Solving integral given the value of another integral

Question

I would like to understand the step 4/4 of this exercise from Khan Academy:

Let $~f~$ be a continuous function with the property that $\displaystyle~~\int^{6}_{0}f(x)dx=8$. Determine the value of $\displaystyle~\int^{2}_{0}f(3x)dx$.

1/4 | Let $~u=3x$.

2/4 | Then $~\dfrac{du}3=dx$. Also, when $~x=0$, then $~u=0~$ and when $~x=2~$ then $~u=6$.

3/4 | Then $\qquad\displaystyle\int^{2}_{0}f(3x)dx = \dfrac13\int^{6}_{0} f(u)du$.

4/4 | But we are given the value of $\displaystyle~~\int^{6}_{0}f(x)dx$.

Hence $\displaystyle\qquad\int^{2}_{0}f(3x)dx =\dfrac13\cdot8 = \dfrac83$.

How does the value of $\displaystyle~~\int^{6}_{0}f(x)dx~~$ help here?

Or, more specifically, why $$\int^{6}_{0} f(u)du = \int^{6}_{0}f(x)dx$$?

Thanks in advance.

Answer 1

Its change of variable property. We can change one variable with other.

Here we replaced $u$ with $x$.

For properties of definite integral see this.

Answer 2

This very reason:

$$\text{ Let }u=x \rightarrow du=dx$$That's why $$\int^6 _0 f(u)du=\int^6 _0 f(x)dx $$