Question

Use the chain rule and the Fundamental Theorem of Calculus to prove the integration-by-substitution formula for definite integrals:

${\int }_a^{b}f\text{'}\left(u\right(x\left)\right)u\text{'}\left(x\right)dx=f\left(u\right(b\left)\right)-f\left(u\right(a\left)\right)$

Short answer

After using the chain rule and the Fundamental Theorem of Calculus by the integration-by-substitution formula for definite integrals we proved that${\int }_a^{b}f\text{'}\left(u\right(x\left)\right)u\text{'}\left(x\right)dx=f\left(u\right(b\left)\right)-f\left(u\right(a\left)\right)$

Step-by-step solution

Step 1. Given Information 

Use the chain rule and the Fundamental Theorem of Calculus to prove the integration-by-substitution formula for definite integrals:

${\int }_a^{b}f\text{'}\left(u\right(x\left)\right)u\text{'}\left(x\right)dx=f\left(u\right(b\left)\right)-f\left(u\right(a\left)\right)$

Step 2. To solve taking the left hand side integral. 

$y={\int }_a^{b}f\text{'}\left(u\right(x\left)\right)u\text{'}\left(x\right)dx$

Let

$$\begin{gathered} t=u\left(x\right) \\ \frac{dt}{dx}=u\text{'}\left(x\right) \\ dt=u\text{'}\left(x\right)dx \end{gathered}$$

Step 3. Now the integral after substitution.

$$\begin{gathered} {\int }_a^{b}f\text{'}\left(u\right(x\left)\right)u\text{'}\left(x\right)dx={\int }_a^{b}f\text{'}\left(t\right)dt \\ {\int }_a^{b}f\text{'}\left(u\right(x\left)\right)u\text{'}\left(x\right)dx={\left[f\left(t\right)\right]}_a^b \\ {\int }_a^{b}f\text{'}\left(u\right(x\left)\right)u\text{'}\left(x\right)dx={\left[f\left(u\right(x\left)\right)\right]}_a^b \\ {\int }_a^{b}f\text{'}\left(u\right(x\left)\right)u\text{'}\left(x\right)dx=f\left(u\right(b\left)\right)-f\left(u\right(a\left)\right) \end{gathered}$$