Indefinite Integrals, Substitution Method, and Area Between Curves

Indefinite Integrals and the Substitution Method

The Substitution Rule

The substitution method is a fundamental technique for evaluating integrals, especially when the integrand is a composite function. It is based on reversing the Chain Rule for differentiation.

• Theorem (Substitution Rule): If u = g(x) is a differentiable function whose range is an interval I, and f is continuous on I, then $\int f(g(x))g'(x)\,dx = \int f(u)\,du$

Steps for the Substitution Method

To evaluate an integral using substitution, follow these steps:

1. Substitute u = g(x) and du = g'(x)dx to obtain $\int f(u) du$.

2. Integrate with respect to u.

3. Replace u by g(x) to return to the original variable.

Examples of Substitution

• Example 1: $\int x^3 \cos(x^4 + 2) dx$

  • Let u = x^4 + 2, so du = 4x^3 dx and $x^3 dx = \frac{1}{4} du$.

  • Integral becomes $\frac{1}{4} \int \cos(u) du = \frac{1}{4} \sin(u) + C = \frac{1}{4} \sin(x^4 + 2) + C$.

• Example 2: $\int e^{5x} dx$

  • Let u = 5x, so du = 5 dx and $dx = \frac{1}{5} du$.

  • Integral becomes $\frac{1}{5} \int e^u du = \frac{1}{5} e^u + C = \frac{1}{5} e^{5x} + C$.

Integrals of Trigonometric Functions

Some standard integrals involving tangent, cotangent, secant, and cosecant are:

$\int \tan x\,dx = \ln|\sec x| + C$	$\int \sec x\,dx = \ln|\sec x + \tan x| + C$
$\int \cot x\,dx = \ln|\sin x| + C$	$\int \csc x\,dx = -\ln|\csc x + \cot x| + C$

Substitution in Definite Integrals

Theorem: Substitution in Definite Integrals

When evaluating definite integrals, the limits of integration must be adjusted according to the substitution.

• Theorem: If g' is continuous on [a, b] and f is continuous on the range of g(x), then $\int_a^b f(g(x))g'(x) dx = \int_{g(a)}^{g(b)} f(u) du$

Even and Odd Functions in Integration

Theorem: Integrals of Even and Odd Functions

Special properties arise when integrating even or odd functions over symmetric intervals.

• If f is even, $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$

• If f is odd, $\int_{-a}^a f(x) dx = 0$

Applications: Area Between Curves

Definition: Area Between Two Curves

The area between two curves y = f(x) and y = g(x) from x = a to x = b (where f(x) \geq g(x)) is given by:

• $A = \int_a^b [f(x) - g(x)] dx$

Example: Area Between $y = e^x$ and $y = x$

Find the area bounded above by $y = e^x$, below by $y = x$, and between $x = 0$ and $x = 1$.

• Area: $A = \int_0^1 (e^x - x) dx = [e^x - \frac{1}{2}x^2]_0^1 = (e - \frac{1}{2}) - (1 - 0) = e - \frac{3}{2}$

Example: Area Between Two Parabolas

Find the area enclosed by $y = x^2$ and $y = 2x - x^2$.

• Points of intersection: $x = 0$ and $x = 1$.

• Area: $A = \int_0^1 [(2x - x^2) - x^2] dx = \int_0^1 (2x - 2x^2) dx$

• Evaluate: $[x^2 - \frac{2}{3}x^3]_0^1 = 1 - \frac{2}{3} = \frac{1}{3}$

Area When Curves Cross

If $f(x) \geq g(x)$ for some $x$ and $g(x) \geq f(x)$ for others, split the region into subregions and sum their areas:

• $A = \int_{a_1}^{b_1} |f(x) - g(x)| dx + \int_{a_2}^{b_2} |g(x) - f(x)| dx + \ldots$

Example: Area Between $y = \sin x$ and $y = \cos x$

Find the area bounded by $y = \sin x$, $y = \cos x$, $x = 0$, and $x = \frac{\pi}{2}$.

• Intersection at $x = \frac{\pi}{4}$.

• Area: $A = \int_0^{\pi/4} (\cos x - \sin x) dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) dx$

• By symmetry: $A = 2 \int_0^{\pi/4} (\cos x - \sin x) dx$

Splitting Regions and Changing Variables

When the formula for a bounding curve changes, the area integral becomes the sum of integrals for each region.

Integration with Respect to y

If the region's bounding curves are described by functions of y, use horizontal rectangles and integrate with respect to y:

• $A = \int_c^d [f(y) - g(y)] dy$

Alternative Area Calculations

Sometimes, the area can be found by subtracting the area of a triangle from the area under a curve.

Example: Area Enclosed by a Line and a Parabola

Find the area enclosed by $y = x - 1$ and $y^2 = 2x + 6$.

• Points of intersection: $(-1, -2)$ and $(5, 4)$.

• Integrate with respect to $y$ between $y = -2$ and $y = 4$:

• $A = \int_{-2}^4 [(y + 1) - (\frac{1}{2}y^2 - 3)] dy$

Note on Method Selection

Sometimes, integrating with respect to $y$ is simpler than with respect to $x$, especially when the region would otherwise need to be split into multiple parts.