Evaluate the Riemann-Stieltjes integral

Question

Given that $f$ is continuous and of bounded variation on $[a,b]$, evaluate $\int^b_a f(x)df(x).$

Answer 1

We will use the following theorem

Theorem: Suppose $f$ and $g$ are bounded functions with no common discontinuities on the interval $[a,b]$, and the Riemann-Stieltjes integral of $f$ with respect to $g$ exists. Then the Riemann-Stieltjes integral of $g$ with respect to $f$ exists, and $$\int_{a}^{b} g(x)df(x) = f(b)g(b)-f(a)g(a)-\int_{a}^{b} f(x)dg(x)\,. $$

Now, in your case the function is continuous, which is nice, and of bounded variation that implies boundedness, so the theorem is applicable to our problem and we have

$$ \int^b_a f(x)df(x) = f^{2}(b)-f^{2}(a)-\int^b_a f(x)df(x) $$

$$ \implies 2\int^b_a f(x)df(x)=f^{2}(b)-f^{2}(a) $$

$$ \implies \int^b_a f(x)df(x)=\frac{1}{2}( f^{2}(b)-f^{2}(a)). $$

Answer 2

$$d(f(x))=f'(x)dx$$ $$I=\int_a^bf(x).f'(x)dx$$ $$Put: f(x)=t$$ $$f'(x)dx=dt$$ So, $$I=\int_{f(a)}^{f(b)}tdt=t^2/2\bigg]_{f(a)}^{f(b)}$$ $$I=\dfrac{f^2(b)}{2}-\dfrac{f^2(a)}{2}$$

OR if $f$ has some discontinuities or not differentiable over $(a,b)$,We use By-Parts: $$\int_a^b f(x)d(f(x))=f(x).f(x)\bigg]_a^b-\int_a^b f(x).d(f(x))$$ $$I=\dfrac{f^2(b)}{2}-\dfrac{f^2(a)}{2}$$