Free lesson · Integral calculus

Integration techniques: substitution and integration by parts

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Start with the idea

Substitution reverses the chain rule.
Integration by parts reverses the product rule.
Both techniques transform an integral into a form you already know rather than changing its mathematical value.

Symbols, units & horizon

u=g(x): substituted inner variable, or the chosen factor in integration by parts
du=g′(x)dx: its differential
dv=v′(x)dx: differential of the other factor
[uv]ₐᵇ: u(b)v(b)−u(a)v(a)
a,b: original bounds
g(a),g(b): transformed bounds
f: outer integrand after substitution

When and why to use this

Use substitution when an inner function and its derivative appear together; use integration by parts for products such as a polynomial times an exponential. Verify the transformed domain.

For substitution, choose an inner expression u=g(x) whose derivative appears as another factor.
Change the differential and, for a definite integral, the bounds too.
Do not mix x-bounds with a u-integrand.
For integration by parts, choose u and dv so that differentiating u simplifies it and integrating dv is feasible.
The technique rearranges one unknown integral into a boundary term and another integral.
It does not guarantee the new integral is easier.

\int_{a}^{b} f (g (x)) g^{'} (x) d x = \int_{g (a)}^{g (b)} f (u) d u, \int_{a}^{b} u d v = [u v]_{a}^{b} - \int_{a}^{b} v d u

Calculus: derivation and arithmetic

Integration techniques: substitution and integration by parts

For $\int_{0}^{1} 2 x e^{x^{2}} d x$ , set $u = x^{2}$ and $d u = 2 x d x$ . The transformed endpoints are still 0 and 1.
Substitute and integrate: $\int_{0}^{1} e^{u} d u = [e^{u}]_{0}^{1} = e - 1$ . Differentiate $e^{x^{2}}$ to check the original integrand.
The product rule is $(u v)^{'} = u^{'} v + u v^{'}$ . Integrating yields $[u v]_{a}^{b} = \int_{a}^{b} v d u + \int_{a}^{b} u d v$ . Subtract the first integral to obtain integration by parts.
For $\int_{0}^{1} x e^{x} d x$ , choose $u = x$ , $d v = e^{x} d x$ , so $d u = d x$ , $v = e^{x}$ . Substitute: $[x e^{x}]_{0}^{1} - \int_{0}^{1} e^{x} d x = e - (e - 1) = 1$ .

Work it by hand

Check the indefinite answers by differentiation: derivative of exp(x²) is 2x exp(x²), and derivative of (x−1)exp(x) is x exp(x).

An analogy to remember

Substitution is changing the unit on a ruler while also changing the interval labels. Integration by parts transfers some of the work from one factor to the other.

How this becomes a building block

Substitution underlies transformations between distributions and simplifications of discounting integrals. Integration by parts appears in expectation identities and continuous cash-flow manipulations. These operations often simplify one term within a larger pricing or risk calculation.

Python implementation

Self-contained teaching example. Python 3.10+; dependencies and input conventions are shown in the code and notation. Run in your own Python environment.

from math import exp

def integrate_2x_exp_square(a,b):
    return exp(b*b)-exp(a*a)

def integrate_x_exp(a,b):
    primitive=lambda x:(x-1)*exp(x)
    return primitive(b)-primitive(a)

print(integrate_2x_exp_square(0,1),integrate_x_exp(0,1))

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