I am reading the book A garden of integrals and in the context of the Cauchy integral there is the following exercise:
How can I go about part c?
Since the sequence converges uniformly I guess I should use the converge theorem cited in the book as:
If {$f_k$} is a sequence of continuous functions converging uniformly to the function $f$ on $[a, b]$, then $f$ is Cauchy integrable on $[a, b]$ and $C\int_a^b f(x)dx=\lim C\int_a^b f_k(x)dx$.
The thing is that I don't know how to proceed to compute the integral of this piecewise recursive sequence. I am really at a loss here, I do not know how to deal with it. Any help would be appreciated.
The region between the $x$-axis and the of graph of $f_1$ with non-zero content is a triangular region of height $4$ above the base interval $[0,\frac{1}{2}]$ with length $\frac{1}{2}$. The area of this region is
$$\int_0^1 f_1(x) \, dx= \frac{1}{2} \cdot \frac{1}{2}\cdot 4 = 1$$
For $f_2$, the region of non-zero content includes that of $f_1$ along with a triangular region of height $2$ below the base interval $[\frac{1}{2},\frac{3}{4}]$ with length $\frac{1}{4}$. This appended region below the $x$-axis makes a negative contribution to the integral,
$$\int_0^1 f_2(x) \, dx= \int_0^1 f_1(x) \, dx-\frac{1}{2} \cdot \frac{1}{4}\cdot 2 = 1- \frac{1}{4}= \frac{3}{4}$$
Similarly for $f_3$, the integral is the sum of the integral for $f_2$ plus the contribution from a triangular region of height $1$ above the base interval $[\frac{3}{4},\frac{7}{8}]$ with length $\frac{1}{8}$. Whence,
$$\int_0^1 f_3(x) \, dx=\int_0^1 f_2(x) \, dx+ \frac{1}{2}\cdot \frac{1}{8} \cdot 1 = 1 - \frac{1}{4} + \frac{1}{16} = \frac{13}{16}$$
Following this pattern, we see that
$$\int_{0}^1 f_k(x) \, dx = \sum_{j=0}^{k-1} \left(-\frac{1}{4}\right)^j= \frac{1 - \left(-\frac{1}{4}\right)^k}{1 - \left(-\frac{1}{4}\right)}= \frac{4}{5} \left[1 - \left(-\frac{1}{4}\right)^k\right]$$
Thus,
$$\int_0^1 \lim_{k \to \infty}f_k(x) \,dx =\lim_{k \to \infty}\int_0^1 f_k(x) \,dx = \frac{4}{5},$$
where you have already established the left equality by uniform convergence.