Calculus via Summations

Since $f(x)$ is non-negative, the value of the integral is the area between $f(x)$ and the $x$ -axis from $x=0$ to $x=1$ . The graph of $f(x)$ is shown below:

To find the area under the graph we have to add the areas of a series of rectangles.

Starting from the right , the $k$ th rectangle will have width $\dfrac{1}{2^{k-1}} - \dfrac{1}{2^k} = \dfrac{1}{2^k}$ and height $\dfrac{1}{2^{k-1}}$ , so its area will be $\dfrac{1}{2^{2k-1}}$ .

Thus the integral, i.e the area under the graph, will be

$\begin{aligned} \int_{0}^{1} f(x) &= \sum_{r=1} \dfrac{1}{2^{2r-1}} \\ \\ &= \sum_{r=1} \dfrac{1}{2} \left( \dfrac{1}{2^{2r-2}} \right) \\ \\ &= \dfrac{1}{2} \ \sum_{r=1} \dfrac{1}{4^{r-1}} \\ \\ &= \dfrac{1}{2} \cdot \dfrac{1}{1-\frac{1}{4}} = \dfrac{1}{2} \cdot \dfrac{4}{4-1} = \boxed{\dfrac{2}{3}} \end{aligned}$

Note that $f(x)$ is a step function with step height $f(x) = y = \dfrac 1{2^{r-1}}$ and a corresponding step width of $\Delta x = \dfrac 1{2^{r-1}} - \dfrac 1{2^r}$ , where $r \in \mathbb N$ . And $x=1$ when $r=1$ and $x=0$ when $r \to \infty$ .

$\begin{aligned} \int_0^1 f(x) \ dx & = \int_0^1 y \ dx = \sum_{r=1}^\infty y \Delta x = \sum_{r=1}^\infty \frac 1{2^{r-1}} \left(\frac 1{2^{r-1}} - \frac 1{2^r} \right) = \sum_{r=0}^\infty \frac 1{2^r} \left(\frac 1{2^r} - \frac 1{2^{r+1}} \right) \\ & = \sum_{r=0}^\infty \frac 12 \times \frac 1{2^{2r}} = \frac 12 \sum_{r=0}^\infty \left(\frac 14\right)^r = \frac 12 \left(\frac 1{1-\frac 14}\right) = \frac 23 \approx \boxed{0.667} \end{aligned}$

I do have a solution, but its not mine.

$\large\ { \displaystyle \int _{ 0 }^{ 1 }{ f\left( x \right) dx } = \sum _{ r=1 }^{ \infty }{ \int _{ { 2 }^{ -r } }^{ { 2 }^{ -(r-1) } }{ \frac { 1 }{ { 2 }^{ r-1 } } dx } } \\ \displaystyle = \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ { 2 }^{ r-1 } } \left[ { 2 }^{ -(r-1) } - { 2 }^{ -r } \right] } \\\displaystyle = \sum _{ r=1 }^{ \infty }{ { 2 }^{ -2(r-1) } - \sum _{ r=1 }^{ \infty }{ { 2 }^{ -2r+1 } } } \\ \displaystyle = \left( { 2 }^{ 2 } - 2 \right) \sum _{ r=1 }^{ \infty }{ { 2 }^{ -2r } } = \boxed { \frac { 2 }{ 3 } } }$ .

My doubt is how can i convert the definite integral asked into a summation? How can i include summation sign along with integral? Is there any formula to do that?

Basically how i got 2nd step from 1st one?

@Chew-Seong Cheong , @Mark Hennings . Please assist me here.