U-Substitution

U-Substitution - For Indefinite Integrals

• $\int f(\underset{u}{\underbrace{g(x)}})\underset{du}{\underbrace{g^{\prime }(x)dx}}=\int f(u)du$

U-Substitution - For Indefinite Integrals - Examples

Example #1

Compute

• $\int (2x^3+1{)}^7(x^2)dx$

Set:

• $u=2x^3+1$
• $\frac{du}{dx}=6x^2$
• $du=6x^{2}dx\text{in differential form}$

Now:

• $\int (2x^3+1{)}^7(x^2)dx=\frac{1}{6}\int \underset{u^7}{\underbrace{(2x^3+1{)}^7}}\underset{du}{\underbrace{6x^2}}dx$
• $\int (2x^3+1{)}^7(x^2)dx=\frac{1}{6}\int u^{7}du$
• $\int (2x^3+1{)}^7(x^2)dx=\frac{1}{6}(\frac{1}{8}{u}^8)+C$
• $\int (2x^3+1{)}^7(x^2)dx=\frac{1}{48}(2x^3+1{)}^8+C$

where 𝐶 is an arbitrary constant of integration.

U-Substitution - For Definite Integrals

Let 𝑔: [𝑎, 𝑏] → 𝐼 be a differentiable function with a continuous derivative, where 𝐼⊂ℝ is an interval. Suppose that 𝑓: 𝐼 → ℝ is a continuous function, then:

• ${\int }_{x=a}^{x=b}f(\underset{u}{\underbrace{g(x)}})\underset{du}{\underbrace{g^{\prime }(x)dx}}={\int }_{u=g(a)}^{u=g(b)}f(u)du$

where:

• $u=g(x)$
• when 𝑥=𝑎 then 𝑢=𝑔(𝑎)
• when 𝑥=𝑏 then 𝑢=𝑔(𝑏)
• $\frac{du}{dx}=g^{\prime }(x)$
• $du=g^{\prime }(x)dx$

U-Substitution - Proof

U-substitution can be derived from the fundamental theorem of calculus.

https://en.wikipedia.org/wiki/Integration_by_substitution#:~:text=the%20trigonometric%20function.-,Proof,-%5Bedit%5D

U-Substitution - Derivation

• u-substitution is a 1-dimensional case of Variables” Warps Volume