Integral Calculus

LessonAbout 1.5 h

Integral Calculus

Key Ideas

  • Process of finding antiderivatives and total accumulated quantities
  • Applications include plane areas, arc lengths, and volumes
  • Pappus propositions relate rotated centroids to surface and volume

Integral Calculus branch of mathematics dealing with integration and its applications.

Basic Integration Formulas

Power Rule

\[ \int u^n du = \dfrac{u^{n+1}}{n+1} + C \]

Logarithmic Integration Formula

\[ \int \dfrac{du}{u} = \ln|u| + C \]

Exponential Integration Formula

\[ \int e^u du = e^u + C \]

Example. Integrate \(\int (5x^3 + 6x^2) dx\).

Solution. Apply the power rule to each term individually:

\[ \int (5x^3 + 6x^2) dx = \dfrac{5x^4}{4} + \dfrac{6x^3}{3} + C = \dfrac{5x^4}{4} + 2x^3 + C \]

Answer: \(\dfrac{5x^4}{4} + 2x^3 + C\)

Methods of Integration

  • Substitution: simplifies integrands using variable change
  • Integration by Parts: evaluates products of algebraic and transcendental terms
  • Partial Fractions: decomposes rational algebraic functions
  • Wallis Formula: solves definite powers of sine and cosine

Integration by Parts

\[ \int u dv = uv - \int v du \]

Example. Evaluate \(\int x e^{2x} dx\).

Solution. Set \(u = x\) and \(dv = e^{2x} dx\), which gives \(du = dx\) and \(v = \dfrac{1}{2} e^{2x}\):

\[ \int x e^{2x} dx = \dfrac{1}{2} x e^{2x} - \int \dfrac{1}{2} e^{2x} dx = \dfrac{1}{2} x e^{2x} - \dfrac{1}{4} e^{2x} + C \]

Answer: \(\dfrac{1}{2} x e^{2x} - \dfrac{1}{4} e^{2x} + C\)

Wallis Formula

\[ \int_0^{\pi/2} \sin^m \theta \cos^n \theta d\theta = \dfrac{[(m-1)(m-3)\cdots][(n-1)(n-3)\cdots]}{(m+n)(m+n-2)\cdots} \alpha \]

Example. Evaluate \(\int_0^{\pi/2} \sin^7 x dx\).

Solution. Apply Wallis formula with \(m = 7\) and \(n = 0\). Since \(m\) is odd, \(\alpha = 1\):

\[ \int_0^{\pi/2} \sin^7 x dx = \dfrac{(6)(4)(2)}{(7)(5)(3)(1)} \times 1 = \dfrac{16}{35} \]

Answer: \(\dfrac{16}{35}\)

Plane Areas

  • Vertical Strip: uses differential width \(dx\)
  • Horizontal Strip: uses differential height \(dy\)
  • Polar Coordinates: uses differential angle \(d\theta\)

Area by Vertical Strip

\[ A = \int_{x_1}^{x_2} (y_U - y_L) dx \]

Area by Horizontal Strip

\[ A = \int_{y_1}^{y_2} (x_R - x_L) dy \]

Area in Polar Coordinates

\[ A = \dfrac{1}{2} \int_{\theta_1}^{\theta_2} r^2 d\theta \]

Example. Find the area bounded by the parabola \(y = x^2\) and the line \(y = 4x\).

Solution. Find the points of intersection by setting \(x^2 = 4x\), yielding \(x = 0\) and \(x = 4\). Integrate using vertical strips:

\[ A = \int_0^4 (4x - x^2) dx = \left[ 2x^2 - \dfrac{x^3}{3} \right]_0^4 = 32 - \dfrac{64}{3} = \dfrac{32}{3} \]

Answer: \(\dfrac{32}{3}\ \text{square units}\)

Arc Length

Arc Length in Rectangular Coordinates

\[ S = \int_{x_1}^{x_2} \sqrt{1 + \left(\dfrac{dy}{dx}\right)^2} dx \]

Centroid of Plane Areas

Centroid Coordinates

\[ \bar{x} = \dfrac{\int x_c dA}{A}, \qquad\qquad \bar{y} = \dfrac{\int y_c dA}{A} \]

Solids of Revolution and Volumes

  • Circular Disk: rotates area adjacent to axis
  • Washer Method: revolves area with central hollow section
  • Cylindrical Shell: revolves differential strip parallel to axis

Washer Method Volume

\[ V = \pi \int_{x_1}^{x_2} (y_U^2 - y_L^2) dx \]

Cylindrical Shell Volume

\[ V = 2\pi \int_{x_1}^{x_2} x y dx \]

Propositions of Pappus

First Proposition of Pappus

\[ A_s = 2\pi \bar{d} S \]

Second Proposition of Pappus

\[ V = 2\pi \bar{d} A \]

Example. A circular area enclosed by \(x^2 + y^2 = 16\) is revolved about the line \(x = 5\). Find the volume generated.

Solution. The radius is \(r = 4\), so area \(A = \pi r^2 = 16\pi\). The centroid is at \((0,0)\), giving distance \(\bar{d} = 5\) to the axis \(x = 5\):

\[ V = 2\pi \bar{d} A = 2\pi (5) (16\pi) = 160\pi^2 \]

Answer: \(160\pi^2\ \text{cubic units}\)

Work and Moment of Inertia

Spring Work Formula

\[ W = \int_{x_1}^{x_2} k x dx = \dfrac{1}{2} k (x_2^2 - x_1^2) \]

Example. Find the work done in stretching a spring from \(x = 0\) to \(x = 0.2\ \text{m}\) if the spring constant is \(k = 100\ \text{N/m}\).

Solution. Apply Hooke’s law work formula:

\[ W = \int_0^{0.2} 100 x dx = \left[ 50 x^2 \right]_0^{0.2} = 50 (0.04) = 2\ \text{J} \]

Answer: \(2\ \text{J}\)

Work in Winding Up a Load

\[ W = PL + W_{\text{rope}}\left(\dfrac{L}{2}\right) \]

Work under Variable Force

\[ W = \int_{x_1}^{x_2} F(x) dx \]

Example. A \(60\text{ m}\) cable weighing \(4\text{ kg/m}\) supports a \(500\text{ kg}\) weight at its lower end. Calculate the work required to wind up the last \(20\text{ m}\) of the cable.

Solution. Let \(x\) be the length of cable hanging below the drum. The total force required at distance \(x\) is \(F(x) = 500 + 4x\). Integrate from \(x = 0\) to \(x = 20\):

\[ W = \int_0^{20} (500 + 4x) dx = \left[ 500x + 2x^2 \right]_0^{20} \]
\[ W = 500(20) + 2(20)^2 = 10000 + 800 = 10800\ \text{kg-m} \]

Answer: \(10800\ \text{kg-m}\)

Example. A \(30\text{ m}\) chain weighing \(0.50\text{ kg/m}\) lifts a leaking bucket. At the start the full \(30\text{ m}\) of chain is extended and the bucket holds \(15\text{ kg}\) of fluid. When the chain is fully wound the bucket holds \(7.5\text{ kg}\). Assuming a uniform leak rate, find the work done in winding up the chain.

Solution. Let \(x\) be the length of chain already wound up, so \(x = 0\) at the start and \(x = 30\) when fully wound. The fluid leaks at a rate of \((15 - 7.5)/30 = 0.25\text{ kg/m}\), so the fluid weight is \(15 - 0.25x\). The chain still hanging is \(30 - x\) long and weighs \(0.50(30 - x) = 15 - 0.50x\). Total force is \(F(x) = (15 - 0.25x) + (15 - 0.50x) = 30 - 0.75x\). Integrate from \(x = 0\) to \(x = 30\):

\[ W = \int_0^{30} (30 - 0.75x) dx = \left[ 30x - 0.375x^2 \right]_0^{30} \]
\[ W = 30(30) - 0.375(30)^2 = 900 - 337.5 = 562.5\ \text{kg-m} \]

Answer: \(562.5\ \text{kg-m}\)

Example. A cable weighing \(60\text{ N/m}\) is unwound from a drum. If the bottom end is initially \(15\text{ m}\) below the drum, find the work done by gravity as an additional \(100\text{ m}\) of cable is unwound.

Solution. The weight of the \(100\text{ m}\) section being unwound is \(60 \times 100 = 6000\text{ N}\). This section starts on the drum (depth \(0\)) and ends hanging from \(15\text{ m}\) to \(115\text{ m}\) below the drum, so its centroid ends at a depth of \(15 + 50 = 65\text{ m}\). The centroid therefore drops \(65\text{ m}\):

\[ W = F \times d = 6000 \times (15 + 50) = 390000\ \text{N-m} \]

Answer: \(390000\ \text{N-m}\)

Moment of Inertia of Plane Areas

  • Horizontal Strip Method: uses \(dA = (x_R - x_L)dy\) for x-axis inertia
  • Vertical Strip Method: uses \(dA = (y_U - y_L)dx\) for y-axis inertia
  • Polar Inertia: combines orthogonal second moments about origin

Moment of Inertia about x-axis

\[ I_x = \int y^2 dA \]

Moment of Inertia about y-axis

\[ I_y = \int x^2 dA \]

Example. Find the moment of inertia with respect to the x-axis of the region bounded by the parabola \(y^2 = 4x\), the x-axis, and the line \(x = 1\).

Solution. Use horizontal differential strips of width \(dy\) and length \(1 - x = 1 - \dfrac{y^2}{4}\). The upper limit of \(y\) at \(x = 1\) is \(y = 2\):

\[ I_x = \int_0^2 y^2 \left(1 - \dfrac{y^2}{4}\right) dy = \int_0^2 \left(y^2 - \dfrac{y^4}{4}\right) dy \]
\[ I_x = \left[ \dfrac{y^3}{3} - \dfrac{y^5}{20} \right]_0^2 = \dfrac{8}{3} - \dfrac{32}{20} = 1.067\ \text{units}^4 \]

Answer: \(1.067\ \text{units}^4\)

Example. Determine the moment of inertia with respect to the y-axis of the plane region bounded by \(x^2 = 4y\), the line \(x = 4\), and the x-axis.

Solution. Use vertical differential strips of width \(dx\) and height \(y = \dfrac{x^2}{4}\). Integrate from \(x = 0\) to \(x = 4\):

\[ I_y = \int_0^4 x^2 \left(\dfrac{x^2}{4}\right) dx = \dfrac{1}{4} \int_0^4 x^4 dx \]
\[ I_y = \dfrac{1}{4} \left[ \dfrac{x^5}{5} \right]_0^4 = \dfrac{1024}{20} = 51.2\ \text{units}^4 \]

Answer: \(51.2\ \text{units}^4\)

Example. Find the moment of inertia about the x-axis of the area bounded by \(x^2 = 4y\), the line \(x = -4\), and the x-axis.

Solution. The horizontal strip length is \(x_R - x_L = -2y^{1/2} - (-4) = 4 - 2y^{1/2}\). At \(x = -4\), \(y = 4\). Integrate from \(y = 0\) to \(y = 4\):

\[ I_x = \int_0^4 y^2 (4 - 2y^{1/2}) dy = \int_0^4 (4y^2 - 2y^{5/2}) dy \]
\[ I_x = \left[ \dfrac{4y^3}{3} - \dfrac{4y^{7/2}}{7} \right]_0^4 = \dfrac{256}{3} - \dfrac{512}{7} = 12.19\ \text{units}^4 \]

Answer: \(12.19\ \text{units}^4\)

Formula Summary

Formula Name Formula
Power Rule \(\int u^n du = \dfrac{u^{n+1}}{n+1} + C\)
Logarithmic Integration Formula \(\int \dfrac{du}{u} = \ln|u| + C\)
Exponential Integration Formula \(\int e^u du = e^u + C\)
Integration by Parts \(\int u dv = uv - \int v du\)
Wallis Formula \(\int_0^{\pi/2} \sin^m \theta \cos^n \theta d\theta = \dfrac{[(m-1)(m-3)\cdots][(n-1)(n-3)\cdots]}{(m+n)(m+n-2)\cdots} \alpha\)
Area by Vertical Strip \(A = \int_{x_1}^{x_2} (y_U - y_L) dx\)
Area by Horizontal Strip \(A = \int_{y_1}^{y_2} (x_R - x_L) dy\)
Area in Polar Coordinates \(A = \dfrac{1}{2} \int_{\theta_1}^{\theta_2} r^2 d\theta\)
Arc Length in Rectangular Coordinates \(S = \int_{x_1}^{x_2} \sqrt{1 + \left(\dfrac{dy}{dx}\right)^2} dx\)
Centroid Coordinates \(\bar{x} = \dfrac{\int x_c dA}{A}, \quad \bar{y} = \dfrac{\int y_c dA}{A}\)
Washer Method Volume \(V = \pi \int_{x_1}^{x_2} (y_U^2 - y_L^2) dx\)
Cylindrical Shell Volume \(V = 2\pi \int_{x_1}^{x_2} x y dx\)
First Proposition of Pappus \(A_s = 2\pi \bar{d} S\)
Second Proposition of Pappus \(V = 2\pi \bar{d} A\)
Spring Work Formula \(W = \int_{x_1}^{x_2} k x dx = \dfrac{1}{2} k (x_2^2 - x_1^2)\)
Work in Winding Up a Load \(W = PL + W_{\text{rope}}\left(\dfrac{L}{2}\right)\)
Work under Variable Force \(W = \int_{x_1}^{x_2} F(x) dx\)
Moment of Inertia about x-axis \(I_x = \int y^2 dA\)
Moment of Inertia about y-axis \(I_y = \int x^2 dA\)