Linear Function Example in Integral Calculus

Intro

After grasping the basic idea of integrals, let’s start with the simplest case to see how integration works.

Problem

Find the shaded area.

Geometry background

For $y=x$ on $[a,b]$:

• It’s a straight line, slope 1.
• Over $[a,b]$ it forms a trapezoid.
• By geometry: area $S=\dfrac{(a+b)(b-a)}{2} = \dfrac{b^2-a^2}{2}$.

But how about using integrals?

Integral approach

1. Split $[a,b]$ into many thin strips.
2. Area of strip = height × width.
3. Total area = sum of strip areas.

For $y=x$:

• Height of $i$-th strip: $x_i$.
• Width: $\Delta x$.
• Area: $x_i \cdot \Delta x$.

Total area:

$\sum_{i=1}^{n} x_i \cdot \Delta x$

As strips get thinner, the sum approaches the true area:

$\lim_{n \to \infty} \sum_{i=1}^{n} x_i \cdot \Delta x$

Detailed computation

We need one summation formula first:

Summation formula

Sum of natural numbers

$\sum_{i = 1}^{n} i = \frac{n(n + 1)}{2}$

Idea: pair forward and backward lists; each pair sums to $n+1$, with $n$ pairs total.

A classic story

Gauss as a schoolchild famously summed $1+2+\dots+100$ by pairing to get $100\times101/2=5050$.

Then compute the limit (omitted intermediate algebra here; geometry already gave $\dfrac{b^2-a^2}{2}$).

Conclusion

Both geometry and the integral approach agree: the area under $y=x$ on $[a,b]$ is $\dfrac{b^2-a^2}{2}$.

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