials)

  • T: Trigonometric functions

  • E: Exponential functions

4.3 Examples

Example 1: Find xexdx\int x e^x dx

Let u=xu = x, dv=exdxdv = e^x dx Then du=dxdu = dx, v=exv = e^x

xexdx=xexexdx=xexex+C=ex(x1)+C\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C = e^x(x-1) + C

Example 2: Find lnxdx\int \ln x dx

Let u=lnxu = \ln x, dv=dxdv = dx Then du=1xdxdu = \frac{1}{x} dx, v=xv = x

lnxdx=xlnxx1xdx=xlnxdx=xlnxx+C\int \ln x dx = x\ln x - \int x \cdot \frac{1}{x} dx = x\ln x - \int dx = x\ln x - x + C

Example 3: Find exsinxdx\int e^x \sin x dx

Let u=exu = e^x, dv=sinxdxdv = \sin x dx Then du=exdxdu = e^x dx, v=cosxv = -\cos x

exsinxdx=excosx+excosxdx\int e^x \sin x dx = -e^x \cos x + \int e^x \cos x dx

Apply integration by parts again to excosxdx\int e^x \cos x dx: Let u=exu = e^x, dv=cosxdxdv = \cos x dx Then du=exdxdu = e^x dx, v=sinxv = \sin x

excosxdx=exsinxexsinxdx\int e^x \cos x dx = e^x \sin x - \int e^x \sin x dx

Now we have:

exsinxdx=excosx+exsinxexsinxdx\int e^x \sin x dx = -e^x \cos x + e^x \sin x - \int e^x \sin x dx

Bring the integral to left side:

2exsinxdx=ex(sinxcosx)2\int e^x \sin x dx = e^x(\sin x - \cos x)
exsinxdx=ex2(sinxcosx)+C\int e^x \sin x dx = \frac{e^x}{2}(\sin x - \cos x) + C

5. Integration by Partial Fractions

5.1 When to Use

For rational functions P(x)Q(x)\frac{P(x)}{Q(x)} where degree of P(x)P(x) < degree of Q(x)Q(x).

5.2 Method

  1. Factor denominator Q(x)Q(x)

  2. Write as sum of partial fractions

  3. Solve for unknown coefficients

  4. Integrate each term

5.3 Cases for Partial Fractions

Case 1: Distinct Linear Factors

For P(x)(xa)(xb)\frac{P(x)}{(x-a)(x-b)}:

P(x)(xa)(xb)=Axa+Bxb\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}

Example: dxx21\int \frac{dx}{x^2 - 1}

Factor: x21=(x1)(x+1)x^2 - 1 = (x-1)(x+1)

Write: 1(x1)(x+1)=Ax1+Bx+1\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}

Multiply through: 1=A(x+1)+B(x1)1 = A(x+1) + B(x-1)

Solve: Let x=1x=1: 1=A(2)A=121 = A(2) \Rightarrow A = \frac{1}{2}

Let x=1x=-1: 1=B(2)B=121 = B(-2) \Rightarrow B = -\frac{1}{2}

So: dxx21=(1/2x11/2x+1)dx\int \frac{dx}{x^2-1} = \int \left(\frac{1/2}{x-1} - \frac{1/2}{x+1}\right) dx

=12lnx112lnx+1+C=12lnx1x+1+C= \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C

Case 2: Repeated Linear Factors

For P(x)(xa)n\frac{P(x)}{(x-a)^n}:

P(x)(xa)n=A1xa+A2(xa)2++An(xa)n\frac{P(x)}{(x-a)^n} = \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_n}{(x-a)^n}

Case 3: Irreducible Quadratic Factors

For P(x)(ax2+bx+c)\frac{P(x)}{(ax^2+bx+c)} with b24ac<0b^2-4ac < 0:

P(x)ax2+bx+c=Ax+Bax2+bx+c\frac{P(x)}{ax^2+bx+c} = \frac{Ax+B}{ax^2+bx+c}

6. Trigonometric Integrals

6.1 Integrals of Powers of sin and cos

Case 1: sinmxcosnxdx\int \sin^m x \cos^n x dx

Strategy:

  • If mm is odd: Save one sinx\sin x, use sin2x=1cos2x\sin^2 x = 1 - \cos^2 x

  • If nn is odd: Save one cosx\cos x, use cos2x=1sin2x\cos^2 x = 1 - \sin^2 x

  • If both even: Use power reduction formulas

Example: sin3xcos2xdx\int \sin^3 x \cos^2 x dx

Here m=3m=3 (odd), n=2n=2 (even)

sin3xcos2xdx=sin2xcos2xsinxdx\int \sin^3 x \cos^2 x dx = \int \sin^2 x \cos^2 x \sin x dx
=(1cos2x)cos2xsinxdx= \int (1-\cos^2 x) \cos^2 x \sin x dx

Let u=cosxu = \cos x, du=sinxdxdu = -\sin x dx

=(1u2)u2du=(u2u4)du= -\int (1-u^2)u^2 du = -\int (u^2 - u^4) du
=(u33u55)+C=cos3x3+cos5x5+C= -\left(\frac{u^3}{3} - \frac{u^5}{5}\right) + C = -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C

6.2 Integrals of Powers of tan and sec

Case 2: tanmxsecnxdx\int \tan^m x \sec^n x dx

Strategy:

  • If nn is even: Save sec2x\sec^2 x, use sec2x=1+tan2x\sec^2 x = 1 + \tan^2 x

  • If mm is odd: Save secxtanx\sec x \tan x, use tan2x=sec2x1\tan^2 x = \sec^2 x - 1

6.3 Power Reduction Formulas

Useful for even powers:

sin2x=1cos2x2\sin^2 x = \frac{1 - \cos 2x}{2}
cos2x=1+cos2x2\cos^2 x = \frac{1 + \cos 2x}{2}

Example: sin2xdx\int \sin^2 x dx

Using formula:

sin2xdx=1cos2x2dx=12dx12cos2xdx\int \sin^2 x dx = \int \frac{1 - \cos 2x}{2} dx = \frac{1}{2}\int dx - \frac{1}{2}\int \cos 2x dx
=12x14sin2x+C= \frac{1}{2}x - \frac{1}{4}\sin 2x + C

7. Definite Integrals

7.1 Definition: Riemann Sum

The definite integral abf(x)dx\int_{a}^{b} f(x) dx is defined as:

abf(x)dx=limni=1nf(xi)Δx\int_{a}^{b} f(x) dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x

where Δx=ban\Delta x = \frac{b-a}{n} and xix_i^* is a point in the i-th subinterval.

7.2 Properties of Definite Integrals

Basic Properties

  1. aaf(x)dx=0\int_{a}^{a} f(x) dx = 0

  2. abf(x)dx=baf(x)dx\int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx

  3. ab[f(x)+g(x)]dx=abf(x)dx+abg(x)dx\int_{a}^{b} [f(x) + g(x)] dx = \int_{a}^{b} f(x) dx + \int_{a}^{b} g(x) dx

  4. abcf(x)dx=cabf(x)dx\int_{a}^{b} cf(x) dx = c\int_{a}^{b} f(x) dx

  5. abf(x)dx=acf(x)dx+cbf(x)dx\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx for any cc

Comparison Properties

  1. If f(x)0f(x) \geq 0 on [a,b][a,b], then abf(x)dx0\int_{a}^{b} f(x) dx \geq 0

  2. If f(x)g(x)f(x) \geq g(x) on [a,b][a,b], then abf(x)dxabg(x)dx\int_{a}^{b} f(x) dx \geq \int_{a}^{b} g(x) dx

  3. If mf(x)Mm \leq f(x) \leq M on [a,b][a,b], then:

m(ba)abf(x)dxM(ba)m(b-a) \leq \int_{a}^{b} f(x) dx \leq M(b-a)

7.3 Mean Value Theorem for Integrals

If ff is continuous on [a,b][a,b], then there exists cc in [a,b][a,b] such that:

f(c)=1baabf(x)dxf(c) = \frac{1}{b-a} \int_{a}^{b} f(x) dx

This value f(c)f(c) is called the average value of ff on [a,b][a,b].


8. Fundamental Theorem of Calculus

8.1 Part 1

If ff is continuous on [a,b][a,b] and F(x)=axf(t)dtF(x) = \int_{a}^{x} f(t) dt, then:

F(x)=f(x)for a<x<bF'(x) = f(x) \quad \text{for } a < x < b

In words: Derivative of integral with variable upper limit = integrand evaluated at upper limit.

Example: If F(x)=0xsin(t2)dtF(x) = \int_{0}^{x} \sin(t^2) dt, then F(x)=sin(x2)F'(x) = \sin(x^2)

8.2 Part 2 (Evaluation Theorem)

If ff is continuous on [a,b][a,b] and FF is any antiderivative of ff, then:

abf(x)dx=F(b)F(a)\int_{a}^{b} f(x) dx = F(b) - F(a)

Notation: [F(x)]ab=F(b)F(a)[F(x)]_{a}^{b} = F(b) - F(a)

Example: Evaluate 12x2dx\int_{1}^{2} x^2 dx

Antiderivative: F(x)=x33F(x) = \frac{x^3}{3}

12x2dx=[x33]12=8313=73\int_{1}^{2} x^2 dx = \left[\frac{x^3}{3}\right]_{1}^{2} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}

8.3 Examples with FTC

Example 1: Find derivative of g(x)=0x2cos(t)dtg(x) = \int_{0}^{x^2} \cos(t) dt

Let u=x2u = x^2, then g(x)=0ucos(t)dtg(x) = \int_{0}^{u} \cos(t) dt

By Chain Rule and FTC Part 1:

g(x)=cos(u)dudx=cos(x2)2x=2xcos(x2)g'(x) = \cos(u) \cdot \frac{du}{dx} = \cos(x^2) \cdot 2x = 2x \cos(x^2)

Example 2: Evaluate 0πsinxdx\int_{0}^{\pi} \sin x dx

Antiderivative: F(x)=cosxF(x) = -\cos x

0πsinxdx=[cosx]0π=(cosπ)(cos0)\int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi} = (-\cos \pi) - (-\cos 0)
=((1))(1)=1+1=2= (-(-1)) - (-1) = 1 + 1 = 2

9. Applications of Definite Integrals

9.1 Area Under a Curve

Area between y=f(x)y = f(x) and x-axis from x=ax = a to x=bx = b:

A=abf(x)dxA = \int_{a}^{b} |f(x)| dx

Example: Area under y=x2y = x^2 from x=0x = 0 to x=2x = 2

A=02x2dx=[x33]02=83A = \int_{0}^{2} x^2 dx = \left[\frac{x^3}{3}\right]_{0}^{2} = \frac{8}{3}

9.2 Area Between Two Curves

Area between y=f(x)y = f(x) and y=g(x)y = g(x) from x=ax = a to x=bx = b (where f(x)g(x)f(x) \geq g(x)):

A=ab[f(x)g(x)]dxA = \int_{a}^{b} [f(x) - g(x)] dx

Example: Area between y=x2y = x^2 and y=xy = x from x=0x = 0 to x=1x = 1

Find intersection: x2=xx(x1)=0x=0,1x^2 = x \Rightarrow x(x-1) = 0 \Rightarrow x=0,1

Between 0 and 1, xx2x \geq x^2

A=01(xx2)dx=[x22x33]01=1213=16A = \int_{0}^{1} (x - x^2) dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}

9.3 Volume by Slicing (Disk/Washer Method)

Disk Method (rotation around x-axis)

Volume when region under y=f(x)y = f(x) rotated about x-axis:

V=πab[f(x)]2dxV = \pi \int_{a}^{b} [f(x)]^2 dx

Example: Volume of solid formed by rotating y=xy = \sqrt{x} from x=0x=0 to x=4x=4 about x-axis

V=π04(x)2dx=π04xdx=π[x22]04=π8=8πV = \pi \int_{0}^{4} (\sqrt{x})^2 dx = \pi \int_{0}^{4} x dx = \pi \left[\frac{x^2}{2}\right]_{0}^{4} = \pi \cdot 8 = 8\pi

Washer Method

When there's a hole: V=πab([R(x)]2[r(x)]2)dxV = \pi \int_{a}^{b} ([R(x)]^2 - [r(x)]^2) dx where R(x)R(x) = outer radius, r(x)r(x) = inner radius

9.4 Arc Length

Length of curve y=f(x)y = f(x) from x=ax = a to x=bx = b:

L=ab1+[f(x)]2dxL = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx

Example: Arc length of y=x3/2y = x^{3/2} from x=0x=0 to x=4x=4

f(x)=32x1/2f'(x) = \frac{3}{2}x^{1/2}, so [f(x)]2=94x[f'(x)]^2 = \frac{9}{4}x

L=041+94xdxL = \int_{0}^{4} \sqrt{1 + \frac{9}{4}x} dx

Let u=1+94xu = 1 + \frac{9}{4}x, du=94dxdu = \frac{9}{4}dx

When x=0x=0, u=1u=1; when x=4x=4, u=10u=10

L=110u49du=4923[u3/2]110L = \int_{1}^{10} \sqrt{u} \cdot \frac{4}{9} du = \frac{4}{9} \cdot \frac{2}{3}[u^{3/2}]_{1}^{10}
=827(103/213/2)=827(10101)= \frac{8}{27}(10^{3/2} - 1^{3/2}) = \frac{8}{27}(10\sqrt{10} - 1)

10. Improper Integrals

10.1 Types of Improper Integrals

Type 1: Infinite Intervals

  1. af(x)dx=limtatf(x)dx\int_{a}^{\infty} f(x) dx = \lim_{t \to \infty} \int_{a}^{t} f(x) dx

  2. bf(x)dx=limttbf(x)dx\int_{-\infty}^{b} f(x) dx = \lim_{t \to -\infty} \int_{t}^{b} f(x) dx

  3. f(x)dx=cf(x)dx+cf(x)dx\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{c} f(x) dx + \int_{c}^{\infty} f(x) dx

Type 2: Discontinuous Integrands

If ff is discontinuous at aa:

abf(x)dx=limta+tbf(x)dx\int_{a}^{b} f(x) dx = \lim_{t \to a^+} \int_{t}^{b} f(x) dx

If ff is discontinuous at bb:

abf(x)dx=limtbatf(x)dx\int_{a}^{b} f(x) dx = \lim_{t \to b^-} \int_{a}^{t} f(x) dx

10.2 Convergence Tests

Comparison Test

If 0f(x)g(x)0 \leq f(x) \leq g(x) for xax \geq a:

  • If ag(x)dx\int_{a}^{\infty} g(x) dx converges, then af(x)dx\int_{a}^{\infty} f(x) dx converges

  • If af(x)dx\int_{a}^{\infty} f(x) dx diverges, then ag(x)dx\int_{a}^{\infty} g(x) dx diverges

p-integral Test

11xpdx{converges if p>1diverges if p1\int_{1}^{\infty} \frac{1}{x^p} dx \quad \begin{cases} \text{converges if } p > 1 \\ \text{diverges if } p \leq 1 \end{cases}

10.3 Examples

Example 1: 11x2dx\int_{1}^{\infty} \frac{1}{x^2} dx

11x2dx=limt1tx2dx=limt[1x]1t\int_{1}^{\infty} \frac{1}{x^2} dx = \lim_{t \to \infty} \int_{1}^{t} x^{-2} dx = \lim_{t \to \infty} \left[-\frac{1}{x}\right]_{1}^{t}
=limt(1t+1)=1= \lim_{t \to \infty} \left(-\frac{1}{t} + 1\right) = 1

Converges to 1.

Example 2: 011xdx\int_{0}^{1} \frac{1}{\sqrt{x}} dx

Discontinuity at x=0x=0

011xdx=limt0+t1x1/2dx=limt0+[2x]t1\int_{0}^{1} \frac{1}{\sqrt{x}} dx = \lim_{t \to 0^+} \int_{t}^{1} x^{-1/2} dx = \lim_{t \to 0^+} [2\sqrt{x}]_{t}^{1}
=limt0+(212t)=2= \lim_{t \to 0^+} (2\sqrt{1} - 2\sqrt{t}) = 2

Converges to 2.


11. Multiple Integration

11.1 Double Integrals

For function f(x,y)f(x,y) over region RR:

Rf(x,y)dA\iint_{R} f(x,y) dA

11.2 Iterated Integrals

Type I Region (vertically simple):

Rf(x,y)dA=ab[g1(x)g2(x)f(x,y)dy]dx\iint_{R} f(x,y) dA = \int_{a}^{b} \left[\int_{g_1(x)}^{g_2(x)} f(x,y) dy\right] dx

Type II Region (horizontally simple):

Rf(x,y)dA=cd[h1(y)h2(y)f(x,y)dx]dy\iint_{R} f(x,y) dA = \int_{c}^{d} \left[\int_{h_1(y)}^{h_2(y)} f(x,y) dx\right] dy

11.3 Example: Double Integral

Evaluate R(x+2y)dA\iint_{R} (x + 2y) dA where R={(x,y):0x2,0y1}R = \{(x,y): 0 \leq x \leq 2, 0 \leq y \leq 1\}

R(x+2y)dA=0201(x+2y)dydx\iint_{R} (x + 2y) dA = \int_{0}^{2} \int_{0}^{1} (x + 2y) dy dx

First integrate with respect to yy:

01(x+2y)dy=[xy+y2]01=x+1\int_{0}^{1} (x + 2y) dy = \left[xy + y^2\right]_{0}^{1} = x + 1

Then with respect to xx:

02(x+1)dx=[x22+x]02=(2+2)0=4\int_{0}^{2} (x + 1) dx = \left[\frac{x^2}{2} + x\right]_{0}^{2} = (2 + 2) - 0 = 4

11.4 Triple Integrals

For function f(x,y,z)f(x,y,z) over solid EE:

Ef(x,y,z)dV\iiint_{E} f(x,y,z) dV

Example: Volume of box 0xa,0yb,0zc0 \leq x \leq a, 0 \leq y \leq b, 0 \leq z \leq c

V=0a0b0cdzdydx=abcV = \int_{0}^{a} \int_{0}^{b} \int_{0}^{c} dz dy dx = abc

12. Numerical Integration

12.1 When Numerical Methods are Used

  • When antiderivative cannot be found in elementary functions

  • When only data points are available, not function formula

  • For quick approximations

12.2 Midpoint Rule

Approximate abf(x)dx\int_{a}^{b} f(x) dx using midpoints of subintervals:

Mn=Δxi=1nf(xˉi)M_n = \Delta x \sum_{i=1}^{n} f(\bar{x}_i)

where Δx=ban\Delta x = \frac{b-a}{n}, xˉi=xi1+xi2\bar{x}_i = \frac{x_{i-1} + x_i}{2}

12.3 Trapezoidal Rule

Tn=Δx2[f(x0)+2f(x1)+2f(x2)++2f(xn1)+f(xn)]T_n = \frac{\Delta x}{2}[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)]

where Δx=ban\Delta x = \frac{b-a}{n}, xi=a+iΔxx_i = a + i\Delta x

12.4 Simpson's Rule (n must be even)

Sn=Δx3[f(x0)+4f(x1)+2f(x2)+4f(x3)++4f(xn1)+f(xn)]S_n = \frac{\Delta x}{3}[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 4f(x_{n-1}) + f(x_n)]

Error Bounds:

  • Trapezoidal: ETK(ba)312n2|E_T| \leq \frac{K(b-a)^3}{12n^2}

  • Simpson's: ESK(ba)5180n4|E_S| \leq \frac{K(b-a)^5}{180n^4}

where f(x)K|f''(x)| \leq K for Trapezoidal, f(4)(x)K|f^{(4)}(x)| \leq K for Simpson's


13. Important Integration Formulas Summary

13.1 Basic Integration Formulas

  1. xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C (n ≠ -1)

  2. 1xdx=lnx+C\int \frac{1}{x} dx = \ln|x| + C

  3. exdx=ex+C\int e^x dx = e^x + C

  4. sinxdx=cosx+C\int \sin x dx = -\cos x + C

  5. cosxdx=sinx+C\int \cos x dx = \sin x + C

13.2 Substitution Rule

f(g(x))g(x)dx=f(u)du\int f(g(x))g'(x) dx = \int f(u) du where u=g(x)u = g(x)

13.3 Integration by Parts

udv=uvvdu\int u dv = uv - \int v du

13.4 Fundamental Theorem of Calculus

abf(x)dx=F(b)F(a)\int_{a}^{b} f(x) dx = F(b) - F(a) where F(x)=f(x)F'(x) = f(x)

13.5 Common Trigonometric Integrals

  1. tanxdx=lnsecx+C\int \tan x dx = \ln|\sec x| + C

  2. secxdx=lnsecx+tanx+C\int \sec x dx = \ln|\sec x + \tan x| + C

  3. cscxdx=lncscxcotx+C\int \csc x dx = \ln|\csc x - \cot x| + C

  4. cotxdx=lnsinx+C\int \cot x dx = \ln|\sin x| + C


14. Solved Examples

Example 1: Integration by Substitution

Find x1x4dx\int \frac{x}{\sqrt{1-x^4}} dx

Solution: Let u=x2u = x^2, then du=2xdxdu = 2x dx, so xdx=du2x dx = \frac{du}{2}

x1x4dx=11u2du2=12sin1u+C\int \frac{x}{\sqrt{1-x^4}} dx = \int \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{2} = \frac{1}{2} \sin^{-1} u + C
=12sin1(x2)+C= \frac{1}{2} \sin^{-1}(x^2) + C

Example 2: Integration by Parts

Find x2lnxdx\int x^2 \ln x dx

Solution: Let u=lnxu = \ln x, dv=x2dxdv = x^2 dx Then du=1xdxdu = \frac{1}{x} dx, v=x33v = \frac{x^3}{3}

x2lnxdx=x33lnxx331xdx\int x^2 \ln x dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} dx
=x33lnx13x2dx=x33lnx13x33+C= \frac{x^3}{3} \ln x - \frac{1}{3} \int x^2 dx = \frac{x^3}{3} \ln x - \frac{1}{3} \cdot \frac{x^3}{3} + C
=x33lnxx39+C=x39(3lnx1)+C= \frac{x^3}{3} \ln x - \frac{x^3}{9} + C = \frac{x^3}{9}(3\ln x - 1) + C

Example 3: Partial Fractions

Find 2x+3(x1)(x+2)dx\int \frac{2x+3}{(x-1)(x+2)} dx

Solution: Write: 2x+3(x1)(x+2)=Ax1+Bx+2\frac{2x+3}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}

Multiply: 2x+3=A(x+2)+B(x1)2x+3 = A(x+2) + B(x-1)

Let x=1x=1: 5=A(3)A=535 = A(3) \Rightarrow A = \frac{5}{3}

Let x=2x=-2: 1=B(3)B=13-1 = B(-3) \Rightarrow B = \frac{1}{3}

So:

2x+3(x1)(x+2)dx=(5/3x1+1/3x+2)dx\int \frac{2x+3}{(x-1)(x+2)} dx = \int \left(\frac{5/3}{x-1} + \frac{1/3}{x+2}\right) dx
=53lnx1+13lnx+2+C= \frac{5}{3}\ln|x-1| + \frac{1}{3}\ln|x+2| + C

Example 4: Definite Integral with Substitution

Evaluate 01x1x2dx\int_{0}^{1} x\sqrt{1-x^2} dx

Solution: Let u=1x2u = 1-x^2, then du=2xdxdu = -2x dx, so xdx=du2x dx = -\frac{du}{2}

When x=0x=0, u=1u=1; when x=1x=1, u=0u=0

01x1x2dx=10u(du2)=1201u1/2du\int_{0}^{1} x\sqrt{1-x^2} dx = \int_{1}^{0} \sqrt{u} \left(-\frac{du}{2}\right) = \frac{1}{2} \int_{0}^{1} u^{1/2} du
=1223[u3/2]01=13(10)=13= \frac{1}{2} \cdot \frac{2}{3}[u^{3/2}]_{0}^{1} = \frac{1}{3}(1-0) = \frac{1}{3}

15. Common Mistakes and Exam Tips

15.1 Common Mistakes

  1. Forgetting +C in indefinite integrals

  2. Misapplying power rule: xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C, not xn1n1\frac{x^{n-1}}{n-1}

  3. Chain rule in reverse: Forgetting to account for derivative of inner function in substitution

  4. Definite integrals: Forgetting to change limits when using substitution

  5. Partial fractions: Not checking if degree of numerator < degree of denominator first

15.2 Problem-Solving Strategy

  1. Identify type: Which method applies? (substitution, parts, partial fractions, etc.)

  2. Simplify first: Use algebra/trig identities to simplify integrand

  3. Try substitution: Often the first method to try

  4. Check answer: Differentiate your answer to verify

  5. For definite integrals:

    • Consider symmetry properties

    • Check if function is even/odd

    • Consider splitting interval at discontinuities

15.3 Quick Checks

  1. Even functions: aaf(x)dx=20af(x)dx\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx if ff is even

  2. Odd functions: aaf(x)dx=0\int_{-a}^{a} f(x) dx = 0 if ff is odd

  3. Periodic functions: aa+Tf(x)dx=0Tf(x)dx\int_{a}^{a+T} f(x) dx = \int_{0}^{T} f(x) dx if ff has period TT

  4. Area interpretation: Definite integral = net area (above x-axis minus below)

This comprehensive theory covers all aspects of indefinite and definite integration with detailed explanations and examples, providing complete preparation for the entrance examination.